A.p french vibrations and waves solutions pdf download

Solutions Manual for

The Physics of Vibrations and Waves –

6

th

Edition

Compiled by

Dr Youfang Hu

Optoelectronics Research Centre (ORC), University of Southampton, UK

In association with the author

H. J. Pain

Formerly of Department of Physics, Imperial College of Science and Technology, London, UK

SOLUTIONS TO CHAPTER 1

1.

In Figure 1(a), the restoring force is given by:

F=−mgsinθ

By substitution of relation sinθ= lx into the above equation, we have:

F=−mg lx

so the stiffness is given by:

s=−F x=mg l

so we have the frequency given by:

=sm= lg

2 ω

Since θ is a very small angle, i. θ=sinθ= lx , or x=lθ, we have the restoring force

given by:

F=−mgθ

Now, the equation of motion using angular displacement θ can by derived from Newton’s

second law:

F= xm&&

i. −mgθ=mlθ&&

i. θ+ θ= 0 l

g &&

which shows the frequency is given by:

= lg

2 ω

In Figure 1(b), restoring couple is given by −Cθ, which has relation to moment of inertia I

given by:

−Cθ=Iθ&&

i. θ+ θ= 0 I

C&&

which shows the frequency is given by:

=CI

2 ω

i. xAl V

Ax −γp =ρ &&

2

i. + x= 0 l V

pA x ρ

γ &&

which show the frequency is given by:

l V

pA

ρ

γ ω =

2

In Figure 1 (g), the volume of liquid displaced is Ax, so the restoring force is −ρgAx. Then,

Newton’s second law gives:

F=−ρgAx= xm&&

i. + x= 0 m

g A x

ρ &&

which shows the frequency is given by:

ω =gρAm

2

1.

Write solution x=acos(ωt+φ) in form: x=acosφcosωt−asinφsinωt and

compare with equation (1) we find: A=acosφ and B=−asinφ. We can also

find, with the same analysis, that the values of A and B for solution

x=asin(ωt−φ) are given by: A=−asinφ and B=acosφ, and for solution

x=acos(ωt−φ) are given by: A=acosφ and B=asinφ.

Try solution x=acos(ωt+φ) in expression x x

2 &&+ω , we have:

cos( ) cos( ) 0

2 2 2 x&&+ω x=−aω ωt+φ +ω a ωt+φ =

Try solution x=asin(ωt−φ) in expression x x

2 &&+ω , we have:

sin( ) sin( ) 0

2 2 2 x&&+ω x=−aω ωt−φ +ω a ωt−φ =

Try solution x=acos(ωt−φ) in expression x x

2 &&+ω , we have:

cos( ) cos( ) 0

2 2 2 x&&+ω x=−aω ωt−φ +ω a ωt−φ =

1.

(a) If the solution x=asin(ωt+φ) satisfies x=aat t= 0 , then, x=asinφ=a

i. φ=π 2. When the pendulum swings to the position x=+a 2 for the first

time after release, the value of ωt is the minimum solution of equation

asin(ωt+π )2 =+a 2 , i. ωt=π 4. Similarly, we can find: for x=a 2 ,

ωt=π 3 and for x= 0 , ωt=π 2.

If the solution x=acos(ωt+φ) satisfies x=a at t= 0 , then, x=acosφ=a

i. φ= 0. When the pendulum swings to the position x=+a 2 for the first

time after release, the value of ωt is the minimum solution of equation

acosωt += a 2 , i. ωt=π 4. Similarly, we can find: for x=a 2 , ωt=π 3

and for x= 0 , ωt=π 2.

If the solution x=asin(ωt−φ) satisfies x=a at t= 0 , then,

x=asin(−φ)=a i. φ −= π 2. When the pendulum swings to the position

x=+a 2 for the first time after release, the value of ωt is the minimum

solution of equation asin(ωt+π )2 += a 2 , i. ωt=π 4. Similarly, we can

find: for x=a 2 , ωt=π 3 and for x= 0 , ωt=π 2.

If the solution x=acos(ωt−φ) satisfies x=a at t= 0 , then,

x=acos(−φ)=a i. φ= 0. When the pendulum swings to the position

x=+a 2 for the first time after release, the value of ωt is the minimum

solution of equation acosωt += a 2 , i. ωt=π 4. Similarly, we can find: for

x=a 2 , ωt=π 3 and for x= 0 , ωt=π 2.

(b) If the solution x=asin(ωt+φ) satisfies x=−a at t= 0 , then,

x=asinφ=−a i. φ −= π 2. When the pendulum swings to the position

2 10 [ ] [42 ]5 10223108

16

8

0

m nm

c ≈ × = ×

× × ×= =

π −

ω

π λ

Therefore such a radiation is found in X-ray region of electromagnetic spectrum.

1.

(a) If the mass m is displaced a distance of x from its equilibrium position, either

the upper or the lower string has an extension of x 2. So, the restoring force of

the mass is given by: F=−sx 2 and the stiffness of the system is given by:

s′ −= F x=s 2. Hence the frequency is given by a s m s 2 m

2 ω = ′ =.

(b) The frequency of the system is given by: b=sm

2 ω

(c) If the mass m is displaced a distance of x from its equilibrium position, the

restoring force of the mass is given by: F=−sx−sx=− 2 sx and the stiffness of

the system is given by: s′=−F x= 2 s. Hence the frequency is given by

c s m 2 sm

2 ω = ′ =.

Therefore, we have the relation: : : 2 : 2: 4:2:

2 2 2 ωa ωb ωc =s m sm sm=

1.

At time t= 0 , x=x 0 gives:

asinφ=x 0 (1.6)

x&=v 0 gives:

aωcosφ=v 0 (1.6)

From (1.6) and (1.6), we have

tanφ=ωx 0 v 0 and

2 212 0

2 a=(x 0 +v ω )

1.

The equation of this simple harmonic motion can be written as: x=asin(ωt+φ).

The time spent in moving from x to x+dx is given by: dt=dx vt, where vt is

the velocity of the particle at time t and is given by: vt=x&=aωcos(ωt+φ).

Noting that the particle will appear twice between x and x+dx within one period

of oscillation. We have the probability η of finding it between x to x+dx given

by: T

2 dt η= where the period is given by: ω

2 π T= , so we have:

2 2 2 2 cos( ) cos( ) 1 sin ( )

2 2

a x

dx

a t

dx

a t

dx

a t

dx

T

dt

−=− +=+=+= =

πω ω φ π ω φ π ω φ π

ω η

1.

Since the displacements of the equally spaced oscillators in y direction is a sine

curve, the phase difference δφ between two oscillators a distance x apart given is

proportional to the phase difference 2 π between two oscillators a distance λ apart

by: δφ 2 π=xλ, i. δφ= 2 πxλ.

1.

The mass loses contact with the platform when the system is moving downwards and

the acceleration of the platform equals the acceleration of gravity. The acceleration of

a simple harmonic vibration can be written as: sin( )

2 a=Aω ωt+φ , where A is the

amplitude, ω is the angular frequency and φ is the initial phase. So we have:

A sin( t+ )=g

2 ω ω φ

i. sin( )

2 ω ω +φ

=

t

g A

Therefore, the minimum amplitude, which makes the mass lose contact with the

platform, is given by:

[01 ]4 58.4

min 2 2 2 2 2 m f

g g A ≈ × ×

= = =

ω π π

1.

The mass of the element dy is given by: m′=mdyl. The velocity of an element

dyof its length is proportional to its distance y from the fixed end of the spring, and

is given by: v′=yvl. where v is the velocity of the element at the other end of the

spring, i. the velocity of the suspended mass M. Hence we have the kinetic energy

i. θ+ θ= 0 I

&& C

In Figure 1(c), the energy is directly given by:

2 2

2

121

E= mv + sx

The equation of motion is by setting dE dt= 0 , i.:

0212122⎟=⎠ ⎞⎜⎝⎛

xm + sx dt

d &

i. + x= 0 m

s x&&

In Figure 1(c), the restoring force is given by: − 2 Txl, then the stiffness is given

by: s= 2 lT. So the energy is given by:

2 2 2 2 2 2

2

2 121212121

x l

T

x xm l

T

E= mv + sx = xm& + = & +

The equation of motion is by setting dE dt= 0 , i.:

02122⎟=⎠⎞⎜⎝⎛

• x l

T

xm dt

d &

i. 0

2

• x= lm

T

x&&

In Figure 1(e), the liquid of a volume of ρAl is displaced from equilibrium

position by a distance of l 2 , so the stiffness of the system is given by

s= 2 ρgAll= 2 ρgA. So the energy is given by:

2 2 2 2 2 2

2

1221212121

E= mv + sx = ρ xAl& + ρgAx = ρ xAl& +ρgAx

The equation of motion is by setting dE dt= 0 , i.:

02122⎟=⎠⎞⎜⎝⎛

xAl + gAx dt

d ρ & ρ

i. 0

2

• x= l

g x&&

In Figure 1(f), the gas of a mass of ρAl is displaced from equilibrium position by

a distance of x and causes a pressure change of dp=−γpAxV, then, the stiffness

of the system is given by s Adp x pA V

2 −= =γ. So the energy is given by:

V

pAx E mv sx xAl

22 2 2 2

2

121212

1 γ = + = ρ & +

The equation of motion is by setting dE dt= 0 , i.:

02121

22 2 ⎟⎟= ⎠

⎞⎜⎜⎝ ⎛+V

pAx xAl dt

d γ ρ &

i. + x= 0 l V

pA x ρ

γ &&

In Figure 1(g), the restoring force of the hydrometer is −ρgAx, then the stiffness

of the system is given by s=ρgAx x=ρgA. So the energy is given by:

2 2 2 2

2

1212121

E= mv + sx = xm& + ρgAx

The equation of motion is by setting dE dt= 0 , i.:

02 12122⎟=⎠⎞⎜⎝⎛

xm + gAx dt

d & ρ

i. + x= 0 m

A g x

ρ &&

1.

The displacement of the simple harmonic oscillator is given by:

x=asinωt (1.12)

so the velocity is given by:

x&=aωcosωt (1.12)

From (1.12) and (1.12), we can eliminate t and get:

sin cos 1

2 2 22

2

2

2 + = t+ t= a

x

a

x ω ω ω

&(1.12)

which is an ellipse equation of points ,( xx &).

The energy of the simple harmonic oscillator is given by:

2 2 2

sinωtcosφ cosωtsinφ a

y = +

into expression

2

2 1

1 2

2

1 2

2 1

sin sin cos cos ⎟

⎟⎠⎞⎜⎜⎝⎛+ −⎟⎟⎠⎞⎜⎜⎝⎛

φ − φ φ φ a

x

a

y

a

y

a

x , we have:

sin ( )

(sin cos )sin ( )

sin (sin cos sin cos ) cos (cos sin cos sin )

sin sin cos cos

2 1

2

2 1

2 2 2

2 1 2 2 1

2 2 2 1 1 2

2

2

2 1

1 2

2

1 2

2 1

φ φ

ω ω φ φ

ω φ φ φ φ ω φ φ φ φ

φ φ φ φ

= −= + −= − + −⎟⎟⎠⎞⎜⎜⎝⎛ ⎟⎟ + −⎠⎞⎜⎜⎝⎛−

t t

t t

a

x

a

y

a

y

a

x

From the above derivation, we have:

cos( ) sin( )

2

2 1

2 1 2 21

2 2

2

2 1

2 + − φ−φ = φ −φ aa

xy

a

y

a

x

1.

By elimination of t from equation x=asinωt and y=bcosωt, we have:

2 1

2

2

2 + = b

y

a

x

which shows the particle follows an elliptical path. The energy at any position of x,

y on the ellipse is given by:

( )212121

cos 2

1

sin 2

1

sin 2

1

cos 2

121212121

2 2 2

22 22

22 2 22 2 22 2 22 2

2 2 2 2

m a b

ma mb

ma t ma t mb t mb t

E xm sx ym sy

= += += + + += + + +

ω

ω ω

ω ω ω ω ω ω ω ω

& &

The value of the energy shows it is a constant and equal to the sum of the separate

energies of the simple harmonic vibrations in x direction given by

22

2

1

mω a and in

y direction given by

22

2

1

mωb.

At any position of x, y on the ellipse, the expression of m( yx&− xy&) can be

written as:

m( yx − xy )=m(−abωsin ωt−abωcos ωt) −= abmω(sin ωt+cos ωt) −= abmω

2 2 2 2 & &

which is a constant. The quantity abmω is the angular momentum of the particle.

1.

All possible paths described by equation 1 fall within a rectangle of 2 a 1 wide and

2 a 2 high, where a 1 =xmax and a 2 =ymax, see Figure 1.

When x= 0 in equation (1) the positive value of y=a 2 sin(φ 2 −φ 1 ). The value of

ymax=a 2. So yx= 0 ymax=sin(φ 2 −φ 1 ) which defines φ 2 −φ 1.

1.

In the range 0 ≤φ≤π, the values of cos are φi − 1 ≤cosφi≤+ 1. For n random

values of φi, statistically there will be n 2 values − 1 ≤cosφi≤ 0 and n 2 values

0 ≤cosφi≤ 1. The positive and negative values will tend to cancel each other and the

sum of the n values: cos 0

1

∑ →

≠

=

n

ji

i

φi , similarly cos 0 1

∑ →

=

n

j

φj. i.

cos cos 0 1 1

∑ ∑ →

= ≠

=

n

j

j

n

ji

i

φi φ

1.

The exponential form of the expression:

asinωt+asin(ωt+δ)+asin(ωt+ 2 δ)+L+asin[ωt+(n− δ])

is given by:

ω ( +δω ) (ω+δ)2 ω −+ δ])1([ + + + +

ti ti ti nti ae ae ae L ae

From the analysis in page 28, the above expression can be rearranged as:

sin 2

2 sin 2

1

δ

ω δ nδ ae

n ti ⎥ ⎦

⎤ ⎢ ⎣

⎡ ⎟ ⎠

⎜ ⎞ ⎝

+⎛ −

with the imaginary part:

sin 2

sin 2

2

1

sin δ

δ ω δ

n n a t ⎥ ⎦

⎤⎢⎣⎡⎟⎠⎞⎜⎝⎛ −+

which is the value of the original expression in sine term.

SOLUTIONS TO CHAPTER 2

The system is released from rest, so we know its initial velocity is zero, i.

0

0

=

dtt=

dx

(2.1)

Now, rearrange the expression for the displacement in the form:

tqp tqp e

F G

e

F G

x

( ) ( )

2 2

+− − −−++=

(2.1)

Then, substitute (2.1) into (2.1), we have

() () 0220

( ) ( )

0

⎥ =⎦⎤⎢⎣⎡ −+ − −+= − +

=

+− −−

= t

tqp tqp

t

e

F G

e p q

F G

p q dt

dx

i.

qG=pF

(2.1)

By substitution of the expressions of q and p into equation (2.1), we have the ratio given by:

()

2 21 r 4 ms

r

F

G−=

The first and second derivatives of x are given by:

()

rt m A Bt e m

r x B

2

2

− ⎥ ⎦

⎤⎢⎣⎡&= − +()

rt m A Bt e m

r

m

rB x

2 2

2

4

− ⎥ ⎦

⎤⎢⎣⎡&&= − + +

We can verify the solution by substitution of x, x& and x&& into equation:

xm&&+ xr&+sx= 0

then we have equation:

() 04

2

⎟⎟ + = ⎠

⎞⎜⎜⎝⎛

− A Bt m

r s

which is true for all t, provided the first bracketed term of the above equation is zero, i.

04

2 − = m

r s

i. r m =sm

2 2 4

The initial displacement of the system is given by:

( ) φ

ω ω 1 2 cos

2 x e eC Ce A

rt m ti ti = + =

− ′ − ′ at t= 0

So:

C 1 +C 2 =Acosφ

(2.3)

Now let the initial velocity of the system to be:

() () ω ω ω φ

ω ω sin 2 2

2 2

2 1 i Ce A m

r i eC m

r x

timr timr ⎟ =− ′ ⎠

⎞⎜⎝⎛⎟ + − − ′⎠⎞⎜⎝⎛= − + ′

− + ′ − − ′ & at t= 0

i. cosφ ω()ω sinφ 2

A i C 1 C 2 A m

r − + ′ − =− ′

If rm is very small orφ≈π 2 , the first term of the above equation approximately equals zero,

so we have:

C 1 −C 2 =iAsinφ

(2.3)

From (2.3) and (2.3), C 1 and C 2 are given by:

( )

()φ

φ

φ φ

φ φ

i

i

e

A i A C

e

A i A C

−

−==+=2 2

cos sin

2 2

cos sin

2

1

Use the relation between current and charge, I=q&, and the voltage equation:

qC+IR= 0

we have the equation:

qR&+q C= 0

solve the above equation, we get:

RCt q eC

− = 1

The frequency of a damped simple harmonic oscillation is given by:

2

2 2 0

2

4 m

r ω′ =ω − => 2

2 2 0

2

4 m

r ω′ −ω = => () 0

2

2

0 4

-

ω ω

ω ω ω ′+

Δ = ′ =

m

r

Use ω′≈ω and r

m Q 0

ω

we find fractional change in the resonant frequency is given by:

()

2 1 2 0

2

2

0

0

0

88

− ≈ =

′−=ΔQ

m

r

ω ω

ω ω

ω

ω

See page 71 of text. Analysis is the same as that in the text for the mechanical case except that

inductance L replaces mass m, resistance R replaces r and stiffness s is replaced by

1 C, where C is the capacitance. A large Q value requires a small R.

Electrons per unit area of the plasma slab is given by:

q=−nle

When all the electrons are displaced a distance x, giving a restoring electric field:

E=nex/ε 0 ，the restoring force per unit area is given by:

0

22

ε

xn le F=qE −=

Newton’s second law gives:

restoring force per unit area= electrons mass per unit area× electrons acceleration

i.

nlm x

xn le F −= = e×&& 0

22

ε

i. 0

0

2

• x= m

ne x ε

&&

From the above equation, we can see the displacement distance of electrons, x, oscillates with

angular frequency:

0

2 2

ε

ω e

e m

ne

As the string is shortened work is done against: (a) gravity (mgcosθ) and (b) the centrifugal

force ( )

2 2 mv r=mlθ& along the time of shortening. Assume that during shortening there are

many swings of constant amplitude so work done is:

A=−(mgcos +ml )Δl

2 θ θ&

where the bar denotes the average value. For small θ, cos 1 2

2 θ= −θ so:

( 2 )

2 2 A −= mgΔl+ mgθ −mlθ&

The term −mgΔl is the elevation of the equilibrium position and does not affect the energy of

motion so the energy change is:

ΔE=(mg 2 −ml )Δl

2 2 θ θ&

Now the pendulum motion has energy:

1( cos ) 2

22 = lθ +mgl − θ

m E & ,

that is, kinetic energy plus the potential energy related to the rest position, for small θ this

becomes:

2 2

22 2 mlθ mglθ E= +

&

which is that of a simple harmonic oscillation with linear amplitude lθ 0.

Taking the solution θ=θ 0 cosωt which gives 2

2 0

2 θ =θ and 2

2 0

2 2 θ& =ω θ with

ω= lg we may write:

2 2

2 0

2 0

22 mlωθ mglθ E= =

and

l

ml l

ml ml E Δ −= Δ⋅ ⎟

⎟⎠⎞⎜⎜⎝⎛Δ = −4 2 4

2 0

2 2 0

2 2 0

2 ωθ ωθ ωθ

so:

l

l

E

E Δ=−Δ21

Now ω= 2 πν= lg so the frequency ν varies with

− 21 l and

EE

l

l Δ =

Δ=−Δ21

ν

ν

so:

=constant ν

E