Pain Solution Manual

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

SOLUTIONS TO CHAPTER 1

1.1

In Figure 1.1(a), the restoring force is given by:

θ

sin

mg F =−

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

l x mg F =−

so the stiffness is given by:

l mg x F

s=− =

so we have the frequency given by:

l g m s = = 2 ω

Since θ is a very small angle, i.e. θ =sinθ = x l, or x=lθ, we have the restoring force given by:

θ

mg F =−

Now, the equation of motion using angular displacement θ can by derived from Newton’s second law: x m F = && i.e. −mgθ =mlθ&& i.e. θ + θ =0 l g &&

which shows the frequency is given by:

l g = 2

ω

In Figure 1.1(b), restoring couple is given by −Cθ, which has relation to moment of inertia I

given by: θ θ I && C = − i.e. θ + θ =0 I C &&

which shows the frequency is given by:

I C = 2

In Figure 1.1(d), the restoring force is given by:

l x T F =−2

so Newton’s second law gives:

l Tx x m F = &&=−2 i.e. x&&+2Tx lm=0

which shows the frequency is given by:

lm T

2 2 = ω

In Figure 1.1(e), the displacement for liquid with a height of x has a displacement of x 2 and a mass of ρAx, so the stiffness is given by:

Ag x Axg x G s 2ρ 2ρ 2 = = =

Newton’s second law gives:

x m G= && −

i.e. − 2ρAxg= ρAlx&&

i.e. +2 x=0

l g x&&

which show the frequency is given by:

l g 2 2 =

ω

In Figure 1.1(f), by taking logarithms of equation pVγ =constant, we have:

constant ln lnp+γ V = so we have: + =0 V dV p dp _γ i.e. V dV p dp=−γ

The change of volume is given by dV = Ax, so we have:

V Ax p dp=−γ

The gas in the flask neck has a mass of ρAl, so Newton’s second law gives:

x m Adp= &&

i.e. Alx V x A p ρ && γ = − 2 i.e. + x=0 V l pA x ρ γ &&

which show the frequency is given by:

V l pA ρ γ ω2 =

In Figure 1.1 (g), the volume of liquid displaced is Ax, so the restoring force is −ρgAx. Then, Newton’s second law gives:

x m gAx F =−ρ = && i.e. + x=0 m A g x&& ρ

which shows the frequency is given by:

m A gρ

ω2 =

1.2

Write solution x=acos(ωt+φ) in form: x=acosφcosωt−asinφsinωt and compare with equation (1.2) we find: A=acosφ and B=−asinφ. We can also find, with the same analysis, that the values of A and B for solution

) sin(ω −φ

=a t

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

cos(ω −φ

=a t

x are given by: A=acosφ and B=asinφ. Try solution x=acos(ωt+φ) in expression x&&+ω2x, we have:

0 ) cos( ) cos( 2 2 2 =− + + + = +ω x aω ωt φ ω a ωt φ x&&

Try solution x=asin(ωt−φ) in expression x&&+ω2x, we have: 0 ) sin( ) sin( 2 2 2 _{=− − + − =} +ω x aω ωt φ ω a ωt φ x&&

Try solution x=acos(ωt−φ) in expression x&&+ω2x, we have: 0 ) cos( ) cos( 2 2 2 =− − + − = +ω x aω ωt φ ω a ωt φ x&&

1.3

(a) If the solution x=asin(ωt+φ) satisfies x=aat t =0, then, x=asinφ =a i.e. φ =π 2. When the pendulum swings to the position x=+a 2 for the first time after release, the value of tω is the minimum solution of equation

2 )

2

sin( t a

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

3

π

ωt = 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.e. φ =0. When the pendulum swings to the position x=+a 2 for the first time after release, the value of tω is the minimum solution of equation

2

cos t a

a ω =+ , i.e. 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, a

a

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

a

x=+ for the first time after release, the value of tω is the minimum solution of equation asin(ωt+π 2)=+a 2, i.e. 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, a

a

x= cos(−φ)= i.e. φ =0 . When the pendulum swings to the position 2

a

x=+ for the first time after release, the value of tω is the minimum solution of equation acosωt=+a 2, i.e. ωt =π 4. Similarly, we can find: for

2

a

x= , 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, a

a

2 a

x=+ for the first time after release, the value of tω is the minimum solution of equation asin(ωt−π 2)=+a 2, i.e. ωt=3π 4. Similarly, we can find: for x=a 2, ωt =2π 3 and for x=0, ωt =π 2.

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

a

x= cosφ =− i.e. φ =π . When the pendulum swings to the position 2

a

x=+ for the first time after release, the value of tω is the minimum solution of equation acos(ωt+π)=+a 2, i.e. ωt =3π 4. Similarly, we can find: for x=a 2, ωt =2π 3 and for x=0, ωt =π 2.

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

a

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

a

x=+ for the first time after release, the value of tω is the minimum solution of equation asin(ωt−π 2)=+a 2, i.e. ωt =3π 4. Similarly, we can find: for x=a 2, ωt =2π 3 and for x=0, ωt =π 2.

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

a

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

a

x=+ for the first time after release, the value of tω is the minimum solution of equation acos(ωt−π)=+a 2, i.e. ωt=3π 4. Similarly, we can find: for x=a 2, ωt =2π 3 and for x=0, ωt =π 2.

1.4

The frequency of such a simple harmonic motion is given by:

] [ 10 5 . 4 10 1 . 9 ) 10 05 . 0 ( 10 85 . 8 4 ) 10 6 . 1 ( 4 1 16 31 3 9 12 2 19 3 0 2 0 − − − − − ⋅ × ≈ × × × × × × × × = = = rad s m r e m s e e πε π ω

] [ 42 ] [ 10 2 . 4 10 5 . 4 10 3 2 2 8 16 8 0 nm m c _{≈ × =} × × × × = = π − ω π λ

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

1.5

(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:

2

s x F

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

ω .

(b) The frequency of the system is given by: b =s m 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=−2sx and the stiffness of the system is given by: s′=−F x=2s. Hence the frequency is given by

m s m s c 2 2 = ′ = ω .

Therefore, we have the relation: ωa2:ωb2:ωc2 =s 2m:s m:2s m=1:2:4

1.6 At time t=0, x= gives: x0 0 sin x a φ = (1.6.1) 0 v x&= gives: 0 cos v aω φ = (1.6.2) From (1.6.1) and (1.6.2), we have

0 0

tanφ =ωx v and a=(x₀2+v₀2 ω2)12

1.7

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 v_t , where v is _t the velocity of the particle at time t and is given by: v_t =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 dt 2 =

η where the period is given by:

ω π 2 = T , so we have: 2 2 2 ) ( sin 1 ) cos( ) cos( 2 2 2 x a dx t a dx t a dx t a dx T dt − = + − = + = + = = π φ ω π φ ω π φ ω ω π ω η 1.8

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 2π between two oscillators a distance λ apart by: δφ 2π =x λ, i.e. δφ =2πx λ.

1.9

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: a= Aω2sin(ωt+φ), where A is the amplitude, ω is the angular frequency and φ is the initial phase. So we have:

g t Aω2sin(ω + )φ = i.e. ) sin( 2 ω φ ω + = t g A

Therefore, the minimum amplitude, which makes the mass lose contact with the platform, is given by:

] [ 01 . 0 5 4 8 . 9 4 2 2 2 2 2 min m f g g A ≈ × × = = = π π ω 1.10

The mass of the element dy is given by: m′=mdy l. 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′= yv l. where v is the velocity of the element at the other end of the spring, i.e. the velocity of the suspended mass M . Hence we have the kinetic energy

of this element given by: 2 2 2 1 2 1 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ′ ′ = v l y dy l m v m KE_dy

The total kinetic energy of the spring is given by:

∫

∫

∫

The total kinetic energy of the system is the sum of kinetic energies of the spring and the suspended mass, and is given by:

(

)

which shows the system is equivalent to a spring with zero mass with a mass of

3

m

M + suspended at the end. Therefore, the frequency of the oscillation system is given by: 3 2 m M s + = ω 1.11

In Figure 1.1(a), the restoring force of the simple pendulum is −mgsinθ , then, the stiffness is given by: s=mgsinθ x=mg l. So the energy is given by:

2 2 2 2 2 1 2 1 2 1 2 1 x l mg x m sx mv E= + = & +

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

0 2 1 2 1 2 2 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ₊ x l mg x m dt d & i.e. + x=0 l g x&&

In Figure 1.1(b), the displacement is the rotation angle θ , the mass is replaced by the moment of inertia I of the disc and the stiffness by the restoring couple C of the wire. So the energy is given by:

2 2 2 1 2 1 _{θ θ} C I E = & +

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

0 2 1 2 1 2 2 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ _{θ + θ} C I dt d _&

i.e. θ + θ =0 I C &&

In Figure 1.1(c), the energy is directly given by: 2 2 2 1 2 1 sx mv E= +

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

0 2 1 2 1 2 2_⎟= ⎠ ⎞ ⎜ ⎝ ⎛ _{+ sx} x m dt d & i.e. + x=0 m s x&&

In Figure 1.1(c), the restoring force is given by: −2Tx l, then the stiffness is given by: s=2T l. So the energy is given by:

2 2 2 2 2 2 2 1 2 2 1 2 1 2 1 2 1 x l T x m x l T x m sx mv E= + = & + = & +

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

0 2 1 2 2 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ _{+ x} l T x m dt d & i.e. +2 x=0 lm T x&&

In Figure 1.1(e), the liquid of a volume of Alρ is displaced from equilibrium position by a distance of l 2 , so the stiffness of the system is given by

gA l

gAl

s=2ρ =2ρ . So the energy is given by:

2 2 2 2 2 2 2 1 2 2 1 2 1 2 1 2 1 gAx x Al gAx x Al sx mv E= + = ρ & + ρ = ρ & +ρ

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

0 2 1 2 2_⎟= ⎠ ⎞ ⎜ ⎝ ⎛ _{+ gAx} x Al dt d _{ρ ρ} & i.e. +2 x=0 l g x&&

In Figure 1.1(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=−γpAx V, then, the stiffness of the system is given by s=−Adp x=γpA2 V . So the energy is given by:

V x pA x Al sx mv E 2 2 2 2 2 2 1 2 1 2 1 2 1 _{+ = ρ +} γ = _&

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

0 2 1 2 1 2 2 2 = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + V x pA x Al dt d _ρ γ & i.e. + x=0 V l pA x ρ γ &&

In Figure 1.1(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 1 2 1 2 1 2 1 gAx x m sx mv E= + = & + ρ

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

0 2 1 2 1 2 2 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ₊ gAx x m dt d _ρ & i.e. + x=0 m g A x&& ρ 1.12

The displacement of the simple harmonic oscillator is given by:

x=asinωt (1.12.1) so the velocity is given by:

t a

x&= ωcosω (1.12.2) From (1.12.1) and (1.12.2), we can eliminate t and get:

1 cos sin2 2 2 2 2 2 2 = + = + t t a x a x _{ω ω} ω & _(1.12.3) which is an ellipse equation of points (x,x& . )

2 2 2 1 2 1 sx x m E = & + (1.12.4) Write (1.12.3) in form x&2 =ω2(a2−x2) and substitute into (1.12.4), then we have:

2 2 2 2 2 2 2 1 ) ( 2 1 2 1 2 1 sx x a m sx x m E = & + = ω − +

Noting that the frequency ω is given by: ω2 =s m, we have: 2 2 2 2 2 1 2 1 ) ( 2 1 sa sx x a s E = − + =

which is a constant value.

1.13

The equations of the two simple harmonic oscillations can be written as:

) sin( 1 =a ωt+φ

y and y₂ =asin(ωt+φ+δ)

The resulting superposition amplitude is given by:

) 2 cos( ) 2 sin( 2 )] sin( ) [sin( 2 1+ = ω +φ + ω +φ+δ = ω +φ+δ δ = y y a t t a t R

and the intensity is given by:

) 2 ( sin ) 2 ( cos 4 2 2 2 2 = δ ω +φ+δ =R a t I i.e. I ∝4a2cos2(δ 2)

Noting that sin2(ωt+φ+δ 2) varies between 0 and 1, we have: ) 2 ( cos 4 0≤I ≤ a2 2 δ 1.14 ) cos( 2 ) cos cos sin (sin 2 ) cos (sin ) cos (sin cos cos 2 cos cos sin sin 2 sin sin cos cos sin sin 2 1 2 1 2 2 2 2 1 2 2 1 2 1 2 1 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 1 1 2 2 2 2 2 2 2 1 2 2 2 1 1 2 2 1 2 2 1 φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ − − + = + − + + + = − + + − + = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − + ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − a a xy a y a x a a xy a y a x a a xy a x a y a a xy a y a x a x a y a y a x

On the other hand, by substitution of :

1 1 1 sin cos cos sinωt φ ωt φ a x _{= +}

2 2 2 sin cos cos sinωt φ ωt φ a y _{= +} into expression 2 2 1 1 2 2 1 2 2 1 cos cos sin sin _⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − + ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − φ φ φ φ a x a y a y a x , we have: ) ( sin ) ( sin ) cos (sin ) sin cos sin (cos cos ) cos sin cos (sin sin cos cos sin sin 1 2 2 1 2 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 φ φ ω ω φ φ φ φ ω φ φ φ φ ω φ φ φ φ − = − + = − + − = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − + ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − t t t t a x a y a y a x

From the above derivation, we have:

) ( sin ) cos( 2 1 2 2 2 1 2 1 2 2 2 2 1 2 φ φ φ φ − = − − + a a xy a y a x 1.15

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

1 2 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:

) ( 2 1 2 1 2 1 cos 2 1 sin 2 1 sin 2 1 cos 2 1 2 1 2 1 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 b a m mb ma t mb t mb t ma t ma sy y m sx x m E + = + = + + + = + + + = 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 2 2

2 1 a mω and in y direction given by 2 2 2 1 b mω .

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

ω ω ω ω ω ω ω ω t ab t abm t t abm ab m x y y x

m( &− &)= (− sin2 − cos2 )=− (sin2 +cos2 )=− which is a constant. The quantity abmω is the angular momentum of the particle.

1.16

All possible paths described by equation 1.3 fall within a rectangle of 2a₁ wide and

2

2a high, where a₁= x_max and a₂ = y_max, see Figure 1.8.

When x=0 in equation (1.3) the positive value of y = a₂sin(φ₂−φ₁). The value of 2

max a

y = . So y_x=0 y_max =sin(φ₂−φ₁) which defines φ₂−φ₁.

1.17

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

values of φ_i, statistically there will be n 2 values −1≤cosφ_i ≤0 and n 2 values 1

cos

0≤ φ_i ≤ . The positive and negative values will tend to cancel each other and the sum of the n values: cos 0

1 →

∑

∑

∑

∑

The exponential form of the expression:

] ) 1 ( sin[ ) 2 sin( ) sin( sinωt+a ωt+δ +a ωt+ δ + +a ωt+ n− δ a _L is given by: ] ) 1 ( [ ) 2 ( ) (ω δ ω δ ω δ ωt₊ i t+ ₊ i t+ _{+ +} i t+n− i ae ae ae ae _L

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

2 sin 2 sin 2 1 δ δ δ ω _n ae n t i _⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − +

with the imaginary part:

2 sin 2 sin 2 1 sin δ δ δ ωt n n a _⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − +

1.19

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

2 sin 2 sin ) 1 ( 2 ( 1) δ δ ω δ δ δ ω n ae e e e ae z= i t + i + i +_L in− = i t

The conjugate of z is given by:

2 sin 2 sin * δ δ ω n ae z = −i t so we have: 2 sin 2 sin 2 sin 2 sin 2 sin 2 sin 2 2 2 * δ δ δ δ δ δ ω ω n a n ae n ae zz = i t ⋅ −i t =

SOLUTIONS TO CHAPTER 2

2.1

The system is released from rest, so we know its initial velocity is zero, i.e. 0 0 = = t dt dx (2.1.1)

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

x F Ge( p q)t F Ge( p q)t 2 2 − − + − ₊ − + = (2.1.2)

Then, substitute (2.1.2) into (2.1.1), we have

(

)

(

)

By substitution of the expressions of q and p into equation (2.1.3), we have the ratio given by:

(

)

The first and second derivatives of x are given by:

(

)

(

)

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

0 = + +rx sx x m&& &

then we have equation:

(

)

0 4 2 = − m r s i.e. r2 4m2 =s m 2.3

The initial displacement of the system is given by:

(

)

( ) _ω ( ) _{ω φ} ω ω ω sin 2 2 2 2 2 1 i C e A m r e C i m r x r m i t ⎟ r m i t =− ′ ⎠ ⎞ ⎜ ⎝ ⎛_{− − ′} + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛_{− + ′} = − + ′ − − ′ & at t =0

i.e. cosφ ω

(

)

2mA i C1 C2 A r _′ − = − ′ + −

If r m is very small orφ ≈π 2, the first term of the above equation approximately equals zero, so we have:

C1−C2 =iAsinφ

(2.3.2) From (2.3.1) and (2.3.2), C₁ and C₂are given by:

(

)

(

)

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

0 = + IR

C q

we have the equation:

0 = + Cq q R&

solve the above equation, we get:

RC t e C q= ₁ −

where C₁ is arbitrary in value. Use initial condition, we get C₁=q₀, i.e. q=q₀e−tRC

which shows the relaxation time of the process is RC s.

2.5

(a) ω₀2-ω′2 =10-6ω₀2 =r2 4m2 => ω₀m r=500

The condition also shows ω′≈ω₀, so:

500 0 = ≈ ′ = m r m r Q ω ω Use ω π τ ′ = ′ 2 , we have: 500 2 π π ω π τ δ = = ′ = ′ = Q m r m r

(b) The stiffness of the system is given by:

] [ 100 10 1012 10 1 2 0 − − ₌ × = = m Nm s ω

and the resistive constant is given by:

2 10 [ ] 500 10 10 7 1 10 6 0 − − − ⋅ × = × = = N sm Q m r ω

(c) At t =0and maximum displacement, x_&=0, energy is given by:

] [ 10 5 10 100 2 1 2 1 2 1 2 1 2 4 3 max 2 2 J sx sx x m E= & + = = × × − = × −

Time for energy to decay to e−1of initial value is given by:

] [ 5 . 0 10 2 10 7 10 ms r m t = × = = − ₋

(d) Use definition of Q factor:

E E Q Δ − =2π

where, E is energy stored in system, and −ΔE is energy lost per cycle, so energy loss in the first cycle, −ΔE₁, is given by:

] [ 10 2 500 10 5 2 2 5 3 1 J Q E E E − − × = × × = = Δ − = Δ − π π π 2.6

The frequency of a damped simple harmonic oscillation is given by: 2 2 2 0 2 4m r − = ′ ω ω => ₂ 2 2 0 2 4m r = − ′ ω ω =>

(

)

Q=ω0 _{we find fractional change in the resonant frequency is given by:}

( )

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

C

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

2.8

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

nle q=−

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

0 /ε nex

E = ，the restoring force per unit area is given by:

0 2 2 ε l e xn qE F = =−

Newton’s second law gives:

on accelerati electrons area unit per mass electrons area unit per force restoring = × i.e. x nlm l e xn F =− = _e× && 0 2 2 ε i.e. 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 = 2.9

As the string is shortened work is done against: (a) gravity (mgcosθ) and (b) the centrifugal force (mv2 r =mlθ&2) along the time of shortening. Assume that during shortening there are

many swings of constant amplitude so work done is:

l ml mg

A=−( cosθ + θ&2)Δ

where the bar denotes the average value. For small θ, cosθ =1−θ2 2 so:

) 2 (mgθ2 mlθ&2 l mg A=− Δ + −

The term −mgΔl is the elevation of the equilibrium position and does not affect the energy of motion so the energy change is:

l ml mg

E= − Δ

Δ ( θ2 2 θ&2)

Now the pendulum motion has energy:

) cos 1 ( 2 2 2θ + − θ = ml mgl E & ,

that is, kinetic energy plus the potential energy related to the rest position, for small θ this becomes: 2 2 2 2 2θ θ mgl ml E= & +

which is that of a simple harmonic oscillation with linear amplitude lθ₀.

Taking the solution θ =θ0cosωt which gives 2

2 0 2 θ θ = and 02 2 2 2 ω θ θ& = with l g = ω we may write: 2 2 2 0 2 0 2 2ω θ θ mgl ml E = = and l ml l ml ml E _⎟⎟Δ =− ⋅Δ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − = Δ 4 2 4 2 0 2 2 0 2 2 0 2θ ω θ ω θ ω so: l l E E ₌₋ Δ Δ 2 1

Now ω =2πν = g l so the frequency ν varies with l−12 and

E E l l Δ = Δ − = Δ 2 1 ν ν so: constant = ν E

SOLUTIONS TO CHAPTER 3

3.1

The solution of the vector form of the equation of motion for the forced oscillator:

t i e F s r mx&&+ x&+ x= ₀ ω is given by: ) sin( ) cos( ) ( 0 ω φ ω φ ω ω ω φ ω − + − − = = − t Z F t Z iF Z e iF m m m t i x

Since F₀eiωt represents its imaginary part: F₀sinωt, the value of x is given by the imaginary part of the solution, i.e.:

) cos(ω φ ω − − = t Z F x m

The velocity is given by:

) sin(ω −φ = = t Z F x v m & 3.2

The transient term of a forced oscillator decay with e−rt2m to e-kat time t , i.e.:

k m rt =−

− 2

so, we have the resistance of the system given by:

r=2mk t (3.2.1) For small damping, we have

m

s 0 = ≈ω

ω (3.2.2) We also have steady state displacement given by:

) sin(

0 ω −φ

=x t

x where the maximum displacement is:

2 2 0 0 ) ( m s m r F x − + = ω ω (3.2.3)

By substitution of (3.2.1) and (3.2.2) into (3.2.3), we can find the average rate of growth of the oscillations given by:

0 0 0 2kmω F t x = 3.3

Write the equation of an undamped simple harmonic oscillator driven by a force of frequency ω in the vector form, and use F₀eiωt to represent its imaginary part

t F₀sinω , we have: t i e F s mx&&+ x= ₀ ω (3.3.1) We try the steady state solution x=Aeiωt and the velocity is given by:

x A x& =iω eiωt =iω so that: x x x&&= i2ω2 =−ω2

and equation (3.3.1) becomes:

t i t i e F e s m ω ω ω2 ₀ ) (−A +A =

which is true for all t when

0 2 F s m+ = −Aω A i.e. m s F 2 0 ω − = A i.e. ei t m s F _ω ω2 0 − = x

The value of x is the imaginary part of vector x , given by: t m s F x ω ω2 sin 0 − = i.e. ) ( sin 2 2 0 0 ω ω ω − = m t F x where m s = 2 0 ω

Hence, the amplitude of x is given by:

) ( 02 2 0 ω ω − = m F A

and its behaviour as a function of frequency is shown in the following graph:

By solving the equation:

0 = + sx

x m&&

we can easily find the transient term of the equation of the motion of an undamped simple harmonic oscillator driven by a force of frequency ω is given by:

t D t C

x= cosω₀ + sinω₀

where, ω0 = s m, C and D are constant. Finally, we have the general solution

for the displacement given by the sum of steady term and transient term: t D t C m t F x 2 2 0 0 0 0 sin cos ) ( sin _{ω ω} ω ω ω _{+ +} − = (3.3.2) 3.4 In equation (3.3.2), x=0 at t=0 gives: 0 sin cos ) ( sin 0 0 0 2 2 0 0 0 ⎥ = = ⎦ ⎤ ⎢ ⎣ ⎡ + + − = = = _m A t B t A t F x t t ω ω ω ω ω (3.4.1) In equation (3.3.2), x&=0 at t=0 gives:

0 ) ( cos sin ) ( cos 0 2 2 0 0 0 0 0 0 0 2 2 0 0 0 = + − = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + − − = = = B m F t B t A m t F dt dx t t ω ω ω ω ω ω ω ω ω ω ω ω i.e. ) ( 02 2 0 0 ω ω ω ω − − = m F B (3.4.2) By substitution of (3.4.1) and (3.4.2) into (3.3.2), we have:

⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − = t t m F x ₀ 0 2 2 0 0 sin sin ) ( 1 _ω ω ω ω ω ω (3.4.3)

By substitution of ω =ω₀+Δω into (3.4.3), we have: 2 0 0 ω m F A 0 ω ω 0

⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − Δ + Δ Δ + − = t t t t t m F x ₀ 0 0 0 0

0 _{sin cos sin cos sin}

) ( ω ω ω ω ω ω ω ω ω ω (3.4.4)

Since Δω ω0 <<1 and Δ tω <<1, we have:

0

ω

ω ≈ , sinΔωt≈Δωt, and cosΔ tω ≈1

Then, equation (3.4.4) becomes:

⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − Δ + Δ − = t t t t m F x 0 0 0 0 0 0 sin cos sin 2 ω ω ω ω ω ω ω ω i.e. _⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ Δ + Δ − Δ − = t t t m F x 0 0 0 0 0 cos sin 2 ω ω ω ω ω ω ω i.e. _⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − = t t t m F x ₀ 0 0 0 0 cos sin 2 ω ω ω ω

i.e. (sin cos )

2 02 0 0 0 0 t t t m F x ω ω ω ω − =

The behaviour of displacement x as a function of ω0t is shown in the following

graph:

3.5

The general expression of displacement of a simple damped mechanical oscillator driven by a force F0cosωt is the sum of transient term and steady state term, given

x t 0 ω 0 0 2mω t F 0 0 2mω t F − 2 0 0 2 ω π m F 0 2 0 0 ω π m F 2 0 0 2 ω π m F 2 0 0 3 ω π m F 2 0 0 2 3 ω π m F 02 0 2 5 ω π m F 02 0 2 7 ω π m F

by: m t i t i m rt Z e iF Ce i ω φ ω ω ( ) 0 2 − + − − = x

where, C is constant, Z_m = r2+(ωm−s ω)2 , ω_t = s m−r2 4m2 and ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = − r s m ω ω φ 1

tan , so the general expression of velocity is given by:

) ( 0 2 2 φ ω ω ω _⎟ − + ₊ − ⎠ ⎞ ⎜ ⎝ ⎛_{− +} = = i t m t i m rt t e Z F e i m r C i x v &

and the general expression of acceleration is given by:

) ( 0 2 2 2 2 4 φ ω ω ω ω ω − + ₊ − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − = i t m t i m rt t t e Z F i e m r i m r C i v& i.e. 2 0 ( ) 2 2 2 2 ω − +ω ω _ω−φ + ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − = i t m t i m rt t e Z F i e m r i m ms r C i v& (3.5.1)

From (3.5.1), we find the amplitude of acceleration at steady state is given by:

2 2 0 0 ) (ω ω ω ω s m r F Z F v m + − = = &

At the frequency of maximum acceleration: =0

ω d v d& i.e. 0 ) ( 2 2 0 = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ − + ω ω ω ω _{r m s} F d d i.e. 2 2 ₂ 0 2 2− + = ω s ms r i.e. ₂ 2 2 2 2 r sm s − = ω

Hence, we find the expression of the frequency of maximum acceleration given by:

2 2 2 2 r sm s − = ω

The frequency of velocity resonance is given by: ω= s m, so if r= sm, the acceleration amplitude at the frequency of velocity resonance is given by:

m F sm sm sm F m s s m r F v sm r 0 0 2 2 0 ) ( ) ( = − + = − + = = _{ω ω} ω &

The limit of the acceleration amplitude at high frequencies is given by: m F s m r F s m r F v 0 2 2 2 2 0 2 2 0 _lim ) ( lim lim = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − + = − + = ∞ → ∞ → ∞ → ω ω ω ω ω ω ω ω & So we have: v v&_{r sm} & ∞ → = =_ωlim 3.6

The displacement amplitude of a driven mechanical oscillator is given by:

2 2 0 ) (ω ω ω r m s F x − + = i.e. 2 2 2 2 0 ) ( m s r F x − + = ω ω (3.6.1)

The displacement resonance frequency is given by:

2 2 2m r ms − = ω (3.6.2) By substitution of (3.6.2) into (3.6.1), we have:

2 2 2 2 2 0 2 2 ⎟⎟_⎠ ⎞ ⎜⎜ ⎝ ⎛ + ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − = m r m r m s r F x i.e. 2 2 0 4m r m s r F x − =

which proves the exact amplitude at the displacement resonance of a driven mechanical oscillator may be written as:

r F x ω′ = 0 where, 2 2 2 4m r ms − = ′ ω 3.7

(a) The displacement amplitude is given by:

2 2 0 ) (ω ω ω r m s F x − + =

At low frequencies, we have: s F s m r F s m r F x 0 2 2 2 2 0 0 2 2 0 0 0 lim _{( )} lim _{( )} lim = − + = − + = → → → ω _{ω ω ω} ω _{ω ω} ω

(b) The velocity amplitude is given by:

2 2 0 ) (ωm s ω r F v − + =

At velocity resonance: ω= s m, so we have:

r F sm sm r F s m r F v m s r 0 2 2 0 2 2 0 ) ( ) ( = − + = − + = = ω ω ω

(c) From problem 3.5, we have the acceleration amplitude given by:

2 2 0 ) (ω ω ω s m r F v − + = & At high frequency, we have:

m F s m r F s m r F v 0 2 2 2 2 0 2 2 0 ) ( lim ) ( lim lim = − + = − + = ∞ → ∞ → ∞ → _{ω ω ω ω} ω ω ω ω &

From (a), (b) and (c), we find x 0 lim →

ω , vr and _ωlim are all constants, i.e. they are all _→∞v&

frequency independent.

3.8

The expression of curve (a) in Figure 3.9 is given by:

2 2 2 2 2 0 2 2 2 0 0 2 0 ) ( ) ( r m m F Z X F x m m a _{ω ω ω} ω ω ω − + − = − = (3.8.1) where ω₀ = s m a

x has either maximum or minimum value when =0

ω d dx_a i.e. 0 ) ( ) ( 2 2 2 2 2 0 2 2 2 0 0 = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + − − r m m F d d ω ω ω ω ω ω i.e. ( ) 02 2 0 2 2 2 0 2 − − = r m ω ω ω

Then, we have two solutions of ω given by:

m r 0 2 0 1 ω ω ω = − (3.8.2)

and m r 0 2 0 2 ω ω ω = + (3.8.3) Since r is very small, rearrange the expressions of ω and ₁ ω , we have: ₂

m r m r m r m r 2 4 2 2 0 2 2 0 0 2 0 1 ⎟ − ≈ − ⎠ ⎞ ⎜ ⎝ ⎛ ₋ = − = ω ω ω ω ω m r m r m r m r 2 4 2 2 0 2 2 0 0 2 0 2 ⎟ − ≈ + ⎠ ⎞ ⎜ ⎝ ⎛ ₊ = + = ω ω ω ω ω

The maximum and the minimum values of x can found by substitution of (3.8.2) _a and (3.8.3) into (3.8.1), so we have:

when ω =ω1: r F m r r F xa 0 0 2 0 0 0 2 2ω −ω ≈ ω = which is the maximum value of x , and a

when ω =ω2: r F m r r F xa 0 0 2 0 0 0 2 2ω +ω ≈− ω − = which is the minimum value of x . a

3.9

The undamped oscillatory equation for a bound electron is given by: t m eE x x ω ( 0 )cosω 2 0 = − + && (3.9.1) Try solution x= Acosωt in equation (3.9.1), we have:

t m eE t A ω ω ω ω ) cos ( )cos ( 0 2 0 2+ = − −

which is true for all t provided:

m eE A ₀ 2 0 2 ) (−ω +ω =− i.e. ) ( ₀2 2 0 ω ω − − = m eE A

So, we find a solution to equation (3.9.1) given by: t m eE x ω ω ω )cos ( 02 2 0 − − = (3.9.2) For an electron number density n, the induced polarizability per unit volume of a

medium is given by: E nex e 0 ε χ =− (3.9.3) By substitution of (3.9.2) and E =E₀cosωt into (3.9.3), we have

) ( 02 2 0 2 0 ε ω ω ε χ − = − = m ne E nex e 3.10

The forced mechanical oscillator equation is given by: t F sx x r x

m&&+ &+ = ₀cosω which can be written as:

t F x m x r x

m&&+ &+ ω₀2 = ₀cosω (3.10.1) where, ω0 = s m. Its solution can be written as:

t Z X F t Z r F x m m m ω ω ω ω sin 2 cos 0 2 0 − = (3.10.2) where, X_m =ωm−s ω , Z_m = r2+(ωm−s ω)2, ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = − r s m ω ω φ 1 tan

By taking the displacement x as the component represented by curve (a) in Figure 3.9, i.e. by taking the second term of equation (3.10.2) as the expression of x, we have: t r m m F t Z X F x m m ω ω ω ω ω ω ω ω ( ) cos ) ( cos _{2 2 2 2 2} 0 2 2 2 0 0 2 0 + − − = − = (3.10.3)

The damped oscillatory electron equation can be written as:

mx&&+ &rx+mω₀2x=−eE₀cosωt (3.10.4) Comparing (3.10.1) with (3.10.4), we can immediately find the displacement x for a damped oscillatory electron by substituting F0 =−eE0 into (3.10.3), i.e.:

t r m m eE x ω ω ω ω ω ω cos ) ( ) ( 2 2 2 2 2 0 2 2 2 0 0 + − − − = (3.10.5) By substitution of (3.10.5) into (3.9.3), we can find the expression of χ for a damped oscillatory electron is given by:

] ) ( [ ) ( 2 2 2 2 2 0 2 0 2 2 0 2 0 m r m ne E nex ω ω ω ε ω ω ε χ + − − = − = So we have: t r m m ne r ε ω ω ω ω ω ω χ ε cos ] ) ( [ ) ( 1 1 _{2 2 2 2 2} 0 2 0 2 2 0 2 + − − + = + = 3.11

The instantaneous power dissipated is equal to the product of frictional force and the instantaneous velocity, i.e.:

) ( cos ) ( ₂ 2 2 0 ω −φ = = t Z F r x x r P m & &

The period for a given frequency ω is given by:

ω π

2 = T

Therefore, the energy dissipated per cycle is given by:

2 2 0 2 2 0 2 0 2 2 0 0 2 0 2 2 2 0 2 2 )] ( 2 cos 1 [ 2 ) ( cos m m m T m Z rF Z rF dt t Z rF dt t Z F r Pdt E ω π ω π φ ω φ ω ω π ω π = = − − = − = =

∫

∫

∫

The displacement is given by:

) sin( 0 ω φ ω − = t Z F x m so we have: m Z F x ω 0 max = (3.11.2) By substitution of (3.11.2) into (3.11.1), we have:

2 max x r E =π ω 3.12

The low frequency limit of the bandwidth of the resonance absorption curve ω1 satisfies the equation:

r s

m− ₁=−

1 ω

ω

which defines the phase angle given by:

1 tan 1 1 1 =− − = r s m ω ω φ

The high frequency limit of the bandwidth of the resonance absorption curve ω₂ satisfies the equation:

r s

m− 2 =

2 ω

ω

which defines the phase angle given by:

1 tan 2 2 2 = − = r s m ω ω φ 3.13

For a LCR series circuit, the current through the circuit is given by

t i e I I = 0 ω The voltage across the inductance is given by:

LI i e LI i e I dt d L dt dI L = ₀ iωt = ω ₀ iωt = ω

i.e. the amplitude of voltage across the inductance is: 0 LI

V_L =ω (3.13.1) The voltage across the condenser is given by:

C iI e I C i dt e C Idt C C q i t i t ω ω ω ω _{= =−} = =

∫

∫

i.e. the amplitude of the voltage across the condenser is: C

I V_C

ω0

= (3.13.2) When an alternating voltage, amplitude V is applied across LCR series circuit, ₀ current amplitude I is given by: 0 I0 =V0 Ze, where the impedance Z is given e

by: 2 2 1 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ₋ + = C L R Zm ω _ω

At current resonance, I has the maximum value: ₀

R V

I 0

0 = (3.13.3) and the resonant frequency ω0 is given by:

0 1 0 0 − = C L ω ω or LC 1 0 = ω (3.13.4) By substitution of (3.12.3) and (3.12.4) into (3.12.1), we have:

0 0 V R L V_L =ω

By substitution of (3.12.3) and (3.12.4) into (3.12.2), we have:

0 0 0 0 0 0 0 V R L R V LC L R V C L V RC LC RC V V_C ω ω = = = = =

Noting that the quality factor of an LCR series circuit is given by: R L Q=ω0 so we have: 0 QV V V_L = _C = 3.14

In a resonant LCR series circuit, the potential across the condenser is given by: C I V_C ω = (3.14.1) where, I is the current through the whole LCR series circuit, and is given by: I =I₀eiωt (3.14.2) The current amplitude I is given by: ₀

e Z V I 0 0 = (3.14.3)

where, V is the voltage amplitude applied across the whole LCR series circuit and is ₀ a constant. Z is the impedance of the whole circuit, given by: _e

2 2 1 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ₋ + = C L R Ze ω _ω (3.14.4)

From (3.14.1), (3.14.2), (3.14.3), and (3.14.4) we have:

t i C t i C e V e C L R C V V ω ω ω ω ω 0 2 2 0 1 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ₋ + =

which has the maximum value when 0 =0

ω

d dVC

0 1 2 2 0 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ₋ + C L R C V d d ω ω ω ω i.e. 1 2 2 ₂1 ₂ 0 2 2 + − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ₋ + C L C L R ω ω ω ω i.e. 2+2 2 2−2 =0 C L L R ω i.e. 2 0 02 2 2 1 1 2 1 Q L R LC − = − = ω ω where LC 1 2 0 = ω , R L Q 0 0 ω = 3.15

In a resonant LCR series circuit, the potential across the inductance is given by: V_L =ωLI (3.15.1) where, I is the current through the whole LCR series circuit, and is given by: I =I₀eiωt (3.15.2) The current amplitude I is given by: ₀

e Z V I 0 0 = (3.15.3)

where, V is the voltage amplitude applied across the whole LCR series circuit and is ₀ a constant. Z is the impedance of the whole circuit, given by: _e

2 2 1 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ₋ + = C L R Ze ω _ω (3.15.4)

From (3.15.1), (3.15.2), (3.15.3), and (3.15.4) we have:

t i L t i L e V e C L R LV V ω ω ω ω ω 0 2 2 0 1 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ₋ + =

which has the maximum value when 0 =0

ω

d dVL

0 1 2 2 0 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ₋ + C L R LV d d ω ω ω ω i.e. 1 2 2 ₂1 ₂ 0 2 2 − + = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ₋ + C L C L R ω ω ω ω i.e. 2+ ₂2 ₂ −2 =0 C L C R ω i.e. 2 0 0 2 0 2 2 0 2 2 2 2 1 1 2 1 1 2 1 1 1 2 1 Q L R L C R LC C R LC − = − = − = − = ω ω ω ω where LC 1 2 0 = ω , R L Q 0 0 ω = 3.16

Considering an electron in an atom as a lightly damped simple harmonic oscillator, we know its resonance absorption bandwidth is given by:

m r =

δω (3.16.1) On the other hand, the relation between frequency and wavelength of light is given by:

λ

c

f = (3.16.2) where, c is speed of light in vacuum. From (3.16.2) we find at frequency resonance:

δλ λ δ ₂ 0 c f =−

where λ₀ is the wavelength at frequency resonance. Then, the relation between angular frequency bandwidth δω and the width of spectral line δλ is given by: δλ λ π δ π δω ₂ 0 2 2 f = c = (3.16.3) From (3.16.1) and (3.16.3) we have:

Q m r cm r 0 0 0 2 0 2 λ ω λ π λ δλ = = =

So the width of the spectral line from such an atom is given by:

] [ 10 2 . 1 10 5 10 6 . 0 14 7 6 0 m Q − − × = × × = = λ δλ

3.17

According to problem 3.6, the displacement resonance frequency ωr and the

corresponding displacement amplitude xmax are given by:

2 2 2 0 2m r r = ω − ω r F Z F x r m ω ω _{ω ω} = ′ = = 0 0 max where, Zm = r2+(ωm−s ω)2 , 2 2 2 0 4m r − = ′ ω ω , m s = 0 ω

Now, at half maximum displacement:

r F x Z F m ω ω = 2 = 2 ′ 0 max 0 i.e. ω r2+(ωm−s ω)2 =2ω′r i.e. ₂ 2 2 2 2 2 4 4 r m r m s s m r _⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ₋ + ω ω ω i.e. ( 2 ) 4 ₄ 0 4 3 2 2 2 2 2 2 4+ − + − + = m r m sr m s m sm r _ω ω i.e. 0 4 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 4 = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − + ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − m r m s m r m r m s m r m s _ω ω i.e. ( ) 3 0 2 2 2 2 2 = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ′ − −ω ω ω m r r i.e. ω2−ω2 =± 3 ω′2 m r r (3.17.1)

If ω1 and ω2 are the two solutions of equation (3.17.1), and ω2 >ω1, then:

2 2 2 2 3 _ω ω ω − = ′ m r r (3.17.2) 12 2 2 3 _ω ω ω − =− ′ m r r (3.17.3)

Since the Q-value is high, we have: 1 0 >> = r m Q ω i.e. ₂ 2 2 0 m r >> ω i.e. ωr ≈ω′≈ω0 Then, from (3.17.2) and (3.17.3) we have:

2 1 2 1 02 3 2 ) )( (ω ω ω ω ω m r ≈ + − and ω₁+ω₂ ≈2ω₀

Therefore, the width of displacement resonance curve is given by:

m r 3 1 2 −ω ≈ ω 3.18

In Figure 3.9, curve (b) corresponds to absorption, and is given by: t r m r F t Z r F x m ω ω ω ω ω ω ω sin 2( ₀2 2)2 2 2sin 0 2 0 + − = =

and the velocity component corresponding to absorption is given by: t r m r F x v ω ω ω ω ω cos ) ( ₀2 2 2 2 2 2 2 0 + − = = &

For Problem 3.10, the velocity component corresponding to absorption can be given by substituting F₀ =−eE₀ into the above equation, i.e.:

t r m r eE x v ω ω ω ω ω cos ) ( ₀2 2 2 2 2 2 2 0 + − − = = & (3.18.1) For an electron density of n, the instantaneous power supplied equal to the product of the instantaneous driving force −neE₀cosωt and the instantaneous velocity, i.e.:

t r m r E ne t r m r eE t neE P ω ω ω ω ω ω ω ω ω ω ω 2 2 2 2 2 2 0 2 2 2 0 2 2 2 2 2 2 0 2 2 0 0 cos ) ( cos ) ( ) cos ( + − = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + − − × − =

The average power supplied per unit volume is then given by: 2 2 2 2 2 0 2 2 2 0 2 2 0 2 2 2 2 2 2 0 2 2 2 0 2 2 0 ) ( 2 cos ) ( 2 2 r m r E ne t r m r E ne Pdt P_av ω ω ω ω ω ω ω ω ω π ω π ω ω π ω π + − = + − = =

∫

∫

SOLUTIONS TO CHAPTER 4

4.1

The kinetic energy of the system is the sum of the separate kinetic energy of the two masses, i.e.: 2 2 2 2 2 2 4 1 4 1 ) ( 2 1 ) ( 2 1 2 1 2 1 2 1 Y m X m y x y x m y m x m

E_k & & & & & & = & + &

⎥⎦ ⎤ ⎢⎣ ⎡ _{+ + −} = + =

The potential energy of the system is the sum of the separate potential energy of the two masses, i.e.:

2 2 2 2 2 2 2 2 2 2 2 2 2 1 4 1 ) ( ) ( 2 1 ) ( 2 1 2 1 ) ( ) ( 2 1 ) ( 2 1 2 1 ) ( 2 1 2 1 Y s l mg X l mg y x s y x y x l mg y x s y x l mg y x s y l mg x y s x l mg E_p ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ₊ + = − + ⎥⎦ ⎤ ⎢⎣ ⎡ _{+ + −} = − + + = − + + − + =

Comparing the expression of E and _k E_p with the definition of E and _X E given _Y by (4.3a) and (4.3b), we have:

m a 2 1 = , l mg b 4 = , c m 4 1 = , and s l mg d = + 2 Noting that: X m y x m X_q 2 1 2 1 2 ) ( 2 ⎟⎠ ⎞ ⎜ ⎝ ⎛ = + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = and Y c m y x m Yq 2 1 2 1 ) ( 2 ⎟⎠ ⎞ ⎜ ⎝ ⎛ = − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = i.e. Xq m X 2 1 2 − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = and Yq m Y 2 1 2 − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =

we have the kinetic energy of the system given by:

2 2 2 2 2 2 2 1 2 1 2 4 1 2 4 1 q q q q k Y X Y m m X m m Y c X a

E & & & & ⎥ = & + & ⎦ ⎤ ⎢ ⎣ ⎡ + ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ = + = and 2 2 2 2 2 2 2 2 1 2 2 2 4 q q q q p Y m s l g X l g Y m s l mg X m l mg dY bX E ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + + = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ₊ + ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ = + =

which are the expressions given by (4.4a) and (4.4b)

4.2

The total energy of Problem 4.1 can be written as:

2 2 2 2 2 ) ( ) ( 2 1 2 1 2 1 y x s y x l mg y m x m E E E = _k+ _p = & + & + + + −

The above equation can be rearranged as the format:

xy pot y pot kin x pot kin E E E E E E =( + ) +( + ) +( ) where, 2 2 2 2 1 ) ( s x l mg x m E E_{kin pot x} ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ₊ + = + & , 2 2 2 2 1 ) ( s y l mg y m E E_{kin pot y} ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ₊ + = + & ,

and Epot =−2sxy

4.3 a x=−2 , y=0: 0 = x , y=−2a: 4.4

For mass m , Newton’s second law gives: ₁ sx x m₁&&₁ = For mass m , Newton’s second law gives: ₂

sx x m₂ &&₂ =− ≡ x=0 -2a -a -a + -a a Y --X y=0 ≡ -2a -a -a + -a a Y + -X

Provided x is the extension of the spring and l is the natural length of the spring, we have: x l x x₂− ₁ = + By elimination of x and ₁ x , we have: ₂

x x m s x m s && = − − 1 2 i.e. 0 2 1 2 1+ = + sx m m m m x_&&

which shows the system oscillate at a frequency:

μ ω2 = s where, 2 1 2 1 m m m m + = μ

For a sodium chloride molecule the interatomic force constant s is given by:

] [ 120 10 67 . 1 ) 35 23 ( ) 10 67 . 1 ( ) 35 23 ( ) 10 14 . 1 ( 4 ) 2 ( 1 27 2 27 2 13 2 2 2 − − − ≈ × × + × × × × × × = + = = Nm m m m m s Cl Na Cl Na π πν μ ω 4.5

If the upper mass oscillate with a displacement of x and the lower mass oscillate with a displacement of y, the equations of motion of the two masses are given by Newton’s second law as:

) ( ) ( y x s y m sx x y s x m − = − − = && && i.e. 0 ) ( 0 ) ( = − − = − − + y x s y m sx y x s x m && &&

Suppose the system starts from rest and oscillates in only one of its normal modes of frequency ω, we may assume the solutions:

t i t i Be y Ae x ω ω = =

Using these solutions, the equations of motion become: 0 )] ( [ 0 ] ) ( [ 2 2 = − − − = + − + − t i t i e B A s B m e sA B A s A m ω ω ω ω

We may, by dividing through by meiωt, rewrite the above equations in matrix form as: 0 2 2 2 = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − − − − B A m s m s m s m s ω ω (4.5.1) which has a non-zero solution if and only if the determinant of the matrix vanishes; that is, if 0 ) )( 2 ( s m−ω2 s m−ω2 −s2 m2 = i.e. ω4−(3s m)ω2 +s2 m2 =0 i.e. m s 2 ) 5 3 ( 2 = ± ω

In the slower mode, ω2 =(3− 5)s 2m. By substitution of the value of frequency into equation (4.5.1), we have:

2 1 5 2 2 2 − = − = − = s m s m s s B A ω ω

which is the ratio of the amplitude of the upper mass to that of the lower mass.

Similarly, in the slower mode, ω2 =(3+ 5)s 2m. By substitution of the value of frequency into equation (4.5.1), we have:

2 1 5 2 2 2 + − = − = − = s m s m s s B A ω ω 4.6

The motions of the two pendulums in Figure 4.3 are given by:

t t a t t a y t t a t t a x a m a m ω ω ω ω ω ω ω ω ω ω ω ω sin sin 2 2 ) ( sin 2 ) ( sin 2 cos cos 2 2 ) ( cos 2 ) ( cos 2 2 1 1 2 2 1 1 2 = + − = = + − =

where, the amplitude of the two masses, 2acosω_mt and 2asinω_mt, are constants over one cycle at the frequency ω_a.