Ordinary Non-Linear Differential Equations

with period 2?r. to move from the initial position y

9,

= position y To

y Q to the central or equilibrium

and corresponds to the

0,

3.10, the

(10), (11),

about the central

oscillation

The

0.

interval 9

Thus by given by

(0, ITT).

complete period of the motion

is

= 6

where F(X, modulus A.

a complete elliptic integral of the first kind with virtue of the multiplier (a+by%)-*, it follows that the periodic time decreases with increase in a, b, y Q either individually or collectively. If y Q is fixed, an increase in a, 6, or both of them, entails

a

%TT) is

By

and therefore a smaller period T O and higher funda= r^ 1 On the other hand, if a, b are fixed, an co/27r y means greater initial acceleration by virtue of the

stiffer spring,

,

mental frequency increase in

.

enhanced average stiffness! corresponding to In this respect it is of interest to remark that T O OC

or

(a+%o)~* TO

=

The constant values of a, f Stiffness

=

[w/^+^yg)]*

=

which we

{-

byl) or

(mass/average

xa

(mass/average stiffness)*

is 4.F(A, \TT)

(a

shall

stiffness)*,

constant.

^-f 3s 8 t/

2 .

Average

stiffness

=

-I (^-f 35 I

.j

2

s2/

)

dy

(2)

(3)

now determine

6.

=

(s^s^yl).

=

for certain

EQUATIONS INTEGRABLE BY

28

3.12. b =

Numerical

100, then the

A

By

3.111,

(1),

illustration of

3.111. Take*/

(1),

m

CH.

=

1,

=

a

10,

modulus

= [by*/2(a+by*)]* = (100/2 x 110)* = 0-674. ^ = (4/noi)^(0-674, = 4x1-82/10-48 = 0-695,

(1) (2)

(3)

77)

(4)

by aid of elliptic integral tables [173]. It follows that the constant in 3.111 is 4x1-82 * 7-28. In a system with linear spring control it

would be

is,

2ir

The frequency of occurrence of the

cz 6-28.

therefore,

oscillation

= To" ~ 1-44. y = y cosa* 1

(5)

co/277

If

6-0,

(6)

f

so the frequency

is

=

w/2;r

a*/27r

=

10V2?r

~ 0-5.

(7)

3

Thus the

influence of adding the term by to the stiffness control is almost to treble the frequency of the oscillation when y Q 1. The

frequency ratio either increases or decreases, according as 2/

= > or <

1.

Graphical representation. To exhibit the relationship between y and t we use the formula 3.13.

t

=

JP(0-674,9)/110*,

(1)

elliptic

with the numerical values inserted, and a table of 0-674 sin a, integrals of the first kind [173]. Then A

so a

42-5.

i.e.

3.10,

(11),

=

Also y

=

2/

cos9

=

0039.

The data

in

Table

1

were

obtained.

TABLE

By Table

1

plotting the points corresponding to the last 1

we

obtain the

first

two columns

quarter period of the Jacobian

in

elliptic

AND FUNCTIONS

ELLIPTIC INTEGRALS

3.13

=

function cut

0039,

the

i.e.

elliptic cosine.

29

A quarter period of the

and the ordinary or circular cosine of equal period and amplitude is shown by the broken curve for comparison. There is no marked difference between the ordinates of the two curves. Thus although the spring control ay -{-by 3 = lQy-{- 10(h/ 8 is sufficiently curve

is

depicted in Fig. 5,

non-linear to increase the rate of oscillation nearly threefold 1, the motion is almost simple harmonic notwithstanding. yQ

when

=

t

FIG.

(radians)

Graphs of cnJ and

5.

Approximate solution

3.14.

Fig. 5 that the solution curve y

of

=

coat.

3.10.

(1),

~

cn ^

It is evident

may

from

be analysed into

0039 a Fourier series. Its shape is such that in a first approximation we may

write

y

where \A\

>

|4

|.

=A

l

cos

a>t+A 9 cos 3orf,

Substituting (1) into

(1)

3.10,

(1),

with

=

cot

we

get o)

y ay

= a(A

In the

2

l

(^ 1 cos

last line,

cos

(2)

terms involving A% (except two) have been omitted

as negligible.

Equating the

coefficient of cos cu

since

Al ^

0.

\fj

to zero, gives

= a+i6^!+i64 ^

It follows

1

from

(3)

(3)

8,

that the frequency

is

dependent

EQUATIONS INTEGRABLE BY

30

upon the amplitude of the motion, which

OH. in

a characteristic of non-

is

When

dynamical or other physical systems. the system is linear, and o> is independent of

A v this

of such a system. Equating the coefficient of cos 30 to zero,

we have

linear

Substituting for a from

(3),

and omitting the term

=

6

0,

A3 =

0,

being a feature

in

AA\,

yields (5)

Using

(3),

A3

without the term in

,

in (5)

we

get

~ bAl/(32a+ 216.4?) AS!A! = l/[21+(32o/MJ)], which

is

always <^

Since y

=y

provided 6

1,

when

>

|.4 8

|.

Take a

=

10, 6

=

100,

this in (5)

2

~ a+lbyl-lby

yQ

1,

Al

with

c^

1

Al ^

As.

(9) .

we

co

so the periodic time

(1Q)

get

c0-04,

25^3. Substituting into 2

(11)

the various quantities yields

(9) for

^ 8575 X 0-04 =

is

T

=

obtain

and neglect the term in A 3 Then to a

^3 so

we

^ _ 1Q+75 = g5

approximation

Using

(8)

Substituting from (8) into (3) o>

first

0.

= A^+AS,

y with \A\

>

(7)

follows that

0, it

t

(6)

27T/o>

=

0-69 8

82,

(12)

(13)

,

which compares favourably with the more accurate value at (4), 3. 12. By (8), (11), A! = (y -A 3 ) = 1-0-04 = 0-96. Substituting into (1)

we

obtain the approximate solution

y

=

0-96 cos ojt+ 0-04 cos 3co,

so the amplitude of the third harmonic

the motion

only about ^th that of the essentially simple harmonic. Using is

fundamental, the data in the last two columns of Table i.e.

is

(14)

1

in

3

.

1 3,

the reader should

compute the coefficients in the Fourier series for cn, and' compare them with those in (14).

AND FUNCTIONS

ELLIPTIC INTEGRALS

3.15

3.15. Application of

3.13, 3.14.

31

The results therein may be used

A large mass m is dropped from a height h

for the following problem:

whose vertical force-displacement relationship is the two bodies interlocking at the instant of impact. 1 2/+s 3 2/ mass of the shell, (b) the displacement due to mg the Neglecting (a) after impact, (c) damping, determine the subsequent motion of the on a

steel shell

f= s

3

,

system.

The

differential equation is

n^+W+W* = y+ay+by* = 0, s 3 /m. The maximum

or

= sjm, b =

where a

calculated from the energy relation. to the loss in potential energy of ra,

If h

>

2/

,

(1)

(2)

displacement y Q

Thus equating the

/ y dy+s 3 j

we have

y*

dy

=

K2/g+i*32/o

=

9^+yo)-

=y

Q,

the motion

3.13, 3,14, so the y-t curve

is

the

is

identical with that discussed in

elliptic cosine

en t. Since damping to y Q is obtained by

absent, the initial part of the curve from y reflecting the first quadrant of cut in the y-axis, so that

is

behind the

origin.

In

=A Thus if we commence at = for

a>t,

it

comes

we assumed

3.14

y

\TT)

(3)

(3) yields

After reaching y

(cot

may be

strain energy

I/O

1/0

si

0,

l

cos wt+As cos

3^.

(5)

t when the impact occurs, we must write which gives the approximate solution

y

=

^4 1

sin^

A l and A^ are calculated as in

.4 3

sin3co.

(6)

3.14, the value of y Q being that in

(4)

above. 3.160.

The simple pendulum.

of motion

Referring to Fig. 6 A the equation

is

mass X acceleration -f- restoring force or so

m(ld 0/dt*)+mgsm6 =

=

0,

(1)

2

d 2 9/dt 2

+ -sin0 =

0.

0,

(2)

(3)

EQUATIONS INTEGRABLE BY

32

Since

sin0=:0

03 3!

05

OH. in

-)----..., the degree of the equation differs

from

5!

unity, so it is non-linear. Owing to the shape of a sine curve, the control 'stiffness' decreases with increase in (Fig. 6 B).

mq sm0 (restoring)

(8)

placement)

V-mg

cos0

(stiffness)

\

FIG. 6 A. Schematic diagram for simple pendulum. * B. Restoring force and stiffness* curves for simple

pendulum, showing non-linearity.

Write v

dd/dt y g/l

= a, and (3) becomes v dv ... + asm0 = 0, ,

do

f

so

giving If the

v dv v

maximum

= 2

a

sin0 d9-\-%A,

f

2a cos 0+-4

swing (amplitude) t;

=

0,

and

is

6

A=

(4)

(5) (6)

.

= ^, 2acosi/r.

Thus Put

AND FUNCTIONS

ELLIPTIC INTEGRALS

3.160

v

=

(2a)*(cos0 k,

sin|i/r

cos^)*

sin|0 = ksiny,

and

(7)

2

^--sin becomes

2a*(sin

33

2

(7)

|0)*.

the negative sign bei^g chosen, since decreases with increase in as reckoned from the instant of maximum deflexion. Also i cos

W d0 = k cos 9 d^, =

2&COS9 dq>/(l~k sm y)*. the for d0 from (8), (9) gives Equating expressions d0

or

t

=

0,

v

2

-

2

=

Accordingly on integrating

between the

(10)

(9)

2

2

-d9/a*(l-& sin 9)*. and 9 = |TT, and when Z = JTO dt

When

tt

limits

(10)

= =

,

9

0,

9

=

0.

we

(|TT, 0),

obtain the quarter period, so a~*

|T O

\

fc

dy/(I

2

sin 2 9)*

(11)

(l/g)*F(k,%7r),

o

where F(k, k

=

sin

\\fj.

a complete elliptic integral of the first kind, modulus Hence the time for a complete period is given by

|TT) is

From a table of elliptic integrals we find that F increases with increase in k, and, therefore, the greater the amplitude the longer the periodic

Consequently the motion is not isochronous, i.e. the time of swing a function of the amplitude, which is to be expected, since (3)

time.

is

is

non-linear in

3.161.

0.

Approximate solution of

write sin0

^

3

J0

.

3.160. If

(3),

Then the non-linear equation

0+a0+60

3

=

|0|

<

creases with increase in

0.

(3),

Thus the

Then

periodic time

is

is

(1)

g/QL In this case, as before, the

an amplitude

we may

0,

with a ~ g/l and b 3.14, for

^TT,

to be solved

'stiffness'

to a first approximation

de-

we have by

iff,

approximately

being a function of the amplitude 0. 3.17. is

Motion with viscous damping.

introduced into

(1),

3.10,

it

y+2 K y+(ay+by*) 5077

If a term representing this

becomes

n

-

0.

(1)

EQUATIONS INTEGRABLE BY

34

When = AC

by

0,

(7),

3.14,

<

AJAi and the motion

is

CH. in

1,

(2)

nearly cosinusoidal with angular frequency

~

(a+3Mf/4)^

(3)

y Q cos cot, approximately.

(4)

a>

Hence we may write y

Thus when K

>

0,

=

but small enough, and

(2) is satisfied,

the equation

to be solved takes the approximate equivalent linear form

Sf+2fcy+a>fy

For the initial conditions y = V = y

T/ O ,

an approximate solution of

is

Q

y e-

=

= Kt

0.

at

0, it

follows that

cosa)t,

Since

(1).

t

(5)

=

(6)

A ^ yz~

Kl

l

>

from

(3)

we

obtain (7)

so

co

-> a* as

t

-> +00.

qualitatively when to motion is oscillatory at hold. that the (6) Suppose the start with by% :> ay Q When the amplitude decreases to a value y,

The behaviour of the system may be considered

K

is

too large for

.

inonosay, the subsequent motion will be non-oscillatory, and y -> a as t -> -{-co. if the control For still /c, tonically (ay -{-by*) is larger

monotonically from the start.

inadequate to promote oscillation, y -> If

we change

the signs of K and

y

and

o>

^

= yQe

b, (6), (7) Kt

become, respectively,

cosa>t,

(a-3fo/ge

(8)

2

(9)

'/4)*.

Hence as

t increases from zero, the amplitude builds up, but the rate of oscillation decreases by virtue of the decrease in stiffness with increase in amplitude. The function ay by 3 has a zero value when K* y (a/6)*, so to avoid instability, y Q e (a/b)* in (8), (9).

=

<

Equation

(1)

may

be solved more accurately using the perturbation

method exemplified in

4. 1 1 0, 4. 1 3 1 ,

and

this is left as

an exercise for

the reader. 3.180.

that (2)

The

elastica. In the linear theory of elasticity

(1) stress is proportional to strain,

i.e.

it is

Hooke's law

assumed

is

obeyed; the deflexions are small so that in the formulae for the strain,

squares and products of the deflexions and their derivatives may be neglected. The elastica is a thin strut deflected beyond the Euler

ELLIPTIC INTEGRALS

3.180

AND FUNCTIONS

35

critical or is

buckling load. Here assumption (1) is maintained, but (2) Consequently the differential equation for the strut is a non-

not.

linear type.

In Fig.

7

A the

axis of the strut

CDEF is bent in the plane of the

Y

di

(B)

FIG.

7.

A.

Schematic diagram for the elastica CEF.

c, D.

Bent form

of elastica.

paper as shown, by two equal but opposite forces OX. Let

P acting along the

axis

I

P

= axial length of strut assumed constant, forces applied along the axis

s = arc length measured from 6 = angles of tangent to strut at y = ordinate at D, B El = flexural rigidity,

OX

as in Fig. 7 A,

(7,

i/r,

The

/

moment

E

modulus of elasticity.

(7,

and

Z>,

respectively,

of inertia of cross-section,

differential equation is [206]

(i)

ds or ds

(2)

36

EQUATIONS INTEGRABLE BY P/B. From Fig. 7 B, dy/ds = sin0, so (2) becomes .

with a

=

CH.

0,

which

identical with

is

we take

(10),

3.160,

3.160

(3),

and make ds

At C

in Fig. 7 A, y

substitutions in

=

write s for

& 2 sin 29)*.

=

3.160. Also

COS

if

-

is

(4)

so 9 |?r, while at any point on the 1 sin~ (sin|0/sin|j/f) by virtue of the

W'l/

o"(7

ckpjds

which

Consequently

t.

=

0,

E, q^

(3)

this change, so

dy/a*(l

0,

C and

between

strut

=

we

if

m

,

negative since 6 decreases with increase in

s.

Hence the

Then

negative sign in (4) is correct. 1

s

=

r

a~*

i?r

r

L n 6

(6) J

n o

F

represents an elliptic integral of the at E, (7) gives

where, as usual,

= ?1

I

Maximum

3.181.

(5)

q>i-,

=-'/-/ When

& 2 sin 29)*

^9/(l

=

2a~*F(k,

deflexion

3.160,

y

max

2/

.

kind.

(8)

JTT).

From

(1),

3.180,

= --- = -2a*icos9, r-

and

(8),

(1)

a

a as

first

(2)

When 9 =

ymax -

0,

-

2te-

M/F(k,

(3)

ITT).

Relationship between load and maximum deflexion. is about to buckle, $ = 0, k = 0, yraax = 0. Then in = = F a Pc/, the Euler critical load is 3.180, |TT, so taking

3.182.

When (8),

the strut

given by the well-known formula

P= c

Writing

(8),

3.180, in the

2

7T*B/l

~

f

2

10jB//

.

(1)

form

P=

4BF*/l

2 ,

(2)

ELLIPTIC INTEGRALS

3.182

we

= ijF /^ ym^l = kjF. 72

PJPC

get

by

Also,

3.181,

(3),

AND FUNCTIONS

37

2

(3)

.

(4)

Using tables of F, we plot PJPC against ym&Jl for various values of k = sin Ji/r. The graph is shown in Fig. 8 A [206]. The maximum value of 2/max/' corresponds to the configuration of the strut (see Figs. 7 o, D) when the ends C and F meet. For greater values of $ and, sin |^, the ends are thrust apart as in Fig. 7 D, so therefore, of k

=

decreases with increase in

P/PC The .

reader should plot the

2-0

P/Pc

1-5

Eul

/ ra

0-1

0-2

03 04

02

05

bical

ymavt

tTci

04

0-6

MA

08

1-0

(A) (B) Relation between (load/critical load) and (ymaT /J) for elastica. B. Relation between (load/critical load) and (l-d)jl for elastica, from critical load where strut buckles, to that where the two ends meet.

FIG.

8. A.

curve of Fig. 8 A and also bend a thin strip of metal to the shapes shown. 3.183.

The distance

d in Fig, 7

A.

This

is

given by

if

d Since

cos0= l-2sin 2 ^

f

=

dx

=

2

f

cos B ds.

l-2& 2 sin 29, from

(1)

(1)

and

(4),

3.180,

we

^

obtain

-

2a~*

2

J [2(l-Psin 9)i-l/(l-A:

2

sin 29)i]

*p

(3)

o

=

2a^[2E(k^7r)-F(k ^)l

(4)

y

E are complete elliptic integrals of the first and second kinds, = [contraction/ respectively. We can now compute the ratio (ld)ll

where F,

original length] for various values of k.

may

Then the graph of Fig. 8 B P/PC and (ld)/l,

be obtained showing the relationship between

EQUATIONS INTEGRABLE BY

38

CH. in

between load and contraction [206]. The curved nature of the graph is accounted for by (3), 3.180, being a non-linear

i.e.

in effect

equation.

Put

= dy/dx and

v

+ (dy/dx)*]* =

d 2y/dx 2 +2ay[l

3.19. Solve

0.

(1)

becomes

(1)

)t

=

0,

(2)

80

and, therefore,

(l+i;

Thus

v

?+? =

where

z

= =

=

2

)*

-ay 2 +A.

2

(A

(4)

1

ay*)

(5)

(4-l)(l-a!y)(l-j8fy),

2a^/(4

2

-l),anda?$ = a*/(A

2

-l).

(6)

From (6) we get

Vi a

dy/[(i-Jy )(i-^y")]*+5, having arbitrarily chosen the positive

sign,

and taken y

(7)

=

(0,

yj

as

the limits in the integral.

Write u

a^y and

(7)

becomes i?/i

2 af

where

=-7Tr-fTt a ll^ ~~ *)

sin a

Let ^

fii/oii

<

f rf/[(l-tt

)(l-BinM]+5,

1.

sin0, then d^

=

u 2 )*

(1

the form

difj,

so (8)

may

^(^L'-l)*

oc

l

yl

=

sin 9,

and k

=

(9)

J

sin a.

A,

B are

+ 12y = 2

3.20. Solve

Write v

be expressed in

<

=

with

(8)

J

^p| dy/dx, dx

|(1

= di//v,

and

)

(1)

constants of integration. 0.

(1)

becomes (2)

ELLIPTIC INTEGRALS

3.20

so

J

vdv

=

4 J

v2

giving

g=

Thus

(l

AND FUNCTIONS

39

+ 12y*) dy+lA,

(3)

= y+4y*+A. v

=

(4)

(y*+y+A)*,

(5)

from which we get (6)

We

now express the solution in terms of the Weierstrassian function elliptic fi(u) defined by the following integral [173] shall

00

=

or

J dyKW-ffty-ga)*,

y

= g>(u,g ,gj.

(7)

a

p(u)

g2 <73 are called the invariants of @(u). The expansion for terms of the invariants is ,

To express Then

(6) in

terms of p we have g 2

=


=

^4,

=

(o;+JS).

we

(9)

If in (1) the

(8).

in (6) the coefficient of

is

\

y

is

(10)

zero, so g 2

=

0,

and the solution

,-A).

+ 4y-a-y =

3.21. Solve

where a and

term

get

6,^0. Thus

^

2/-p(o:+5,-l,-^)

whose expansion may be obtained from omitted,

1,

$>(u) in

c are positive

is

(11)

c,

(1)

non-zero constants.

we transform the equation to eliminate the term 4j/, by writing (w+2a). Then

First

y ^

= a

_-_w =

d 2w

Writing v

=

2

.

(

.

.

C -4a)

= 6. ,

c,

(2)

7

_

v

(3)

dw/dx yields .

(4)

ELLIPTIC INTEGRALS

40 so

f

vdv= I

and, therefore,

v2

or

6av 2

with g 2

AND FUNCTIONS

(b+^\dw+^A

A-{-2bw-\--

3 a

12ab, g 3

~

~

4w 3

g2 w

g2 so

I

dw/(4:W^

g%w

^3)^

^

(7)

,

wg

sign, (8)

3 )*

=

^ = p^,^^)= t^+2, the solution is given by y (72)^3) + 2a, fc)(u,

and

g3

(9)

it,

(6&)^

and, therefore,

^4

(6)

,

=

J

Since y

(5)

y

6aA. Hence choosing the positive ttx

CH. in

being constants of integration.

(

10 )

(H)

CHAPTER

IV

EQUATIONS HAVING PERIODIC SOLUTIONS 4.10.

Self -oscillatory thermionic valve circuit. This

schematically in Fig. 9 A,

is

by a short

replaced

is

shown

but the applied potential difference

circuit.

The sum of the

potential differences

Eg grid potential

(A) FIG.

(B)

diagram for thermionic valve oscillator, with oscillatory circuit between grid and cathode, together with 'driving' potential difference J57 cos( 0^4-9) applied in series therewith. B. Assumed cubical-parabolic relation between anode current and grid potential. The working part of the curve lies between P and Q O is the centre of oscillation. The 'actual' curve beyond P, Q takes the form indicated by the broken lines, i.e. the current tends to zero at P, and to a saturation value at Q. 9.

A. Circuit

;

round the grid

circuit

must vanish, so

the differential equation

in the absence of grid current

is

Idt-M I

0.

dt

(1)

M

The signof is minus, this being a known condition for self-oscillation. Assume that the anode current is given by Ia

=

<,[Eg -E*l*Et},

(2)t

the transconductance of the valve, and Es is the grid potential corresponding to the anode saturation current. The

where a

is

f This relationship applies when the grid potential is adjusted so that the valve operates about the point of inflexion of the curve (see Fig. 9s). Otherwise a term in 0, which E% must be included. (5) then takes the form y 6(1 y*~ b^^y + ay f introduces unnecessary complication into the analysis.

EQUATIONS HAVING PERIODIC SOLUTIONS

42

relationship at (2)

is

EJE = u, since Eg = 8

shown graphically in Fig. (1/0) J / dt, (I) becomes

9 B.

CH. iv

If

we

write

= 0, tibu+c-- + au =

or

with

u

=

a=l/LC, (6/3c)ty, (4)

=

b

t

(4)

=

aM/3LC.

Writing

takes the form [122]

y-b(l-y*)y+ay Replacing

0,

c

(aM/LC)-(R/L),

(3)

by a~H, and a~^b by

(5)

e,

in (4)

0.

(5)

becomes

y-e(l-y 2 )iH-y For a sustained self-oscillation, 4.311 where \I Q > 0),

=

=

0.

(6)

we must have

b

>

(see (8),

\

aM/LC >

so

R/L,

(7)

M > CR/a.

or

Thus the mutual inductance between the exceed a certain

The

(8)

grid

and anode

coils

must

critical value. 2

the damping coefficient, which ) represents alternates in sign during the steady oscillatory state. When positive, energy is dissipated in the circuit, but when negative, energy is factor

y

6(1

supplied from the external source and compensates exactly for the the motion is periodic. If in (5) a~*6 <^ 1, as shown below

loss, since

y

^ 2sino>

where

,

eu

=

-b(l-y so the

which gives

a*,

~

2 )

-6(2cos2aV-l),

sign at twice the frequency of the steady increased sufficiently to make 1 <^ a~*6, the circuit exhibits relaxation oscillations (see Fig. 63), which will be

damping changes

When

oscillation.

M

is

considered later. At the

moment we

shall take

obtain an approximate periodic solution of (5). Let y Asmaj Q t, then substituting into

=

and integrating from

t

=

21T/OJO

f

sini/rcosi/f dt

to

27r/co

,

(5),

with

< a~^b

<^

1,

and

multiplying by

=

co

,

we

get

27T/OJO

co Q

bA

\

(l

+aA

smifjcoa^dt

=

0,

(9)

EQUATIONS HAVING PERIODIC SOLUTIONS

4.10

The

first

and third

integrals vanish,

43

and the second may be written

-

-bAir(l -A*/).

(10)

o

For

we must have A

this to vanish,

=

2.

Repeating the above procedure but multiplying by sinoj

,

we

obtain 27T/CUO

27T/0>o

dt-co Q bA J sm*t

>l

(l-^ 2 sin 2
j

sinV*

J

The middle

integral vanishes, the first

is

O>Q

a rough approximation. Hence

solution j,

=

0,

=a

(13)

we obtain the approximate periodic

2sina.

becomes

(14)

dy

dv-e

v

and third

by

=

0.

(16)

virtue of periodicity, so with

(16) gives

27T

J

(15 )

J (!-T/> dy+ lydy

integrals vanish

v dt, y = A sin,

it

27T

00

J first

dy/dt

and

nearly, so

27T

27T

The

=

= 2rr

(11)

(12)

Alternative method of determining A. In (6) write v

The steady motion has period

0.

and third giving

Aw^TT/WQ+aAiT/ujQ so

=

27T

(1-^2 ^ ^

Hence

7r(l-^

^2

2

/4)

=

r

(l-^2gi n 2^ cos 2^ 0,

so .4

=

2,

^=

0>

(17)

(18)

The physical meaning of (17) is that over a period 2-n the due to circuital damping is equal to the energy obtained energy from the unidirectional supply source, the valve acting as an auto-

as before.

loss

matic regulator.

method applied to (5), 4.10. We employ method, which was used extensively by Lindstedt and Poincar6

4.110. Perturbation this

for the solution of perturbation problems in celestial mechanics, to

EQUATIONS HAVING PERIODIC SOLUTIONS

44

CH. iv

4.10 [90, 193]. obtain a better approximation than (14), that the periodic solution may be expressed in the form

= 2/o+fyi+&

y

where

t/

,

2/

lv

..,

2 2/ 2

+-

<

6 <^

a

1 is

=

ao

the perturbation parameter. Also

+6 ai +M* 2 +...,

- ^(0) -

...

(1), (2)

-

,

a>

0,

into

shall take the initial

< b ^ 6 it follows that y (0) - Q A, &(0) = y (0) -

Since these are to hold for

(5),

a

...

-

0.

4.10, leads to

..

coefficients of b

Equating

r

to zero for r

6:

2/o+o2/o

so

with ^4

=

=

=

0, 1, 2,...,

i/r

obtain (4)

a)

oj

we

(3)

0,

y = ^ sinco +J5 cosco Since y (0) = 0, y Q = Q A, it = so with co^.

A,

let (2)

We

a ,a lv .., being constants to be determined. conditions

Substituting

(1)

are continuous twice-differentiable functions of t to be

determined, and

2/o(0)

Assume

(5)

,

follows that

BQ =

0,

,

(6)

(9)

cos3i/f.

The

particular integrals of (9) corresponding to the terms in sin
solution

is

to be periodic,

it

follows that the coefficients of sin iff, cos $

must vanish independently. Hence ttl

so

= 2/

Thus

(9)

becomes

A =

0,

=

2,

(11)

2sini/r.

=

2

% cos

(10)

3lA >

(

x

2)

EQUATIONS HAVING PERIODIC SOLUTIONS

4.110

of which the complete solution

yl

and since ^(0) Thus we have

45

is

= A smi/j+B cosi/t (l/4oo )cos30, = ^(0) = 0, we find that A = 0, B = ,,,, w nl y = (l/4a> )(cos0 cos 30). l

(13)

1

l

l/4o>

l

.

....

.

,

(14)

l

(16)

Substituting from (11), (14) into the right-hand side of (15) gives 2/2+ a o2/2

=

2a 2 sin0+(

4siri 20)

Jsin0+f sin30)(l

-sin20(cos0

=

(2a 2

)sin0

Since the solution

Vani8h

'

f sin

(17)

=

(16) (17)

5(/(.

to be periodic, the coefficient of sin^

is

S 2

Then

3^+| sin

cos 30),

must (18)

1/8.

becomes

=

2/2+ a o2/2

f sin

of which the complete solution

30+| sin 50,

(19)

is

(20)

Since y 2 (0)

y 2 (0)

=

=

7^-^ sin r

0,

we

find that

A2 =

29/96a)g, J? 2

=

0,

and,

therefore, 2/2

Whence from of

29

3

(1), (11), (14), (21),

=

(5), v h S* 4.10,' is

296

2

(

5/

sin

I

sin 50. r

to the second order in ,

/

fe,

(21)

the solution

o 2

,

/0 o\

(22)

6

6

\

2

5

+ T7r-oSin30

2

+ 4co-(cos 0- cos 30) + ---2 (3 sin 30-| sin 50). 16o>5

(23)

Also by

(2), (10), (18),

to order

a so

From

= o;

(23), (25) it is

two

oo+ifi

2

in 6,

-

cai(l-6

evident that

" 2 +i& 2

(24)

>

2

(25)

/16a).

when

6 is small

enough and a

large

enough, the approximate solution (14), 4.10, is satisfactory. (25) shows that the influence of damping is to reduce the angular frequency of the oscillation by a small amount. In this respect it is of interest to compare (25) with the angular frequency obtained from the linear

equation

y+by+ay

=

0.

(26)

EQUATIONS HAVING PERIODIC SOLUTIONS

46

= cog

Here so

o>

The reduction It should

in

a>

(a-J6

OH. iv

2

(27)

),

6 2/8a).

^a*(l

due to damping in

(28)

(28) is twice that in (25).

be observed that the method of solution

is

one where

successive approximations are found by solving linear differential equations, such that the initial conditions are satisfied.

4.111. (5),

Remarks on amplitude

4.10, 6 being

a real constant,

limitation. Writing 0y for y in we obtain

y-6(l-0y)jH-ay

= =

0.

(1)

Thus (5), 4.10, is reproduced only if |0| 1. It follows, therefore, that the solution of this non-linear differential equation cannot be multiplied by an arbitrary constant. For given a, 6, the coefficients of the various terms in the solution are unique. When the differential

equation has the form oscillation is nearly 2,

amplitude of the fundamental 0-1. If the equation were (1)

4.10, the

(5),

< a~*b <

above, the solution would be (23), 4.110, multiplied by be derived from (4), 4.10, by writing 6 2 3c/6.

0.

(1)

=

may

Amplitude limitation may also be considered in the following way. During growth, when the amplitude is small, the c term in (4), 4.10, is negligible, so the equation becomes

#

bu+au e iW

=

0.

(2)

The solution of (2) takes the form Under this condition the anode current-grid

sin(co+0), with

o>

=

(a

J6

2

)*.

potential relationship of unlimited characteristic -E linear. a linear is Moreover, (Ia g) -> -> t oo as extent would entail an amplitude which +00, by virtue

of the negative resistance property of the valve. Owing to curvature of the characteristic (Fig. 9 B), amplitude limitation occurs. By aid of the differential equations (4), (5), 4.10, we see that non-linear 2 damping represented by the term by y is introduced as a consequence

of the curved characteristic.

Thus

stabilization

is

effected

by the

inherent non-linear property of the valve.

Frequently if a system, electrical, mechanical, etc., is unstable, it may be stabilized by introducing some form of nonlinear element, e.g.

the systems described in

damping

is

non-linear.

7.13, 7.14, 7.420, 7.422

The action of a non-linear

element in a mechanical system

may

where the

spring-control

be compared with the detuning

EQUATIONS HAVING PERIODIC SOLUTIONS

4.111

47

of a resonant mass-spring arrangement, when either the mass or the 7.20-7.22, 7.410. spring stiffness varies with amplitude, e.g. 4.12.

Self-oscillations with electric motor-generator comThe schematic diagram of Fig. 10 A shows an electric

bination.

motor with separately excited electrically to a

dynamo

field, coupled mechanically and a series field. If / is the circuital having

series

separate

generator!

C supply

generator

motor

coupling

E(non-lmear)

shaft

Go>

FIG. 10. A. Schematic diagram of series-wound D.C. generator separately excited D.C. motor. B. Circuit diagram for A.

current, the potential difference at the generator terminals

mately *

TVT

ti

=

T

!/

CX

and

is

TQ 3

3

/

approxi/1X (1)

,

sum of the potential differ(x 3 being positive constants. Since the ences round a closed circuit is zero, we must have ax

,

Ldl/dt+RI+Gu-E =

0,

(2)

where L, R are the total circuit inductance and resistance, respecis the back e.m.f. at the motor terminals per unit angular tively, and velocity co. Inserting (1) in (2) and differentiating with respect to t gives

LtflldflfaR)

dl/dt+oiz

d(P)/dt+G

da>/dt

=

0.

(3)

Apart from ohmic loss, the power to the motor is the product of back e.m.f. and current, i.e. Gcol. In absence of mechanical loss, this is equal to the rate of change of the kinetic energy of the rotating system. Thus if I is the moment of inertia,

so

efco/eft

=

GIjl.

EQUATIONS HAVING PERIODIC SOLUTIONS

48

CH. iv

Substituting from (5) into the last term of (3), and dividing throughout by L, yields the equation

l-bl+cd(l*)/dt+al with a

=

G

2

=

/IL,b

(a^BJ/L.c =

a 3 /L.

=

0,

(6)

(6) is

identical in form with

> <

>

Hence if 6 R, 0, i.e. a x the system will be self-oscillatory, but non-oscillatory if cx x R. When the former condition is satisfied, the motor-generator combina(4),

4.10, for a thermionic valve circuit.

For a relaxation type large enough and a small

tion rotates periodically in each direction. oscillation (sudden reversal),*)* b

must be

enough. 4.130. Solve

y+ay+by*

-

(a

>

>

b

0,

(1)

0).

This equation pertains to an undamped mass-spring system (see Fig. 2) in which the spring control J takes the asymmetrical form ay-{-by

2 ,

If the spring

illustrated in Fig. 11 A.

=

extended and the

We write

mass released, the ensuing motion is periodic. y+co*y

is

K=

(

1)

in the

form

2

(aj*-col)y-by*

a)

(2)

and use the method of successive approximation (iteration). First we assume the right-hand side to be negligible, so with = cut, the ifj

complete solution

is

y

Taking the

initial

= ^4

1

approximation

y

(3)

cosi/r-fJ5 1 sinj/f.

=y

conditions y

y

Q,

=

0,

we

obtain the

= y.cos^.

first

(4)

Substituting this in the right-hand side of (2) yields 2

y+co y

=

Since the solution vanish, so

2

(a)

is

cog)2/

to be periodic, the coefficient of cos */r

or2

'

The complete y

i&y(l+cos2^r).

cos
solution of (5)

2

cog is,

=

~

(5)

must /*\ (6)

a.

therefore,

= 4co8^+B8in^ + MooB20-M.

(7)

A

relaxation oscillation is a self-excited type in which a physical quantity, e.g. f current, velocity, alternates abruptly between two levels, where it remains approxicurve in Fig. 63. mately constant for a relatively long time. This is illustrated

bj^the

md

=

2 z 2 3.10, (1) is a suitable form for the equation 0, y/dt ~{-s l y-+-8 z y 8 1 jm, b with a s 2 /m. Although the actual spring control is *!?/4-s 2 2/ 2 , it is con2 venient in analytical work to refer to ay -f by as the spring control, and this is done

J

As

in

throughout the text. It

is

the control corresponding to unit mass.

EQUATIONS HAVING PERIODIC SOLUTIONS

4.130

Now y =

2/

,

y

0,

at

t

AHence with

co

y 2

=

co

=

0,

(2/o+

so

follows that

it

W"

2 ),

and

jB

=

0.

(8)

=_ =

49

(9)

If

a.

&2/

/6a

<

the coefficient of cos2to will be

1,

tH

stiffness

displacement

y

(B) FIG. 11. A. Asymmetrical curve showing combination of linear and parabolic spring controls 2 ay, by respectively: ay -{-by* is neither an odd nor an even function of y. B. Stiffness-displacement curve for case (A). ,

small in comparison with that of cos cut. The remark at the end of 4.110 regarding the solution of linear differential equations applies here also.

The constant term (byl/2a) which the

implies that the central point about

oscillation occurs is displaced

language of the radio engineer, there

is

by

this

amount.

In the

a rectification effect, since in an

would correspond to a unidirectional current. In 3.111, 3.14 we found that, in the non -linear system the considered, frequency of the oscillation was dependent upon the electrical circuit

(

by%/2a)

amplitude, whereas in (6) this is not mation we neglected the term in 6. better approximate value of 6077

00/277

so, since in

We

our

first

shall, therefore,

approxiobtain a

using the method of perturbation. TC

EQUATIONS HAVING PERIODIC SOLUTIONS

50

4.131.

More accurate

solution of Q

(i)

2

l

2

and with

4.130. Let

(1),

= y +by +b*y +... a = w*+ bwl+b a>l+... conditions y A, y = 0. From these y

CH. iv

initial

(2) it

follows as in

4.110 that 2/

(0)

=

A, Vl (0)

=

=y

2/ 2 (0)

Substituting for y and a from

we

-

8 (0)

=

...

(1), (2)

0,

(<>)

-

2/i(<>)

=

...

-

0.

into the differential equation,

get

2 2 2 +by l +b 2y 2 +...)+(^ +bco +b aji+ +6 2 2 +26 2 / o2/1 +6 3 (2/ 12 +2i/ i/ 2 )... 2 Equating the coefficients of 6, 6,6 ,... to zero, we have

(y Q

/

b:

#o+"o2/o

so

^

0.

(3)

0,

A cos oo

2/0

-

2

(4)

(5)

;

(7)

],

To avoid a

by (5). must have

o>

2

=

0,

A

non-periodic term in the particular so the complete solution of (7) is

A

2

integral,

we

2

(8)

.

a>Q

With the above

A =^ Vl

-

(^

2

2

/a>

find that

2

2

l

so

we

initial conditions, /3co

J?!

,

=

0,

(9)

)[-^H|cosa; ^+icos2a;

(10)

^.

l 2/o+ 2 2/o 2/i)

A* 3oj [A*

J3

1

(12)

To avoid a

non-periodic term in the particular integral of (12), the coefficient of cos o> 1 must vanish, so o>|

Then the complete 2/2

=

=

5A 2 l6ajl

solution of (12)

is

(13) f

1

n 9

A3

1

cog

48

A* co

OS3c 0^ (14)

EQUATIONS HAVING PERIODIC SOLUTIONS conditions above, we obtain

4.131

Using the

A2 = Hence by of

(!),

y

-

51

initial

(1), (5),

4.130,

29,4 3 / 144w o>

and

JB 2

=

0.

(15)

(10), (14), (15) to the second order in 6, the solution

is

-[

o

^

(16)

bA /6a)%, i.e. bA/6a <: 1, the second and third harSo long as A monics will be small in comparison with the fundamental. Also by = 0, to the second order in b (2), (13), since cof

>

2

cog

If a

>

56 2 4 2 /6<4

=

(

a -56M 2/6w 2 ).

(17)

we obtain

~ a*(l-56M

co

2

/12a

2

(18)

).

from (18) that the frequency decreases with increase in which is just the opposite to the result in 3.14. At first amplitude, sight the term by* in the equation might be expected to ensure an increase in the spring control with increase in amplitude. The stiff2 3.111 and Fig. 11 B), ness, however, is (d/dy)(ay-{-by ) = a-}-by (see so it increases and decreases by equal amounts for y. But the oscillation centre is now at the point defined by the constant term in (16), It follows

so the effective decrease in control exceeds numerically the increase, as the amplitude increases (see Fig. HA). Thus the frequency

reduction

is

Fig. 11 A, the

and

explained by the 'rectification' effect. Referring to two spring control terms ay, by 2 are plotted separately

(QPR). For the latter curve, we see that the oscillamoves to the negative side of 0, where the slope is less

collectively

tion centre

than that at 0. Also

increases with increase in

\c\

A

owing to the

curve QPR being asymmetrical. Thus a> Q decreases with increase in A. For a spring control ay+by*, the stiffness is (a+36?/ 2 ) and increases with increase in y, whether i.e. it is an even function of y (see ,

3 Fig. 12 c). The graph ofay-\-by is anti-symmetrical, so cu increases with increase in amplitude. The minimum ordinate in Fig. 1 1 A occurs when y a/26, and for stability the negative swing must be

less

than

a/6, i.e. |ymin

|

< a/6.

apposite to remark in connexion with the foregoing analysis that a tone of double frequency (the octave) is audible when a tuningIt

is

EQUATIONS HAVING PERIODIC SOLUTIONS

52

CH. iv

fork vibrates with a large amplitude. As the latter decreases, the pitch of the fundamental tone rises. This is in accord with the analysis.

EXAMPLE. The reader should solve (1), 3.10, by the perturbation method using the initial conditions y = A, y = 0. The result to order three in 6

is:

bA 3

-- ---^^* ,

4-

IbA*

\

,

1

1

,

cos 3o)

.

-.

(20)

4.132. Effect of viscous

Write y

= e~'

c/

-u(^),

damping. The equation

y+Zxy+ay+by* and we get 2


is

small enough, by

=

an approximate solution of

0.

W=

K

4.131, writing be~

to be solved is

Ki

(1)

0.

for

(2) 6,

and

O>Q

for

(a/c 2 ),

(1) is

-

16^ 2e-^\

bA 2e-' ,

+^3

2

cog

6

/

cog

(3)

When

=

0, y =

A,y =

KA which is small enough to be neglected.

A better approximate solution of (1) may be obtained using the perturbation method. This

is left

as an exercise for the reader.

4.140. Solve

y+ay+by* =fcosa)t

(a

>

0,

b

%

0).

(1)

the equation for a loss-free mass-spring system, with control proportional to ay+by*, driven by a force/cos a>t as shown schemati-

This

is

cally in Fig. 12 A.

The

of the driving force at t

=

inevitable transient follows the application 0, but we may suppose that there is sufficient

damping to extinguish this and enable the periodic gftate to be attained after a short time interval. The damping is supposed to be small enough to have negligible influence on the amplitude of the motion. When a, 6, co, / have appropriate values, a subharmonic of

4.140

EQUATIONS HAVING PERIODIC SOLUTIONS

Fcoswt

53

(A)

m

A. Schematic diagram of mass and non-linear driven by an alternating force /cosorf. B. Antisymmetrical curve showing combination of linear and

FIG. 12. spring

s,

cubical springs ay, 6i/ 8 , respectively. The restoring force is an odd function of y, o. Stiffness-displacement curve for case B: a-f36t/ a is an even function of y.

frequency CO/GTT occurs. This aspect is considered in 4.190 et seq. For the time being we shall assume that the conditions for the existence of a subharmonic are not satisfied. In analytical work herein,

it is

tacitly

presumed that the reaction of the system on the

EQUATIONS HAVING PERIODIC SOLUTIONS

54

CH. iv

driving agent may be neglected. Thus the amplitude and functional form of the applied force is invariable.

For a periodic solution we may assume a Fourier series. Since ay -{-by* is an odd function of y, the corresponding force-displacement graph in Fig. 12 B is anti-symmetrical about the force axis. It follows that the solution may take the form

where

=

ifj

ay

=

We shall restrict ourselves

a)t.

stituting into

(1),

we

to

two terms

in

(2).

Sub-

get

a(A l cosi/j-{-A 3 cos3ifj)

by* J == (3)

Equating the have

or

coefficients of cos

on each side of

(1),

*

or*

=

by

aid of

=/,

(a+lbAl-f/A l )+lbAl

+ 2Yl

(3),

we (4)

(5)

This may be regarded as an approximation to the amplitude-frequency relation.

For the

coefficient of cos

3
we

obtain

(a-$a>*)Az+lbAl+lbA\Az+lbAl so

In

.4 3

(5)

=

assume \A^A^\

0,

lbAH($a>*-a-lbAl-lbAl).

<

1,

w2

Then

=

so that

=

we may neglect

(6) (7)

the third member.

(a+^bAl-f/AJ,

(8)

a condition which must be satisfied (approximately) for the solution of (1) to be periodic. If/ and A l are fixed, aj is given by (8). For fixed

and /, the amplitude of the fundamental vibration may be tained if (8) is written in the form

a)

|64J+(a-VMi-/=0. If b

=

0,

we

f

ascer-

(9)

obtain the well-known formula for a linear system,

A^fKa-v*).

(10)

EQUATIONS HAVING PERIODIC SOLUTIONS

4.HO (9) is

a cubic equation for

have three values:

A

l

55

which will, therefore, for fixed a, 6, co, /, (ii) one real, two complex conjugate.

all real,

(i)

The consequences of A l being multivalued will be discussed later. Next we consider (7). Substituting therein for to 2 from (8), and neglecting f bA\, leads to (11)

A 3 /A, ~ l/[21+{32a-36/MJM],

Thus

(12)

|21 + (32a-36//^ )/6^f > \A*IAi\ <1, which includes the two cases b > 0, b < 0. Hence if (13)

so

if

1

|

1,

(13)

is satisfied,

the forced motion of the system is nearly cosinusoidal, the displacement of the fundamental being in phase with the driving force. In the foregoing procedure, the non-linear equation is solved directly, whereas

the methods of iteration and perturbation entail the solution of linear differential equations.

Energy considerations. Writing

4.141.

takes the form so

v dv/dy +ay+b v dv-\-a

I

By

4.

(2),

j

y dy-\-b

dy/dt, (1),

^ = /cos ^ =/

3

\

=

v

y dy

J

cos cut dy.

4.140, (1) (2)

140,

dy Substituting

(3)

=

into

=

oj(A l sin

(3) ^+3^4 3 sin ty+...)dt. the right-hand side of (2), and integrating over a

(0, ^TT/OJ), the left-hand side vanishes by virtue of periodiand the right-hand side by virtue of orthogonality of the circular city, functions. Hence during steady motion, no energy is supplied to the system from the driving mechanism, as we should expect, since there is no dissipation.

period

t

4.142. Equivalent linear differential equation. By virtue of the motion being almost simple harmonic, the original equation may be replaced by what tion.

Then by

may be regarded as an equivalent linear approxima-

(8)

4.140, if f/A l

is

we have

negligible,

y+(a+I64|)y=/cosarf,

(1)

the particular integral being

y =/cosa>^/(a+|6^4f

co

2 )

A

l

cos wt.

(2)

Experimental illustration of analysis in 4.140 [92]. The apparatus shown schematically in Fig. 13 A, B has been used to obtain

4.15.

EQUATIONS HAVING PERIODIC SOLUTIONS

56

wave forms (1),

4.140.

for

CH. iv

a system akin to one represented symbolically by

A beam is mounted on a fulcrum so that oscillation occurs

about the latter in a vertical plane. The mechanical construction is such as to reduce friction to a small amount. One end of the beam is constrained by a non-linear type of spring while a mass m rotates at Rotating Unbalance

Fulcrum Timing motor Slotted time disc

Parabolic mirror fastened to ax le v

Axle fastened to beam

-s

s

\

Point sources ,

of

light

;yxNs.

Springs and supports rfot

Shown

Base/ Beam Concrete pier (approximate weight 1-5 ton)

Revolving

drum Slotted isc which rotates with unbalance

Fio. 13. A. Schematic diagram of apparatus for investigating system with non-linear restoring force. B. Plan view of apparatus.

radius r with angular velocity o> about the other end of the beam, thereby causing unbalance. If I is the moment of inertia of the oscillating parts about the fulcrum, f(0) the spring-control torque,

distance of the centre of rotation of

of motion

I

the

m from the fulcrum, the equation

is

2 where m(a) r cos ojt) is the accelerational force due to the rotation of m, and the right-hand side of (1) the corresponding driving torque.

The motion of the lever was recorded photographically, and a record is 14. The torque-deflexion curve /(#) for the spring is depicted in Fig. 15, being such that f(0) = 9), i.e. it is an odd /( reproduced in Fig.

4.15

EQUATIONS HAVING PERIODIC SOLUTIONS

57

1y

3o sec.

FIG. 14. Record of wave form obtained from apparatus in Fig. 13 using load-deflexion curve of Fig. 15. 60

h

50

O

0_

-30 &>

3

cr

10

-02

01

Deflection

-03 in

-04

radians

FIG. 15. Load-deflexion curve pertaining to Fig. 14.

function of

0,

and the curve

is

anti-symmetrical about the torque

Despite appreciable departure from linearity everywhere in Fig. 15, the wave form in Fig. 14 is almost a simple harmonic type. axis.

4.16.

Amplitude-frequency relation for

y+2 Ky+ay-\-by* = fcos(ajt+y). This equation is for a dynamical system of the type considered in 4.140, but with a term 2/cy representing viscous damping. The

68

driving

EQUATIONS HAVING PERIODIC SOLUTIONS force will now be out of phase with the displacement

CH. iv

corre-

sponding to the fundamental vibration, so to simplify the analysis have introduced the constant phase angle 9. may assume as

We

adequate approximation that with

=A

y

= 2/cy

Then while

an

a)t

iff

cost/j+A 3 cos3i/j.

(I)

2a)K(A l sin *fj+3A 3 sin

/cos(o)+9)

.

l

we

= /(cos 9 cos

(2)

3
sin 9 sin 0).

(3)

Using (2), (3), (3), 4.140, and equating the coefficients of cosi/r, sin^ on each side of the above differential equation, we obtain

and

2a)KA l

terms involving

A3

smallness (see

4.140).

,

/sin 9

virtue of their relative

A\ being neglected by

A\,

Squaring and adding,

;

(4) yields

[(a-^+|6^)2+4a,V]Jf=/2

(5)

(

which is the amplitude-frequency relation for the non-linear system with viscous damping. When K 4.140. 0, (5) degenerates to (8),

Amplitude-frequency curves. These are obtained by plotting the relationship between co and \A\ in (5), 4.16, for given values of a, The general trend of the AC, and different values of/.

4.170.

fe,

curves will be gleaned from Figs. 16 A, B, c, which correspond to 6 0. The curve through o> 0, 6 0, and 6 a*, \A^\ 0, is

>

=

<

given by

4.16,

(5),

when/ =

K

=

0, i.e.

for6>0, and

1^1

It is evident

from

=

(-
2

for b

)

<

(1)

0.

(2)

Figs. 16 A, B, c that the curves for b

=/

two

complex conjugate. If

roots of

(5),

4.16, are

3,

the roots are real, and \A\

<

pairs at co t co 2 If &> 2 are complex conjugate. ,

.

<

>

or

<

sheared over to the right or

correspond to those for 6 tively. In Fig. 16 A with/

if a)

<

a)

v

|^4 1

is |

left, respecsingle-valued so

co

l

^

co

^

o> 2 , all

triple-valued, there being coincident Mil ^ s ag ai n single-valued, and two roots is

The curves of

Fig. 16

similarly, but of course with the proviso that

B oo

may

be considered

>

When

0.

K

=

0,

EQUATIONS HAVING PERIODIC SOLUTIONS

4.170

59

the amplitude-frequency relation is that at (8) or (9), 4.140. The curves resemble those in Figs. 16 A, B except that they are not closed at their upper ends. They approach the curves (1), (2) for

>

< 0. When

2

;> a, the amplitude-frequency curves for b are asymptotic to the straight line I-4J b

0,

to

=

>

(B)

FIG. 16.

Amplitude-frequency curves for driven mass-spring system, with spring characteristic of the

form

ay-}- by

3 .

For

6

<

use

|6|

in formulae.

Consider the mechanism of Fig. 13 whose motion is represented (mutatis mutandis) by the differential equation in 4.16. Starting at P in Fig. 17 A with b 0, let the frequency be

Jump

phenomenon.

>

increased gradually. The operating point will travel to Q, where \A\ drops suddenly to R and continues to move along to S with increase in

From S suppose the frequency is decreased gradually On reaching T \A!\ jumps suddenly to U and then follows the curve down to P.

a>.

.

y

Thus a form of hysteresis

when b

is

exhibited.

<

A similar behaviour is evinced TQ

cannot be 0, as indicated in Fig. 17 B. Operation along realized in practice, and the explanation is as follows: In Fig. 16 A

>/ >/

The point there are three amplitude curves for forces /3 x 2 a If is lies on such w an increase constant, f'2 P! that/3 2 curve/g

>

>/

.

.

EQUATIONS HAVING PERIODIC SOLUTIONS

60

M

(B)

f = O.K =

V

b>o

CH. IV

b
Sti

Ui

|A,|

FIG. 17 A, B. Diagrams illustrating 'jump' phenomenon in driven system having spring characteristic of the form ay+by*. See [11] respecting B.

OOZ4

0020

I

CO

4

0-016

2 03

-02

-01

.

.

01

-02

Deflection

*

.

in

-03

0-012

"5

radians

-y

10

"|-0-008 -20

0-004

'

50

150

100

Frequency

200

250

\n rad./sec.

(A) (B) FIG. 18. A. Load-deflexion curve pertaining to amplitude-frequency curve in FIG. 18s. B. Amplitude-frequency curve obtained from apparatus of Fig. 13 using loaddeflexion curve of Fig. 18 A.

in /g entails anincreasein^l 1 sod|^4 1 |/d/ ,

But

P

>

0,

and the system is stable.

constancy of co entails a decrease in and the system is unstable. so d\Ai\fdf at

2>

\A^\

when /increases,

<

Using the apparatus o^ Fig. 13, and a spring whose control-torque graph is shown in Fig. ISA, the curve depicted in Fig. 18 B was obtained. The jump phenomenon is exExperimental

illustration.

hibited distinctly [92],

EQUATIONS HAVING PERIODIC SOLUTIONS

4.171

61

4.171. Loci of vertical tangents in Figs. 16 A, B. The values of 4.16, are parallel to the \A\ co, \Ai\ for which the tangents of (5), 0.

axis, satisfy daj/dA l

A we l9

Differentiating

(5),

with respect to

4.16,

get

/

}=

0,

=
When

K

=

= 0, or a-oj +|6^f = 0, or > 0. When 6 < 0, write

0,

(3)

= [(4/3&)(a> -a)]*, = [(4/96) -^)]*, 2

j^l

corresponding to

(7)

(6),

\A^\ (a

(7)

(eo

a>

2 )

for

2

a).

(a>

marked accordingly

are

(6)

2

2

with 6

=

0.

gives

0, (5)

a-o> 2 +|6^|

and

]

(1)

The graphs

in Figs. 16 A, B.

There are no vertical tangents corresponding to the curve represented by (6), since the amplitude-frequency curves never cross it. However,

we may consider the case to be that for K

-> when the tangents exist. the of take the form of the broken line curves in 0, (5) graphs 16 back to A, B. Referring 4.170, we may infer that the region Figs. within the broken curve in Fig. 16 A (also 16 B) is one where the motion

For K

>

of the system

is

unstable. This

region in Fig. 38

4.18. I

J

between

b1

may

be compared with the unstable

%

which corresponds to equation

and

Energy considerations. Proceeding

vdv+a J ydy+b I

\

j

y* dy+2Ka)*

=

Integrating over a period t left-hand side of (1) vanish 27T/CO

2777W

\

J

4.141,

we

obtain

2 (A l sinJj-}-3A 3 smty+...) dt

(0, 27r/o>),

by

as in

the

first

three integrals on the

virtue of periodicity.

Thus

EQUATIONS HAVING PERIODIC SOLUTIONS

62

we

so

CH. iv

obtain the relationship

)Mi+(5co)M|+...]

The expression on the

=

77/^ lS

m9

left-hand side of (3) represents the

.

(3)

work done

in overcoming viscous loss per period. The right-hand side represents the energy supplied by the driving mechanism. Each term on the left-hand side, represents the loss for the corresponding harmonic.

The energy is supplied by the driving mechanism at the fundamental frequency o>/27r, since the only contribution from the right-hand side of (2) comes from the term involving (sin^) 2 This must be so from a .

physical viewpoint, since the driving force has

no higher harmonic

components.

Owing (3)

to the approximate nature of the analysis, the two sides of 2co/c^4 1 // To demonstrate this, substitute sin 9

=

are not equal.

from

4.16,

(4),

and we get 2rrcoKAl [I

for the left-hand side,

+ (3A^/A^+ (5A,/A

l )*+...]

and (5)

2rra)KAl for the right-hand side. Since

tion at (3)

4.190.

is

in keeping

Harmonic

(4)

[(2r+l)^ 2r+1 /^4 1 ]

2

<

1,

the approxima-

with that in the rest of the analysis.

of order

J.

Under

certain conditions a periodic

solution of the differential equation

y+ay+by*=fcosa>t

(a,6,/>0)

(1)

has a term involving coso>. Since (1) is unaltered if (t-\-2n7r/a>) replaces t, it follows that if cos ^tot exists, so also do cos \(a}t-\-2ir) and

Thus from a mathematical viewpoint, there are three subharmonics of order 3 having the same amplitude, but differing in cos

J(co^-f-47r).

phase by

The

2?r/3 radians.

analysis given below does not prove the existence of a sub-

harmonic. It

is

based on the hypothesis that (1) has a subharmonic 1. An experimental demonstration for (1) with a

solution if b/a <^

small damping term 2/ct/, andfc/a^> 1, is reported in [926]. Moreover, it may happen that (2)-(6), (8), (9) are usable when b/a is much larger

than we have contemplated.

As a

first

approximation we assume that

y

=A

cos ^ajt-\-A 1 cos wt.

(2)

EQUATIONS HAVING PERIODIC SOLUTIONS

4.190

63

Substituting (2) into (1) and equating the coefficients of cos a)t on each side, leads to

=

[(a~l^)+lb(Al+A k A l +2Al)]A^ and o>

between

A ^

provided

(3), (4)

l

l

k

0, for

hold.

When A

J

AJ

for

=

0,

=

by

(4)

there 2

-^!-/ =

0.

(7)

must be compatible for the subharmonic Solving (6) for A$ gives the two values

from

real, it follows
2

>

to exist.

that

(8)

9(a+214f 6/16),

Substituting into

(5),

(9)

\A V when the subharwe obtain

the equality sign corresponding to A^ == starts.

(6)

the single relationship (writing

is

(4), (6)

be

(5)

4//6,

only then does the equation

feJf-fco

A must

(4)

gives

(2lAl+2lA A +5lAl+^2a/b) =

(a-fa*)+&(Al+AiAi+2Al)

monic

(3)

(a-o> )^ 1 +|6(^f+Mf^ 1 +3^?) =/. 2

AlA

Since

0,

2

Eliminating

t

cos^ootf,

after reduction

^p6^5+8a^ 1 +/=0, which has only one first term gives which

is

real root, since a, 6,

Inserting

A\

< 256a/3436,

=

positive.

Neglecting the

^ ~ -//to,

an adequate approximation

A\

/ are

(10)

3

(//8a)

if 2

or

into (10)

(11)

0-02&/ /a

and

3

<

1.

(12)

solving, yields the second

approximation 2

A*1

~ -/.(l- 3436/ ^ 16384a \

which

is

the value

oj c

8a

/

when the subharmonic

v

starts.

^ 9(a+216/ /1024a 3a^, provided 2l6/ /1024a < co

so

-/.(l-Q.026/ ^

3

8a\

2

2

Using the equality sign in the equation

3

2

(9)

3

2

/a

;

), h

(13); V

Also

2

(14)

),

1.

and substituting into

^bAl+Sa^+f =

0.

(7),

we

obtain (15)

EQUATIONS HAVING PERIODIC SOLUTIONS

64

From

CH. iv

that to the degree of approximation enis alteration in A l when the subharmonic starts. there little visaged, The value of A l at (1 1) indicates that it is almost on curve in Fig. 19. (10), (15) it follows

Solving

gives

(6)

r

i

1

MI

39

(16)

If in

0.

provided

(5), (6)

we put A t

=

0,

(17)

FIG. 19. Curves showing amplitude of fundamental and J harmonic in the periodic solution of y-}-2y-}-Q'05y 9 2coso>.

A when A =

=

A when A ^ 0. When
l

A^,

;

t

l

curve

(J).

(J)

corresponds to the curve in Fig. 17 A to the right of S, where A r (not \A t \) is negative.

the negative sign being inadmissible, since

Al =

0,

and

A = k

and the solution of

2

=

not

satisfied.

When

gives

(4//fc)i, (6) o>

(4) is

9a+27(6/

2

(18)

/4)*,

then

(1) is

y=(4//6)*costarf.

.

(19)

These results enable us to obtain an equation which has a subharmonic solution A cos at. Writing 3o> for w, 4//6 = A* in (18), gives a = ( so (1) becomes ,

3

= /cos 3^.

(20)

The other subharmonic solutions are A cos(a>+277/3),4 cos(a>^+ 477/3).

A

similar equation is given in [27]. a subharmonic solution of 2

2/+(4o> /3.4

3

2 )i/

=

If 3/

= 4o>

2 ,

then

A Goscot

is

(21)

EQUATIONS HAVING PERIODIC SOLUTIONS

4.191

4.191.

a

=

2, b

Example. To illustrate the analysis in 4.190, we take 0-05, f = 2. When the subharmonic starts

A l ~ -f/8a = Using

(9),

4.190, co

2

=

given in Table

2.

A = -\A^ =

and

-0-125,

we

Assigning values to

A^

and using

4-24

~

(1)

4.190,

(5), (6),

3a*.

we

(2)

obtain the data

These are exhibited graphically in Fig.

A =

Data

is

relationship between tion of the case when b

<

4.192. Effect of viscous

19.

The

inserted also.

relating to

The

2

J harmonic

A l and co is sensibly linear. An examinais left

as

an exercise

for the reader.

damping. The differential equation is now

y+2Ky+ay+by* approximation we assume y = A

=

w

and

18-004,

TABLE

first

0-0625.

find that

graph corresponding to

For a

65

=fcosojt.

(1)

that

cos

l

cos wt-\-

B

l

sin cot

(2)

(3)

In

(3)

= C = C} l

X

(A\~\-B\)*, the

amplitude of the subharmonic

(Al+Bl)*, the amplitude of the fundamental of" ), the phase angle due to viscous the subharmonic ing;

== tan- 1 (j8 the phase angle of 1 /^4 1 ),

compare

damp(2),

4.190.

the fundamental Substituting either sections,

we

(2)

or

(3)

into (1)

ultimately find that if 6

the J harmonic

is

and proceeding as

>

0,

K

>

in previous are not too large,

present provided

its Referring to Fig. 19, curve <3), with K small, as co increases beyond value at R, the denominator of (4) increases due to reduction in A v 5077

CH. iv EQUATIONS HAVING PERIODIC SOLUTIONS Ultimately a value of CD is reached when (4) is not satisfied, so the sub-

66

harmonic ceases, and A jumps to its value on curve e.g. to S. Since an alteration in the motion is associated with a change of energy, the 'jumps' must be interpreted accordingly. As co approaches a critical value, there will be a transition stage during which the ,

amplitude of vibration either increases or decreases at a finite rate. In other words the diagrams showing 'vertical jumps' refer to the steady state. .Diagrams having time bases are needed to show the nature of the vibration at either a 'jump', or at the initiation or the cessation of a subharmonic. Oscillograms for the two latter cases are reproduced in [926].

4.193.

Experimental illustration of subharmonics of higher

order. It seems likely that when the 'spring control is representable b by a polynomial of the form ay-\-by*-{-cy -\-..., odd in y, there will be

harmonics of order

is illustrated by the show \ and \ harmonics They which were obtained with the apparatus of Fig. 13 and a control

\, J,... (see [38, p. 113]).

This

oscillograms reproduced in Figs. 20 A, B.

form depicted in Fig. 20 c. In these experiments was a small amount of damping, so K ^ 0. The fundamental

characteristic of the

there

present in both cases, being of smaller amplitude than the subharmonic, and this agrees qualitatively with the data in

vibration

is

Table 2. Subharmonics of this type may occur in aeroplane structures, the rate of vibration being a sub-multiple of the engine speed.

Subharmonics down to ^th the fundamental frequency have been demonstrated with the apparatus described in reference [92 a]. The oscillating member is a bar of I-section pivoted at its lower end (see Fig. 4 A, c).

A metal strip, anchored as shown, passes between two

shaped blocks which cause a variation in its effective length when In this way a non-linear spring control may be

the bar oscillates.

obtained, whose characteristic is determined by the shape of the inner faces of the blocks. The oscillation is maintained by a harmonic force

due to the rotation of an unbalanced mass near the upper end

of the bar.

4.194. Solution of

y+ay+by 2

= fcoawt. We make the same

basic

assumptions relating to damping, subharmonics, etc., as in 4.140, while the amplitude must be limited in accordance with the last

paragraph of

4.198.

Writing wt

=


the equation becomes

=/cos
(a,

6

>

0),

(1)

4.194

EQUATIONS HAVING PERIODIC SOLUTIONS

67

501-

"40

:30

20

10

A

-02

01

-03

Deflection

Sec

in

radians

FIG. 20. A. Record illustrating harmonic. B. Record illustrating harmonic, c. Loaddeflexion curve used in apparatus of Fig. 13 to obtain records Fig. 20 A, B.

differentiation being with regard to
which satisfies the now assume that in 0,

where the yr are twice mined; o>

(b)

2

initial

conditions y(Q)

=

diff erentiable functions of

=

A,

iff

cop

=

0.

We

to be deter(2)

2

o>g+6cof+6 o;l+...

being determinable constants. Here we expand in ascending powers of 6 (see 4.110, 4.131).

the

y'(Q)

o>

2

(not a)

EQUATIONS HAVING PERIODIC SOLUTIONS

68

Substituting from

(2)

CH. iv

into (1) gives

(co8+&co|+&i+..0(y;;+^^

+b(y*+2by Q yl +...) Equating the

coefficients of 6

6:

r ,

r =

"g^+a^^O,

For period

27r

s

in

if/,

or

so the complete solution of (4) 2/

2

y"Q +(a/a> )y Q

=

(3)

side of (3) yields: 0.

(4)

=^

= a,

(5)

is

sini/f+JS

(6)

cosi/r.

4.110, the initial conditions entail

in

y

on each

bFcos
we must have cog

As

0, 1, 2,...

=

(0)

- A, yi (0) = y

=

a (0)

so the appropriate solution

"Wi+ayi =

and by

= = y&(0) = y{(0) =

...

= Acosifj.

(7)

.

Joos^-(cof yj+yj),

(8)

=

[(^+cof^)cos0-^

2

(l+cos20)]/a.

To avoid a non-periodic term in the particular integral of coefficient of cos^ must vanish, so

F+&1 A =

0,

giving

Thus the complete solution of yl

BI = A

0,

(5), (7)

2/1+2/1

The

=

is

y b:

...

initial 2

/ 3a,

=4

1

2/j(0)

=

F/A.

(9),

(10)

(9) is

sin0+^ 1 cos0 +

conditions

cof

(9)

the

A2

A2 cos20-

= 2/i(0) =

0,

.

necessitate

(11)

A =

0,

and, therefore, (12)

(13)

and by

(5), (7), (12)

EQUATIONS HAVING PERIODIC SOLUTIONS

4.194

69

To avoid a non -periodic term in the particular integral, we must have

by

(10).

Then the complete

solution of (14)

is

A 9a 2

The

initial

conditions y z (0)

(16)

= y' (0)

necessitate

0,

2

so the appropriate complete solution of (14)

A*

.

is

A 2

\--*-

-* i

j^>^-"^>

f

i

,

n

n v/OS OyJ

(17)

Hence

to the second order in b

(18)' v

2

\2

Also to order two in 6

In virtue of the constant term in 'centre of oscillation'

is

(18),

on the negative

what may be regarded

as the

side of the origin of the force-

2 displacement curve ay-^-by (Fig. 11 A). We showed in Chapter II that the solution of a non-linear differen-

equation depends upon the initial conditions. If (1) is solved such that the coefficient of cos ^ is A for all t, the coefficients of the various

tial

terms will differ from those in y(Q)

= A.

(18),

and the initial condition will not be

Also the last term in (19) will not appear. The reader should

verify these remarks as an exercise.

4.195. Amplitude-frequency relation. (18), at t A, y' subject to the conditions y

=

must be

satisfied also,

=

which means that

=

4.194, 0.

But

was derived (19),

for fixed values of a, 6,

4.194,

and/,

EQUATIONS HAVING PERIODIC SOLUTIONS

70 co

2

CH. iv

complies with this equation. Written in the form

a cubic equation for A, which

it is

The consequences of such a

may be compared with

relation are

examined

4.140.

(9),

4.16 et seq.

in

From the analysis therein it appears that a physical system operating in accordance^ with the equation

y+2 K y+ay+by 2 = /cos wt

(2)

should exhibit the jump phenomenon described in 4. 1 70. Consideration of the signs of the terms in (1), where b 0, shows that it is of the

>

same type K

>

4.140, with b

<

so the (|^4|,co) relationship for in be akin 16 should to that 0, B, 17 B. Figs.

4.196.

as

(9),

Harmonics of order

.

0,

In this case we shall solve the equation

y+2Ky+ay+by 2 =/cos2toZ

(1)

subharmonics of order 2 (period 27r/co), using the method of pere& 2 ,/ = bF, with a > 0, 6, K > turbation. Let wt = z, 2/c small,

for

and

becomes

(1)

oj

2

2

y"+eb y'+ay+by y = y

Assume that

=

2

bF cos 2z.

(2)

2

Q

+by 1 +b y 2 +...,

(3)

the yr being periodic twice differentiate functions of

where the

o> r

are to be determined. Substituting from

z,

and that

(3), (4)

into (2)

gives

^2) + ..-K + %I + 6

2

2/2+...)

+ (5)

Equating

coefficients of b

r

6:

,

r = 0, uW+ay -

1, 2,...,

so for a solution with period

yQ

A,

13,

shall

we must have co

b, /, K, to;

yields (6)

=

a,

and, therefore,

-4sinz4-7? cos z-

(7)

to effect simplification of the

keep them the same throughout.

stage) differ slightly b:

0,

(5),

the equation non-linear, the ultimate initial conditions (unknown at the present

algebra is

=

are functions of a,

we

2?7,

on each side of

from

Sftice

= J5, y'(Q) ~ A. = - 2^^ yl~yl+F cos 2z,

j/(0)

wgyl+ay 1

(8)

EQUATIONS HAVING PERIODIC SOLUTIONS

4.196

71

so

-(l/a){(A*+B*)/2+ABsm2z-[(A*-B*)/2+F]cos2z}. (9) To avoid non-periodic terms in y l9 we must have a) l = 0. Thus with g

- (A*+B

2

)/2,

-

h

F+(A*-B*)/2, we

get

sin 2z

"2 cos z

.

J3 sin

2)+ 2a>

a> 2 (^4

(A/3a)cos 20.

(10) (11)

2/o2/i

sin

2+ JB cos 2)

so

(12)

(13)

where P, Q are given below. To avoid non-periodic terms and Q must vanish, and with A, B, non-zero we have

= 2(o> K) + 2(^/a )-(A/3a Q/B = 2(^> + 2((7/a + (A/3a 2

P/A

2

Subtracting

w2

=

)--(^ /3a

)

2

+ (^^

)~M/co B)

(14),

=

= -

P

0,

(14)

0.

(15)

a^/2, gives

-5(A +B*)/$a+(Ka> Q /b*)(A/B-BJA). (15),

obtain

(16)

and multiplying throughout by

=

(A/B+B/A)

Solving (17) yields

A/B

=

6 2 co /2/c,

=

6//3/ccog

- 0.

(17)

{)8(j3-4)t}/2,

K Mo (A/J3-~BfA)

so

2

2

j/ 2 ,

2

o> /e

we

2

)-(J5 /3a

and multiplying throughout by a/2

(14), (15) 2co Q

2

)

)

Adding

2

2

in

=

(18)

2 {(6//3a) -4*c a}!. 2

(19)

a. = a>2+26 WoWl +& K+2a> a> )+..., By (4) = 0, by aid of (16), (19), we get and since w^ = a, w = a-56 (4 +^ )/6a{(6//3a) -4/c a}i. 2

2

2

(20)


2

2

2

2

2

2

(21)

Thus

Y=

(4*+.B

8

)*

=

2 2 2 2 {(6a/56 )[(a-o> ){(6//3a) -4/c a}*]}i

(22)

which gives the amplitudes. of the subharmonics. To order one in 6 (small) the solution is y = yQ Jrbyv so

y

= .4sin2+Bcosz+(6^5/3a)sin22 = -(6/2a)F2 +rcos(z-6' )-(6/6a)r 1

(6^/3a)cos 2 2

(bg/a)

(23)

cos(2z+6 2 )-(//3a)cos22, (24)

EQUATIONS HAVING PERIODIC SOLUTIONS

72

= AB/{(A -B )/2}, = where tan = A/B, tan when there are tion is valid only for Y real > 0, 2

2

2

2

l

i.e.

The

CH. iv

20^ This solusubharmonics.

term represents a unidirectional displacement, and as in 4.195 the 'centre of oscillation' is on the left of the origin. The second and third terms represent, respectively, the subharmonic and its first overtone, while the last term gives the forced oscillation which first

has the same period as the driving force.

4.197.

Subharmonics and

The motion

stability.

defined to be

is

stable (unstable), if after being subjected to a small disturbance

returns to (moves

away from)

its

former

In [131

state.

a],

by

it

aid of

stability criteria given in [85], it is shown that the upper (lower) internal and the two external signs in (22), 4. 196 correspond to a pair

of stable (unstable) subharmonics. Thus there are four subharmonics, two stable and two unstable. By virtue of the two external signs in 4.196, the

(22),

components of each pair

differ in

phase by

TT

radians

Under

certain conditions specified in [131 a], the forced of oscillation driving frequency becomes unstable (see remarks in

or

7T/o) sec.

4.198). Nevertheless, there is still the oscillation due to the overtone of the subharmonic. The analysis pertaining to stability is beyond our present purview, so we shall merely state what results may be

expected on the basis of reference [131 4.198.

Forms

of a and

to,

of oscillation. For

a].

7 to be real,

whatever the values

we must have /

>

(I)

6i
This gives the threshold of/, below which subharmonics cannot occur. Also for the existence of the stable pair of subharmonics, 2

{(6//3a)

and

-4/c%}i

>

we must have

(w*-a),

(2)

~4* 2a}*.

(3)

for the unstable pair,

(a-aj

2 )

>

2

{(6//3a)

Variation in f.

For

(i)

co

2

<

or

> a,

as

/

is

forced oscillation of amplitude (ii)

>

For a

6/ca f

>w

2 ,

when /

=

increased from zero, there

X C2 //(4co

6/ca*/6

2

a),

is

a stable

but no subharmonics.

the threshold

is

reached, and for

a pair of stable (unstable) subharmonics having amplitude given by (22), 4.196 with the upper (lower) internal sign.

/

When

/6 there is

2

a

[a-co -{(6//3o) -4ic^}]

=

0,

the unstable subharmonics vanish while the stable ones remain.

(4)

At

4.198

EQUATIONS HAVING PERIODIC SOLUTIONS

this point, however, the forced oscillation

73

becomes unstable and the

amplitude jumpsf to that for one of the stable subharmonics, as given by (22), 4.196 with the upper internal sign. Increase in /is accompanied by that of

Y

and

also in the amplitude of the

subharmonic

now

decreased, the stable subharmonic persists but with decreasing F, until the threshold/ 6*a*/& is reached. It then

overtone. If/

is

=

stable forced oscillation; CA = unstable subharmonics; CB A. OA subharmonics. At A, B, /8 = (3a/&){(a-o> 2 ) a -f 4* 2 a}i; at C, ^ = 6*a*/6. For / increasing from O, the forced oscillation becomes unstable at A, and there is a jump to one of the stable subharmonics at B. B. OA = forced oscillation AB = stable forced oscillation which is unstable between B and subharmomcs. c. ABDF ; BC = stable subharmonics DE unstable subharmonics.

FIG. 21.

A

;

D

;

vanishes and the amplitude of the motion jumpsf down to that for the 2 forced oscillation (now stable), namely, a). The cycle c//(4o>

X

is portrayed in Fig. 21 A, and there is a type of 'hysteresis' as in Fig. 16 A. In the first case two frequencies are involved, but only one in the second.

of changes

system were started suddenly by applying a force / (/!
of the initial displacement and velocity. f See remarks on 'jumps' in

Fig. 21 4.192.

A

indicates that

EQUATIONS HAVING PERIODIC SOLUTIONS

74

CH. iv

conditions yielding a relatively small (large) displacement would be initiate the forced oscillation (subharmonic).

needed to

For

(iii)

co

>

2

a,

7

4.196 cannot be real with the lower

in (22),

internal sign, so there are no unstable subharmonics. The stable pair cannot occur unless (1), (2) are satisfied. As /in (2) increases from zero, there is the forced oscillation

f=/

the stable subharmonics

now

with

X ^//(4o>

is

tone. Increase in

/ is accompanied by that

Variation in If (1)

(i)

There

a),

and when

X

commence with amplitude

unstable, but there

forced oscillation

illustrated in Fig. 21 B.

2

is

is

.

The

the subharmonic over-

These remarks are

in Y.

no hysteresis

effect [131 a].

CD.

not

is

satisfied, there are

no subharmonics, but there

is

a

stable forced oscillation. (ii)

If (1)

is

and w 2 > a+{(bf/3a) 2 4* 2a}*, there is a stable but no subharmonics, since Y is imaginary. But

satisfied

forced oscillation

when \aaj 2

<

\

2

{(6//3a)

4/c

2

a}^,

the forced oscillation

is

unstable

and there is one pair of stable subharmonics, provided a is not too nearf

>

2 If a-{(bf/3a) 2 -4:K 2 a}* to there is a stable forced also two of the oscillation, subharmonics, provided a is not too pairs

to

co

2

-{

nearf to

}*.

co

2

+{

}*.

The trend of these remarks

is

illustrated in Fig.

21 c. Subharmonic resonance does not occur.

In general the forced oscillation is unstable when accompanied by a single pair of subharmonics, but remains stable if another pair of subharmonics is present to 'counteract the instability. This remark 5

is

exemplified in Figs. 21 A, B, o [131

a].

In Fig. 11 A the slope of the force-displacement curve changes sign to the left of the minimum, and if

Amplitude

limitation.

Iftninl

>ab l

>

the displacement increases continually with increase in time. Usually the characteristic for a physical system has no minimum, but it may

have a point of inflexion. In the case considered restoring force occurs.

pneumatic, neither a

Example. Imagine a hollow

4.199. length

is

I,

closed

in

4.199,

where the

minimum nor an

circular cylinder, of

inflexion

working

by a rigid disk of mass J m which is driven axially by a

The approximate analysis does not hold if a is too near to these values, j This includes 'accession to inertia' due to the external air [183].

t

EQUATIONS HAVING PERIODIC SOLUTIONS

4.199

75

such that the displacement | is 'finite'. If p Q is the atmospheric pressure, and v the corresponding undisturbed volume,

force /cos

2a)t,

we have

for adiabatic operation

pv7

p Q v%

=

a constant.

If the displacement during expansion

p(Z+f )? so

i>

Thus the

=

jp

IY,

(2)

(i+^)- y ^i>o[i-y^+y(y+i)^/2P-...].

'spring control' per unit cross-sectional area,

within the cylinder,

Accordingly disk

then

is -)-

= PO

(1)

if g/l

is

is

^

air

given by

1,

the approximate equation of motion of the

= /cos 2co,

2

fw"+2*ffaf +6

=

(3)

due to the

=

(5)

yp Q Ajl, b y(y+l)p ^4/2Z ^ being the cross-sectional area of the cylinder. Here the slope of the p-v curve is negative, and what may be regarded as the 'centre of oscillation' moves down this with a

curve (see

minimum large

(18),

,

4.194 with b

value, so

would

2

<

0).

The

adiabatic curve has no

need not be limited as in

entail additional terms involving

A

4.198. 3

4 ,

,...

solution of this equation on the lines of previous sections exercise for the reader.

relatively

The

in (5). is left

as

an

Intermodulation frequencies. Consider a loudspeaker diaphragm of the type illustrated in Fig. 22 and suppose the relationay-\-by* ship between force and axial displacement takes the form F as shown graphically in Fig. 12 B. This relationship has been chosen

4.20.

In practice, especially to be complicated in an likely asymmetrical about the force

in the interest of analytical simplification.

when y

is

large, the relationship

average loudspeaker.

Usually

is

it is

and as in 4.130 there is then a unidirectional displacement of the diaphragm. During the reproduction of speech or music a myriad of frequencies is present in the current which flows in the driving coil. However, to simplify the analysis, we shall consider only two frequencies, namely, a powerfulf low audio frequency 0^/277, and a axis,

comparatively weakf high audio frequency

co 2 /27r, e.g.

100

c.p.s.

and

f These adjectives refer to amplitude of motion, not to the aural sensation, which varies considerably with change in frequency at normal sound levels.

EQUATIONS HAVING PERIODIC SOLUTIONS

76

2,500

The

c.p.s., say.

CH. iv

latter will execute 25 oscillations in the

time as the former executes

1.

same

We may visualize the higher component

on the characteristic of Fig. 12s moves relatively slowly from to N. On the approximately straight portion OM, the 2,500 c.p.s. oscillation of the diaphragm will be nearly sinusoidal, but as the operating point traverses the curved part in action while the operating point

Conical shell

Outer

p ^

^/Electromagnet

/

pole

-Path of main flux

'Energising

coil

|Clamp FIG. 22. Schematic diagram for moving-coil loudspeaker.

of the characteristic the

wave form

gets distorted.

It follows that

sound output is affected accordIn the of one frequency modulates radio-acoustics, terminology ingly. the other, and the effect known as intermodulation occurs. The result alien frequencies are created, so the

in practice, if

many

audio frequencies in a wide band are present,

may be aurally distressing. It is then imperative to reduce the input to the loudspeaker to bring the distortion below the limit of audibility.

An effect of this type may occur in a much more marked degree with a certain class of thermionic valve, since the characteristic curve turns

much more

over at each end of the linear part cated by the term by*. 4.21. free

Simple analysis

system whose

of intermodulation.

differential equation

=f

l

may

abruptly than

is

indi-

We deal with a loss-

be written in the form

eos^+,/2 cos

a>

(1)

EQUATIONS HAVING PERIODIC SOLUTIONS

4.21

with fa

= o^tf,

*/r

2

=

aj 2

t,

;>

u>\

cof,

^>/2

/!

.

As

in

4.140,

77

we assume

(i) that the conditions for subharmonic oscillations are not satisfied, (ii) that the damping is small enough to be neglected, although it

ultimately extinguishes the initial transient. In (1) a)Q/2rr is approximately equal to the natural frequency of the effective diaphragm mass and the axial constraint, so 6 is small. In

a well-designed speaker having a diaphragm 10 in. in diameter, OJ Q should be less than 2?r X 20. Thus we take o> 2 ;> cof ;> co, and seek a solution whose individual terms are periodic. The nature of the

system contemplated angular frequencies

we may assume

aj

v

o> 2 .

as a first approximation that

=A

y

which

is

in

we

(1),

such that there will be components having If/!,/2 and the term by* are not too large,

is

the form of solution

l

cos fa+A 2 cos fa,

=

if 6

(2)

3 Inserting (2) into the term by

0.

obtain the equation

y+u*y =fi cos fa+fa cos fa+ h(cos fa,

cos fa),

(3)

where

[

cos 3 fa

+A

cos

|

3i/r 2)

(4 )

.

Using the inequalities given above, we in that part of the particular integral of

find that a

may

(4),

IT

I

is

this

given by

"!

.

Owclj

P

1(^0

which corresponds to

(3),

The approximate representation of

be neglected.

function (P.I.)

number of terms

^i)

Ki

(

i

o

\oi 4 J

~-(a> 2 +2a> 1 )

wo

Q

/

i

^i) W*2

i

rl/ T"

6AlA 2 cosfa /

2

(cog

c

(5)t

Omitting the question of initial conditions which do not affect the point under discussion, (5) shows that there is a third harmonic 3u) v

also

two additional components having angular frequencies

f We omit consideration of the complementary function of (3) and the P.I. of the terms in flt f therein, as they are irrelevant to the point at issue.

78

EQUATIONS HAVING PERIODIC SOLUTIONS

(a) 2 ~\~2a) 1 )

and

These are classes of sum and difference

20^).

(co 2

If 0^/277

frequencies, respectively.

second, the coefficient

A

l9

A%.

first

CH. iv

frequency

is

=

o> 2 /277

100,

=

2,500 cycles per

2,700 and the second 2,300

The

c.p.s.

called a modulation product since it involves solution also contains components of 4,900, 5,100,

3AlA 2 /[

The

]

is

but their amplitudes are relatively small. Higher approximations to the solution would reveal the presence of other com7,500

c.p.s.,

ponents.

Direction of sound

Source

propagation

FIG. 23.

4.22. find

it

Additional example of mtermodulation. The reader interesting to confirm that an approximate solution of

may

2

i/+uoy+by =/iCosw 1 ^+/2 cosco 2 ^, is

(1)

given by

~

bA 1 A 2

r ~T(

[(<*>!

Al

Here

=/i/(cog

(

Z +

cof),

(<*>lTu>2)* <*>

\2 2)

/

2TT-

(

v

2 Z )

^Oj

4 2 =/ /(o> 2

cof),

|*/

min

|

<

cog/6

= a/6

in

1 1 A, and it is assumed that the conditions for the existence of a subharmonic are not satisfied. There is a constant term as in 4.130, which represents a unidirectional displacement of the diaphragm. Also there are double, sum, and difference frequencies. The above

Fig.

akin to the creation of alien tones in the ear, by virtue of its asymmetrical characteristic curve. This property of the ear was discase

is

years ago by Helmholtz [169].

cussed

many

4.23.

Amplitude and frequency modulation

reproduction.!

Referring to Fig. 23,

in loudspeaker

represent! a very small

f This is a linear problem, but has been included as a matter of interest. If d were large enough, the curvature of the adiabatic characteristic for air would have to be taken into account as in [100, 101], 4.199, Appendix I.

EQUATIONS HAVING PERIODIC SOLUTIONS

4.23

sound source which

79

with frequency CO^TT along the of the motion being x Simultaneously the source executes a motion along OP of frequency co/27r and amplitude line

oscillates bodily

OP, the amplitude

.

co ;> co 1? but <; x Apart from the size of the source, scheme corresponds to the case of a loudspeaker diaphragm. Since <^ g v the distance of the source from P at any instant, is ,

such that

.

this

=

TI

r

^sinco^

==

r(l

esinco^),

(1)

=

with j/r <; 1, provided r ;> g l9 as it would be in practice. The instantaneous velocity potential at P due to a simple source of strength

47?

$ and

frequency

ge

co/27r is

[183]

$e -i(kr -kfi sin ^ t-

- f (fcri -cot)

cot)

csi

r(l

(3)

where

k

z

ly

k

co/c, c

density, the pressure at

the velocity of sound.

Then

if

p

is

the air

P is (assuming 8 to be unity for convenience)

(4)

P

-i

Taking the

p

=

COS

real part of (5),

^-[a> l

cos

we

CO

obtain

w l tcos(krz sin aj^ e

sin

co x t)(za> l

cos o> x ^ ~|-co)sin(fcr

2 sin c^!

^

cu^)].

Now

lo^ceoso^l <^ |(l+sinco 1 ^)(2co 1 cosco 1 <4-a)|, so to a approximation we get

p

-

c

[(l+

But

a>

^>

p

rw

~^[lJr

Putting

c

^coj,

=

A(t)

sin

so (7)

a) l

t)(zw l cos

may

^+co)sin(co^+^ sin coj^

a> 1

fo-)].

(6)

close

(7)

be written

smaj 1 t+(zto 1 /a))cosa) 1 t]sm(tot+zsmw l t-- kr).

(8)

= /c the in (8) becomes = ^^ < = {l+^^l + l/ijr^cos^^-tan-^l/ijr)]}, (9)

^/r,

2COJ/CO

1,

A:

x

co x

[

]

EQUATIONS HAVING PERIODIC SOLUTIONS

80

CH. iv

so amplitude modulation (fluctuation with time) occurs at P, having a frequency co 1 /27r. Using the formula in [184, p. 52, ex. 16], we obtain 00

= 2

sin co^)

Jnft&s^fa+ncojtkr].

>

P (apart from that due to a)

Hence the sound pressure at

(10)

oo

lt

2ir) is

given

00

p

by

= -A(t)

JnWi^tioo+nuJt-kr]. =

(11)

oo

comprises an infinite line spectrum of frequencies extending indefinitely in the range above o>, it

Consequently

finitely

below

co/27r

since

it,

CD <

Amplitude (fluctuation at

When n

>

quencies

0^/277, 2o> 1 /27r,...,

0, aj-\-nci) 1

When n

where

m

but

w

1 /27r)

side frequencies

^)

CO/27T.

2-r,

0.

Frequency the fundamental

two

iisv-

>

<

(n

and there is an unlimited series of frewhich occur at equal intervals 0^/277 above

co,

<

0, there is a similar series terminating at ma) l /27r, the largest integer making (co mcoj) 0. For other n, an unlimited series of interlaced with those there is frequencies mentioned above. The reader should plot a diagram showing the

>

is

frequency spectrum, taking f!

=

0-25 cm., c

from these data 4.310. circuit

=

3-43

it is

X

a>/277

=

2,560

c.p.s., 0^/277

=

50

c.p.s.,

10 4 cm. sec.- 1 For the value of fcf x obtained

accurate enough to take

c^ ($z) n /nl.

Jn (z)

Forced oscillation in self -oscillatory thermionic valve [9, 119]. The arrangement is shown schematically in Fig. 9 A,

the differential equation being

(3),

4.10,

right-hand side. Thus corresponding to

at

FQ =

with

(4),

J?

cos(ca^+9) on the

4.10,

we have

= .F co8(arf+?)

(1)

a, 6, c as before. There are two possible main virtue of the negative resistance property of by the valve, (2) forced due to the applied potential difference

with

a(EQ IE8 ), and

oscillations, (1) free

EQUATIONS HAVING PERIODIC SOLUTIONS

4.311

4.31 1. Solution of

4.310.

(1),

To

81

an approximation adequate

find

to explain salient experimental results, during the steady state, assume thatf u AQ cos ^ A cos

=

with

=

i/r

o>

and forced

cot, eo

2,

,

we

^

Q+

{1)

being the angular frequencies of the free

o>

Substituting from

oscillations, respectively.

(1)

into

(1),

4.310, yields:

= bu = b[a U

C

d(u*) dt

(2)

= a[^4

au

terms involving dently,

,

3i/f

,

etc.,

+J.cos
being omitted.

coefficients of cosi/r,

Equating

we

2(/r

cos
get cu

cosifj:

sin(/r, cosi/r

A+aA =

2

,

sin^ to zero indepen-

-F cos
(3) (4)

= sin
From

6o> ^4

:

(6)

we

f^ga>

0,

(5)

2 $A Q A a) Q c =

c

0.

(6)

0.

(7)

obtain or

A =

In absence of an applied potential difference, tude of the free oscillation is given by

^4

0,

and the ampli-

'\*.

(8)' V

3c/

The

free oscillation vanishes

the forced oscillation

In

(7) since

A Q must

is

when

AQ =

0,

so

by

(7)

the amplitude of

then

be

real,

the forced oscillation exists alone

if

^ |J |/V2. = a = 1/LC,

\A\

From and by

(5), (3), (4) if

f It can be 5077

o>

the free oscillation

shown that no phase angle

is

is

suppressed A

(10)

=

0,

so

required for the free oscillation.

EQUATIONS HAVING PERIODIC SOLUTIONS

82

or

(a>*-ajl)*/aj*b*+(l-3A*C/b)

=

2

Writing (a> -a>g)/a>& the form

-4

-5

2

,

3.4 c/4&

=

-Z

-1

=

(FQ/Aba>)*.

3^gc/46

2/>

x*y+(l-y)*y =

-3

2

CH. iv

3

a>

2

-

(12)

J5,

(12) takes

E)

(13)

2

1

x

4

3

5

FIG. 24. Amplitude-frequency curves for 'driven' thermionic valve oscillator.

which

is

when

valid

there

is

no

free oscillation.

With the above

substitution (6) yields i/

Taking

|o>

co |

<^

3^ 2 c/46 co

,

then

4.312. are

(l-2y),

\x

^

2|ca

T/ O > !f

co

|/6,

>

so if

0.

the central

is

,

The amplitude-frequency curves when A Q

shown

(14)

OJ Q is

nearly proportional to the amount of 2 constant, y oc A the square of the amplitude

frequency or tune point, x detuning. Since 3c/46 is of the forced oscillation.

-

in Pig. 24, being obtained

by

plotting (13),

=

0.

These

4.311, for

EQUATIONS HAVING PERIODIC SOLUTIONS

4.312

83

They resemble the tuning curves of an LCR a from but circuit, physical viewpoint the broken line parts have no significance here. Imagine the curve marked E = 0-2 to be followed various values of E.

for

x increasing from zero.

A

value x l

is

reached where the tangent

22

*- free + forced p

for

oscillation

E= 0-5

*

FIG. 25. Amplitude-frequency curves for driven' thermionic valve oscillator, trating the phenomenon of 'entrainment', and depicting the 'silent zone*. is vertical.

illus-

The locus of the vertical tangents to the family of E curves

represents a transition stage known from experiment to be that where the free oscillation sets in. When \x\ Jo^l, there are both free and

>

forced oscillations in the circuit, soycc (A%-\-A 2 ). The corresponding x-y curves veer away from the z-axis as illustrated in Fig. 25 [119].

Since

|^4

is |

now

>

incorporated in (11),

0,

the term

4.311, so,

^AlAuc in

using

(7),

(4),

4.311,

4.311,

must be

we have (1)

EQUATIONS HAVING PERIODIC SOLUTIONS

84

which is

4.31 1, with 9 written for 3 in the second

(1 1),

corresponding to (13),

we

4.311,

=

2

=

member. Thus

get

x 2y+(I-3y) 2y

When y

CH. iv

=

E.

(2)

2

so the curves corresponding to 0-5, (1 (1 3y) y) 4.311, and (2) above, meet at this point, provided \x\ is outside

(13),

,

the locus of the vertical tangents. The two curves do not meet on the locus, since our approximate solution of (1), 4.310, is not accurate

enough for this to happen. Nevertheless the diagram of Fig. 25 agrees with experiment. Additional information on this topic is given in [28]. Suppose the

circuit of Fig. 9

A were

loosely coupled to a detector

valve followed by an audio-frequency amplifier with loudspeaker. co i.e. x 0-5, if co and, there0, on the curve E Starting at co

=

=

,

=

fore, x were increased, silence would ensue up to the point P. Thereo> )/27r in the after, there would be a beat tone of frequency (co circuit.

loudspeaker

audibility of the

This would be heard

human

PP

ear.

: is

if it

known

were in the range of

as the silent zone corre-

=

0-5, where the free oscillation sponding to E forced one [9, 119].

Locus of vertical tangents. The

4.313.

by

is

suppressed by the

ovals in Fig. 24 are given

4.311. Differentiating with respect to y,

(13),

2

2

2xy(dx/dy)+x +(l-y) -2y(l-y) so if dx/dy =

0,

we

=

get

0,

(1)

we must have

2

x +(I-y)(l-3y)

=

3x 2 +9(y-f) a

or

0;

-

1.

(2)

which represents the ellipse, centre (0, f ), semi-axes 1/V3, locus of the vertical tangents as shown in Figs. 24, 25. Using (13), This

is

an

4.311,

For

and

E=

above,

(2)

= 8/27, y

f,

we find that y is x

=

The parts of the curves y

may

=

triple-valued

when

E<

8/27.

The

latter

1/V3. in Fig. 25 corresponding to

(y Q +y) oc (A%-\-A

be computed by aid of

(2),

2 )

(mean square

4.312,

and

(14),

basis)

4.311.

=

1 ~~2/)> an ^ by the former ( y decreases with increase gives (2/o+2/) in |B |, so (y Q +y) cannot exceed unity.

r

4.40. (1),

Subharmonics and relaxation

oscillations.

A

study of

4.310, taken in the form 2

i/-e(l-y )y+y

=

Fernet,

(I)

EQUATIONS HAVING PERIODIC SOLUTIONS

4.40

has been

made

85

using a differential analyser [53], and curves are

reproduced in Figs. 26, 27. The parameters are given in the captions; y is proportional to the potential difference across the capacitance in Fig. 9 A, is

and the time scale is arbitrary. The phenomenon of resonance

almost absent in the case of a circuit executing relaxation

oscilla-

Time

Time

10

Time

12

14

Time

Time

FIG. 26

FIG. 27

A. Free relaxation oscillation with e 1-25: oscillation of drifting type. c. e 8,

FIG. 26.

F=

=

F=

B. e 0. 8, 25/32, 1-25: oscillation is 25/8, to J harmonic of driving force. FIG. 27. A. c 2, to 2; oscillation is J harmonic of driving force. B. 8, 8, 2 ; oscillation is harmonic of driving force, c. e 2; 8, 160, cu 80, o> oscillation has same period as driving force. aj

=

=

=

F=

F=

8,

F

=

=

F=

= =

which a potential difference .Fsino^ is applied. The main influence of the latter is to alter the time period of the circuital

tions, to

relaxation oscillation. Thus a subharmonicf of very low order

wave form

may

from sinusoidal, the term subharmonic is an integral multiple of that of the applied potential difference Fsincvt. In experiments described in [120, 123], by increasing the frequency of the driving potential difference, all subharmonics as far as the 200th were obtained, i.e. the periodic time of the relaxation oscillation was 200 times that of the driving potential difference. f Since the oscillator

merely

signifies

is

far

that the periodic time of the oscillation

EQUATIONS HAVING PERIODIC SOLUTIONS

86

occur.

CH. iv

An increase in the driving potential difference reduces the order

of the harmonic, but has tion oscillation (Fig. 27).

little effect

Some

on the amplitude of the relaxa5

of the 'driven oscillations are of the

and never attain the periodic synchronized Both the amplitude and the time interval between alternate

'drifting type' (Fig. 26 B) state.

zeros fluctuates (see [28, 29]). relaxation oscillation for which

A

Fig. 26 A.

The

result of applying

e

=

8,

F=

is

depicted in

a sinusoidal potential difference

having a frequency three times that of the relaxation oscillation is shown in Fig. 26 c. The main component of the latter is a synchronized J harmonic. The original paper should be consulted for curves obtained by making a large number and variety of parametric changes.

CHAPTER V

METHOD OF SLOWLY VARYING AMPLITUDE AND PHASE 5.10. Introduction. The approximate method to be described may be applied to non-linear equations when the solution takes the form

y

=

A(t)sm[ojt+
(I)

y

where the amplitude A and the phase angle 9f are functions of t whose rate of variation is small compared with the angular frequency In some cases A(t) may be variable but y(t) constant, and vice a). versa. Also, after a short time interval, i.e.

an

A

(t) may be sensibly constant, -> a or to zero constant, A(t) asymptotically when -> +00. As example of the latter we cite the familiar case of the free oscillation

of a linear electrical

LCR /

Here A(t)

=

Ce~ K* ->

circuit in

=

which the current

is

given by

(7e-*<sin(a>+9).

(2)

-{-co, while 9 is a constant phase angle. Consider a differential equation of the form

as

t

->

2

y+*g(y,y)+< y

=

o,

(3)

which the non-linear term eg(y,y) is relatively small, and g is a function J of both the displacement y (or its equivalent) and its first time derivative y. Neglecting this term, the equation reduces to the in

linear one 2y

of which the complete solution

y

=A

l

=

Q?

(4)

is

coQa>t-{-B 1 sma)t

(5)

i

A!, B! being arbitrary constants. In a sense this may be regarded as a first approximation, although it possesses no non-linear characteristics. (5) is expressible in the form (1), with A (A\ ^i)*, and

=

=

+

1

tsm- (A l /B 1 ) these being constants. 9 Writing (a)t+q>) == x> the solution takes the form )

= Asmx; y = AOJ cos xy

so

(6) (7)

t Strictly 'phase' pertains to sine waves, but it is convenient to use it here. 3 J If terms of the form 6y occur, they are to be included in g, i.e. g comprises damping and non-linear control terms.

METHOD OF SLOWLY VARYING

88

We now tiating (6)

A

suppose that

we

and 9 vary slowly with

CH. t,

so

by

v

differen-

get [77, 175]

y

=

Substituting for y from

Asmx+A(a)+^)GOs x (7)

0.

(9)

we have

(7),

AOJ cos x~ ^4co(co+9)sin ^.

y Substituting from

(8)

into (8) leads to

A sin x+^9 cos X = Differentiating

^

(6), (7), (10)

(10)

into (3) yields

A GOSX~~ ^4
(H)

CO

Multiplying

(9)

by

by cosx and adding, gives

sin^, (11)

A = -- gr(^4sinx, ^4cocosx)cosx-

(12)

CO

Multiplying

(9)

by cosx, 9

(11)

=

sinx and adding,

by

0(4sin x ,4wcos x )sin x

we

get

.

(13)

By hypothesis A and 9 vary but little in a period 27r/co, which latter

=

co^. We shall assume, therefore, corresponds to a period 2?r in that to the degree of approximation contemplated, the mean values i/r

of A and 9 over a period 2n in (12), the mean value of

are adequately accurate.

Then using

27T

A=

^

g(Asmifj, u4cocos0)cos0d/r,

2-7TCO

while by (13), the

9

mean value

=-

-

(14)

J

of

2TT

-j-

27TO)A J

g(Asm*jj, J.cocosi/r)sin^d(/r.

(15)

o

Since x

=

^^+?

it

follows that 27T

^

=

co+

ATTUjA.

g(Asm$,

Aajcoafysinifj d$.

(16)

J

When the restoring or spring control force (or its equivalent) contains terms of the form 2/ 2 ,2/ 3 ,..., A is unaffected, because the integrals corresponding to (14) vanish.

These terms affect 9, however,

AMPLITUDE AND PHASE

5.10

89

since the corresponding integrals do not vanish. For a similar reason but not 9. Hence in the first approxithe damping terms y, y 2 affect non-linear control terms affect the frequency but not the mation,

A

amplitude, whereas damping terms affect the amplitude but not the frequency. In higher approximations either type of term influences

both frequency and amplitude. 5.11. Solve a

yThis

is

(l-y )jH-y

=

(0<

<

c

(1)

1).

the equation for a thermionic valve oscillator (see v(y>y)

Substituting

=

(

l

y

(2) into (14),

=

2

)y

5.10,

(iA with

2

4.10).

Here

siu 2 x )Aa)cosx-

(

2)

for ^, gives

27T

A= I

=

(!-^ 2 sinV)cosV*/r

(3)

l^t(l-i^).

(4)

A=

A =

When

the steady state prevails, 0, so 2, as in 4.10, A a function with c 1. To determine as of t 4.110, <^ during the we have from (4) growth of the oscillation,

<

2AA

=

2 4 Je(44 -.4 )

=

=C :-(l'-2)

If

A

A = 4Be '/(l+Be A Q when = 0, B = ./4/(4 ^4g), A=

and, therefore,

2l

e*

(9)

).

from

so

this

*

*

2

^

i

t

obtain

(5)

].



J

2

giving

8

Je[4_(4*-2)

and

we

(9),

(10)

rf

rf

/[l+i-45(e -l)]*.

It should be observed that if A =

=

at

(11)

= 0, A

0,

which means

that there is no oscillation. Hence it is necessary to start the oscillation

by external means,

so that

AQ >

0.

From

increases monotonically with increase in value 2 as t -> +00. If we suppose that

decay and -> 2 as

t

-> +00.

t,

(11) it follows that

and tends

AQ >

2,

A

to the ultimate

the amplitude will

METHOD OF SLOWLY VARYING

90

By

(2)

above, with

x and

for

v

5.10,

(15),

,

OH.

27T

-

9 == Since

where (11),

o>

cp

t

(l-^ 2 sin 20)cos
is

=

x

1,

is

y

=

0.

(12)

<+.

an arbitrary value of x when

(13) above, the solution of

visaged

as

=

2n Jf

(

=

t

0.

Hence by

(1),

13 )

5.10,

(1) to the degree of approximation en-

= A(t)sin(t+
(14)

)

-> +00.

The graphical representation of (15) is akin to the curve in Fig. 59. The foregoing is inapplicable when > 0-25. It is then necessary to resort to a graphical method as exemplified in 8.12.

y-2 Ky+cf+ay =

5.12. Solve

0,

(1)

K being positive non-zero constants [131]. is the equation for an electrically (not electronically) driven 2/cf/ may be regarded as introducing a tuning-fork. The term a, c,

This

negative resistance, equivalent to a driving force. After the initial 5.10, (starting) transient has ceased, the motion is periodic. As in 5.11

is

we assume

that

Then by

relatively small.

f J

27TO)

= When

given by K

2

$ca>

is

A

periodic

2

f

2770)

=

=

o>

2 ,

27T

A aj cos V dJj --

KA-ICU*A*

the motion

with a

'

27T

A=

5.10,

(14),

A(K-%ca>

A =

^ 3co 3 cosV dJi

(3)

).

(4)

J

A

2

2

0, so the ultimate amplitude

is

0,

*U A=-2 {y. A

i.e.

/2

,

Kv

(5)

ti>\3c/

To determine Thus

A(t)

we have

to solve (4) as a differential equation. (6)

where p

=

2

4/c/3co>

,

so

AMPLITUDE AND PHASE

5.12

and, therefore,

2

2

2

2

)

A = A Q when

t

2Kt

2

.

l

)]

Hence If

91

= 2 Kt-\-B log[A /(2pA (8) = A /(2p A Be (9) = 0, B == ^4J/(2jp .4g). Expressing (9) in the form A = 2pBe /(l + Be (10) 2

"'.

2Kt

),

and substituting

into (10) for

A =A 2

so

A(t)

If

=

AQ

at

t

=A 0,

2

e

Qe

2Kt

p, yields

J5,

+ (3c/8K)a) A A

2

Ki

there

2

2

/[l

/[l-\-(3c/8K)a> is

2Kt

(e

2

-l)],

(11)

-l)]*.

(12)

2Kt

(e

absence of motion, so the fork must be

started initially. By (5) the ultimate amplitude is A 8 (2/
<

is

>

,

similar to that portrayed in Fig. 59.

By

5.10,

(15),

277

2*

27T

T

27TO)A J

=

27TO)A J (13)

0,

so

cp

which

f

C

.

is

=9

(14) ,

a constant,

an arbitrary value of ^ at x

Hence by

(1),

0.

^

(15)

Thus

=

5.10, (12), (16) above,

an approximate solution of (1) V

is

;

l)]^

When

the supply of electrical energy to the form reduces (17)

is

discontinued, K

=

0,

=

and

so the tuning-fork comes to rest. 5.13. Solve

the terms

This

is

y+2xy+ay+by*

(1)

3

being relatively small. the equation for the system mentioned in 2/cy, 6j/

3.10,

but having

METHOD OF SLOWLY VARYING

92

viscous a)

damping represented by

Then from

2*7/.

CH.

5.10,

(14),

v

with

a*, 27T

A=

27T

Aa)cos

^-

-

2 ifj

^4 3 sin 3 0cosj/f

difj

o

=

-KA. r dv4

Thus

^4

logJ.

=

irf-fj^,

(4) (5)

t

Also by


or

IC^+B!,

4 = Ate-*, = 0. is the value of A at

so

(2)

(3)

=

~j-

where

difj

o

5.10,

(15),

=

-

27T

-

I

4o>cosi/fsint/r

difj-\

---

27T

A^smtydifi

(6)

v ; (7)

Thus, on the assumption that

9 and, therefore,

Hence by

by

(16),

is

constant,

-= -3*^

,

ocu

(8)

l-^Po.

5.10,

5.10, the

(1),

A

approximate solution of

-2

so the motion once started dies e -2Ki

i

n fae formula

with increase in reader If b

=

may

find

0, (11)

t.

it

away

(1)

above with

^/8a^,

exponentially.

(11)

By

virtue of

for the angular frequency, the latter decreases

This corresponds to reduction in amplitude. The 3.17. interesting to compare (11) with (6), (7),

reduces to the well-known solution for an oscillatory

dissipative linear system.

5.14. Solve

ff+[2KB+(oi\6\6+p6* where is

a,

/?,

y,

3

y0 )]+o>

2

=yO,

(1)

K are real positive constants such that the part in [ ] This differential equation occurs in the problem

relatively small.

of the rolling of a ship fitted with a stabilizing equipment, and it applies

AMPLITUDE AND PHASE

5.14

when the phase

of the control

is

93

reversed [108], After reversal, the

amplitude of the motion increases, reaching an ultimate value, the oscillation then being periodic. This occurs only if K 0. The term

>

2KB

may be written &

o^ 0, the first part being

due to the

with reversed phase, and the second to rolling resistance,

stabilizer i.e.

in the

language of electronics to negative and positive damping, respectively. If fit a x then K 0, so the damping is negative, and the stabilizer

>

>

,

supplies the energy necessary to maintain the periodic oscillation.

By

5.10,

(14),

27T

O

/*

A =-

Aco cos 2

t/r

difj

--

27T (*

^4 2 co 2 |cos
27T

--

_ f

27TCO

difj

27T

4 8a> 8 cosV*A +

J

f A^sm^GOSifjdtfj

(2)

ZTTO) J

(3)

When

the motion

is

A

periodic

0,

so

by

(3)

^O. Solving

(4),

and choosing the

A If

?Lcp <^

32a

i ; (5)

i

s

positive root, yields

= approximately 16a

.

8: --M-64a

f.

2777 2

J

,

.] 1

2

L

By

(4)

,

,

== 37TK/4aaj. 7

,

ft

(b)

J

5.10,

(15),

27T

f

^ 3 sinVr <Wr =

J

-^-, 8

since the other three integrals corresponding to Thus during the steady state

since

A

is

constant, having the value at

(6),

'

(7)

[ ]

in (1) vanish.

and 9

is

an arbitrary

phase angle.

Hence

x

= ^+9 = (^

c

|

)*+ Po>

(

9)

METHOD OF SLOWLY VARYING

94

and by aid of 6

(6)

=

the periodic solution

The angular frequency of the decreases with increase in

A

v

is 2

2

oscillation

co

(37r/<:/4 a a))sm[a>(l-2777

2

OH.

/c

r /128(x

a;

4

)^+9

=

(10)

].

2

[co

-(3y-4 /8o>)]

.

5.14. To simplify the analysis we shall take 5.15. Solution of (3), the case when K 0, and find the law of amplitude decay for the

=

damped system.

positively

We

have to solve

(2)

where

S

=

Then

16a/97rj3o>.

dA (3)

\ Now

the integral

may

dA

1

T

1 2

4S J

4S

(A+28)

be written

tdA 2

J

1

JV2SJ

At the beginning of the decay,

Substituting

(5)

tdA

I

A*

4S

if 28/^4

<^

2

lQ g

M+2S\

A

\

j

1

2SA'

'

we have approximately

1,

into (3) yields

p = fjfc^+5. If

v

A = A Q when = t

0,

B=

(6)

1/A% and, therefore, (7)

t)*,

so the oscillation decays according to this law if K

the steady state.

(7)

shows that the greater

AQ

is ,

made zerof during

the more rapid the

decay in amplitude. By (9), 5.14, the rate of oscillation increases with decrease in A. Apart from the damping terms, the behaviour is akin to that of a simple pendulum, the restoring force in both cases being due to gravity. The 'stiffness', or derivative of the force with respect to angular displacement, decreases with increase in amplitude. 5.16.

Solve the simultaneous differential equations [107] $-b<j>+c*+a

and

8+b^+a^ =

means that the driving force due neutralized by the resistance to rolling. f This

/c

1

=

' 0,

^+/c 2 ^

(1) 3 ,

(2)

to reversal of phase of the stabilizer

is

AMPLITUDE AND PHASE

5.16

95

constants being positive and non-zero. These equations occur in the theory of ship stabilization, and the problem is to investigate the motion. First we remark that if b 2/c, and y, ( 1 ) is identical with all



equation for an electrically maintained tuning-fork. The periodic solution may be obtained from (17), 5.12, by letting 2 t -> +00 o> 6 2*, 9 taking a |TT (arbitrary). Then 5.12, the

(1),

=

,

=



where A =

= (46/3o> c)*, $ 2

Since K 2 /K 1 <^ for

<j>

from

A cos^,

cot.

we omit the last term in

1

usually, (3) into (2). Thus the latter

e+b l 6+at 6 where

B = AK

function, since

is

=

and substitute

= B sin 0,

(4)

In the solution of

it is

e

cof,

(2),

becomes

(4) we omit the complementary an exponentially damped circular function which -> +00. Then the particular integral of (4) is

I OJ.

tends to zero as

with a x

(3)

t

=

tan- 1 [(cof

periodic with period |0

27r/o>

|

2

Thus the ultimate motion and amplitude co )/b l aj].

Ato Kl /[(a>t-to*)+co*bt]*.

In the self-oscillatory mechanical system to which

(6) (1), (2) refer,

the

term

in (1) represents the driving force entailed by energy b<j> from an external source (the stabilizer); and in electronic supplied it

language resistance

damped

<

0, the equivalent to a negative resistance. If b and the solution of will then a (1) represent positive

is

is

oscillation of decreasing amplitude.

Using the expression

for this in (2) with 2 0, we obtain the equation of a dissipative system driven by an oscillatory force of gradually decreasing amplitude. Hence an oscillation once started will die away with increase /<:

in time. This

is

the desired practical result in the anti -rolling of ships.

Surges in hydro-electric installation [30]. The general arrangement of a surge tank or chamber for a hydro-electric power plant is shown schematically in Fig. 28. The water is supplied from a reservoir at a level h above the turbo-generators. The function of the surge chamber is to relieve surges which travel up the pipeline when the turbine load decreases and the water consumption is reduced, and to supply water when the load and, therefore, the water 5.170.

AMPLITUDE AND PHASE

5.170

$1

consumption increases. If we imagine a quiescent condition corresponding to a constant load on the turbines, the water level in the surge chamber will be substantially constant. But when the water to the turbines

momentum

stopped suddenly by closure of the sluice valves, the of that in the tunnel causes the level in the surge chamber is

to rise, resulting ultimately in a slow damped oscillation of the water in surge chamber, tunnel, and reservoir. When the valves are opened

a quiescent

after

interval, the level in the surge chamber falls and water in the system occurs as before.

a

oscillation of the

damped

Remarks on friction. When water flows in a tunnel or pipe, the whole cross-section being utilized, the friction force has been shown experimentally to be proportional to some power of the velocity u. In an iron or steel pipe this power may be rather less than 2, and in a concrete pipe nearly 2. The loss of water head due to friction, i.e. the height

of a column of water to cause the same pressure, varies as u v There 2 is also a loss of head at the tunnel entrance which varies as u and .

,

this

be represented by ku 2/2g. If

may

length

I

is

Iu 2 /rm

=

v

2,

the loss in a tunnel of

C 2 rm being the hydraulic mean radius of the ,

and C the Chezy constant. Thus the total loss of head due to 2 = (k/2g)-\-(l/rm C 2 ). is/16 where the friction coefficient/

tunnel, friction

,

We

assume that 5.171. Derivation of differential equation [30]. the cross-section of the reservoir is much greater than that of the surge chamber; also that sections.

The

it,

the pipe, and tunnel have uniform cross-

force relationship

is:

=

(force due to water pressure at tunnel en(force due trance) + (component of weight of water in tunnel)

mass X acceleration

to pressure at lower end of tunnel)

Thus

if

w

is

the weight of a cubic foot of water (62-4

acceleration due to gravity (32-2

^^ = A g

(factional force).

dt

t

R,+wlA

t

ft.

see.- 2 ),

lb.),

and g the

we have

*m*-A Pi-wlA t

t

u*lr m

C\

= I (Pz-Pi)+Y -lu /rm C\

(1)

2

or

where

F3 -

Zsina.

stituting these into

(2)

3

(l/g)

Now P2 /w = (2)

(Y2 -ku

2

/2g),

PJw

=

(Ti+y).

gives ,

= 5077

Sub-

-y-M

(3) (*)

METHOD OF SLOWLY VARYING

ag

CH.

v

=

The rate of flow of water into the surge 0. the difference between the respective rates for tunnel and pipe-line. Hence if Q is the rate of flow in the latter, since (^2 +^a"~^i)

chamber

is

A Eliminating y between

namely

This

a

^ = A u-Q.

(5)

t

leads to the equation for the velocity,

(4), (5)

-.

'

may

be written in the form

=

U-\-2KUU-\-au

with

a

K=fg/l,

5.172. Solution of

=A

t

g/A 8 l

d,

(7)

d =

y

Qg/A 8 l.

5.171. Write (u-dja) and the equation transforms to (7),

=

2 x+2K(x+c)x+a> x

where c

= d/a, and

2

(8)

= x, u = x, = x, 11

0,

(1)

In practice if the damping term 2K(x-{-c)x is relatively small,f the method of 5.10 may be applied. The coefficient of the frictional damping is always positive, and we must write the

co

a.

damping term

in the

form

2/c( \x

\

+c)x. Then by

(14),

5.10,

27T

A=

_?^ 27T60

f

2

(,4|sin
2

^ IT

dJt

(2)

27T

7T

=


J

f cos

2

^

d(cos 0)

IT

J

f c J

(3)

(5)

p*],

with 2p

37J-C/4.

Thus J dAI[(A+p)*-p*] .Llog[A

so

and

=

-- +

/(A+2p)]=:-^ +

B,,

B1

(6)

(7)

,

f

(8) A/(A+2p) = Be-&, In certain cases this may not be so. A numerical method of solution is then neces-

f sary (see [30]).

AMPLITUDE AND PHASE

5.172

with

j8

=

If A = A

Kdja.

+2p) and on

- A Q e-P*/[I + (4a/37Td)At(l-e-P<)].

(9)

A(t) (1)

99

B = A Q/(A

Q

we

inserting this in (8),

Also from

when

=

t

0,

y

find that

above and

5.10,

(15),

277

=-


(^|sini/r|-f-c)cosi/fsin
7T

so 9

9

,

d$

=

0,

(10)

J

an arbitrary phase angle.

5.173.

Water

Using

this in (9),

surge chamber. After the water supply to the turbines is stopped by closing the sluice valves, Q = 0, so d = 0. level in

5.172, gives

A(t)

=

= Atl(l+yt),

(1)

The water from the tunnel now (4/c/37r)^4 of the surge chamber, and its velocity is where y

.

Next we have

to determine the

conditions u(0)

= d/a = A

two constants 9

u(0)

l9

=

0.

From

,

AQ

,

enters the base

using the initial

(2)

(3)

and when

t

=

0,

we must have

tocoscpo

With

this value of

= ysin

in (2),

or

,

we

= tan^to/y.

1 (o>2+y2)Vo>.

(4),

5.171, in the

form

y

and substituting from

,

--

(6)

(2), (6), .

Using

(4)

get

^ =^

SO

Hence by

cp

(7),

= leads to

(7)

METHOD OF SLOWLY VARYING

100

CH.

v

This is an approximate expression for the water level in the surge chamber at any time after closure of the sluice valves. When t = 0, y = /4f which is the loss of head between the reservoir and surge ,

chamber, due to calculate

Using the data in

friction.

and plot the y-t curve from

5.175, the reader should

to 210 seconds.

t

The

first

2/max is of order 6 feet.

5.174. Aperiodic case. From physical considerations it is evident that if friction in the tunnel were large enough compared with the gravitational restoring force, i.e. K large enough in comparison with a, there would be no oscillation. The method of solution used in 5.172 is

inapplicable here, since

slowly varying amplitude.

it is

The

based upon oscillatory motion, with differential equation for

=

U+2i
0.

t

>

is

(l)t

The form of aperiodic solution which satisfies the initial conditions u = d/a, UQ = 0, and vanishes when t -> is not obvious. We can, however, determine an upper limit for aperiodicity in the following way: u decreases with increase in time without oscillation, and 2KU < 2Kd/a for t ^ 0. Now if 2/m = 2f
+>

Kd/a

and

this furnishes the

Substituting from

^ a*,

upper

(8),

or

K

^ a*d-

1

(2)

,

limit.

5.171, into (2)

above gives the

criterion that

A 3 ^ AJllfgQ*. Hence

(3)

and rate of flow

for a specified tunnel, friction coefficient,

in

the pipe, the motion will be aperiodic provided A 8 the cross-sectional area of the surge chamber, equals or exceeds a certain magnitude ,

given by the right-hand side of (3). The main reason for this requirement is that the restoring force corresponding to a decreases inversely as

A3

.

Numerical examples illustrating 5.172, 5.174 [30] 1. In a hydro-electric surge-chamber system the tunnel is 3 ft. 707 sq. ft., its length I = 500 ft., /the friction codiameter, A t

5.175.

=

= efficient/

0-027,

and

As =

56-5 sq.

ft.

What is the

quasi-period of

the oscillation ? t Since

u does not change

sign in aperiodic motion, there

is

no need to write

\u\.

AMPLITUDE AND PHASE

5.175

To a

first

101

approximation co .-= a*,

r

so

27T/a>

=

2?r/a*

(1) (2)

70 seconds.

In this ultra high-frequency seems ultra strange!

2. In 1, what

(3)

era, a periodic time of this magnitude

chamber would be required to prevent

size of surge

g(y)

velocity

Fia. 29. Illustrating

Coulomb damping.

oscillation if the rate of flow of water in the pipe

per

were

Q

33*7 cu.

ft.

sec. ?

By

(3),

5.174,

A8 >

7-07 3 x500/0-027 2

=

6,600 square feet.

x32-2x33-7 2

(1) (2)

Hence if the surge chamber were cylindrical, its diameter would have to be at least 92 feet to avoid oscillatory motion. 5.18.

Coulomb damping.

In this case the damping

is

presumed

to be independent of the velocity (magnitude and direction) when it is greater than zero, but vanishes with the velocity. The corresponding force always resists the motion independent of its direction.

These points are illustrated in Fig.

EXAMPLE. Solve where y =

g(y)

=

/a,

2

y+g(y)+<*> y

a constant

as in Fig. 29 [187].

29.

By

when y (14),

=

0,

> 0, y <

(1) 0,

but g(y)

when

5.10, dividing the interval (0, 2ir)

SLOWLY VARYING AMPLITUDE AND PHASE

102

CH.

v

in accordance with Fig. 29,

A=

cos0eM + -^ ^ *^

f

27TCO

27T

ITT

$TT

277co

J

GosJjdJj r y

J +i7T

=

2/i/TTO).

Hence

\A\

=A

2rra>

cosibdJj ^ r

(2); v

J 57T

(3)

(4)

S^t/noi,

being the initial displacement. Now \A cannot be negative, so the motion ceases when \A vanishes, 7ro>^4 /2/x. Thus the solution is given by i.e. t ^4

[

\

=

y

=

==

=

^(<)sin(a>^+9

)

(0

Since