Vibration and Waves solution manual PDF

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this is a solution manual to vibrations and waves by AP French...

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Solutions Manual for The Physics of Vibrations and Waves – 6th 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

© 2008 John Wiley & Sons, Ltd

SOLUTIONS TO CHAPTER 1

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

F = −mg sin θ By substitution of relation sin θ = x l into the above equation, we have:

F = −mg x l so the stiffness is given by:

s = − F x = mg l so we have the frequency given by:

ω2 = s m =g l Since

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

given by:

F = − mg θ Now, the equation of motion using angular displacement θ can by derived from Newton’s second law:

F = m&x& i.e.

− mgθ = mlθ&&

i.e.

θ&& + θ = 0

g l

which shows the frequency is given by:

ω2 = g l In Figure 1.1(b), restoring couple is given by − Cθ , which has relation to moment of inertia I given by:

− C θ = I θ&& i.e.

C I

θ&& + θ = 0

which shows the frequency is given by:

ω2 =C I

© 2008 John Wiley & Sons, Ltd

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

F = − 2T x l so Newton’s second law gives:

F = m&x& = − 2Tx l x&& + 2Tx lm = 0

i.e. which shows the frequency is given by:

ω2 =

2T lm

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

s=

G 2ρ Axg = = 2ρ Ag x2 x

Newton’s second law gives:

− G = m&x& − 2 ρ Axg= ρ Alx&&

i.e.

&x& +

i.e.

2g x= 0 l

which show the frequency is given by:

ω 2 = 2g l

In Figure 1.1(f), by taking logarithms of equation pV

γ

= constant, we have:

ln p + γ ln V = constant so we have:

i.e.

dp dV +γ =0 V p dp = −γ p

dV V

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

dp = − γp

Ax V

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

Adp = m&x& © 2008 John Wiley & Sons, Ltd

− γp

i.e.

A2 x = ρ Al&&x V

γpA x =0 x&& + l ρV

i.e. which show the frequency is given by:

ω2 =

γpA l ρV

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

F = −ρgAx = m&x& g ρA x=0 x&& + m

i.e. which shows the frequency is given by:

ω 2 = gρ A m 1.2

Write solution x = a cos(ωt + φ ) in form: x = a cos φ cos ωt − a sin φ sin ωt and compare with equation (1.2) we find: A = a cos φ and B = −a sin φ . We can also find, with the same analysis, that the values of A and B for solution x = a sin(ωt − φ ) are given by: A = − a sin φ and B = a cos φ , and for solution x = a cos(ωt − φ ) are given by: A = a cosφ and B = a sin φ . Try solution x = a cos(ωt + φ ) in expression x&& + ω 2 x , we have: &x&+ ω2 x = − a ω2 cos( ωt + φ) + ω2 a cos( ωt + φ) = 0

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

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

© 2008 John Wiley & Sons, Ltd

1.3

(a) If the solution x = a sin(ω t + φ ) satisfies x = a at t = 0 , then, x = asin φ = 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 ωt is the minimum solution of equation a sin(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 = a cos(ωt + φ ) satisfies x = a at t = 0 , then, x = a cos φ = a i.e. 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 a cos ωt = + a

2 , i.e. ω t = π 4 . Similarly, we can find: for x = a 2 , ωt = π 3

and for x = 0 , ωt = π 2 . If

the

solution

x = a sin(ω t − φ )

satisfies

x= a

at

t =0 ,

then,

x = a sin( −φ ) = 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 ωt is the minimum

solution of equation a sin(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 = a cos(ω t − φ )

satisfies

x= a

at

t=0 ,

then,

x = a cos( −φ ) = a i.e. 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 a cosωt = + 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 .

(b) If

the

solution

x = a sin(ωt + φ)

satisfies

x = −a

at

t =0 ,

then,

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

© 2008 John Wiley & Sons, Ltd

x= +a

2 for the first time after release, the value of ωt is the minimum

solution of equation a sin(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 = a cos(ωt + φ )

satisfies

x = −a

at

t=0 ,

then,

x = a cos φ = − a i.e. φ = π . 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 a cos(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 = a sin(ωt − φ )

satisfies

x = −a

at

t =0 ,

then,

x = a sin( −φ ) = − 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 ωt is the minimum

solution of equation a sin(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 = a cos(ωt − φ )

satisfies

x = −a

at

t=0 ,

then,

x = a cos( −φ ) = − a i.e. φ = π . 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 a cos(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:

ω0 =

s e2 (1.6 ×10 −19 )2 16 −1 = = −12 −9 3 −31 ≈ 4 . 5 × 10 [ rad ⋅ s ] 3 me 4πε 0r me 4× π × 8.85× 10 × (0.05× 10 ) × 9.1× 10

Its radiation generates an electromagnetic wave with a wavelength λ given by:

© 2008 John Wiley & Sons, Ltd

λ=

2 πc

ω0

=

2 ×π ×3 ×10 8 8 ≈ 4.2 ×10− [ m] = 42[nm] 16 4.5× 10

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: s′ = − F x = s 2 . Hence the frequency is given by ωa2 = s′ m = s 2m .

(b) The frequency of the system is given by: ω 2b = s m (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

ω 2c = s′ m = 2s m . Therefore, we have the relation: ω2a : ωb2 : ω2c = s 2m : s m : 2s m = 1 : 2 : 4 1.6

At time t = 0 , x = x0 gives: a sinφ = x0

(1.6.1)

x& = v 0 gives: aω cosφ = v0

(1.6.2)

From (1.6.1) and (1.6.2), we have tan φ = ωx 0 v 0 and a = ( x20 + v02 ω2 )1 2 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 vt , where v t is the velocity of the particle at time t and is given by: vt = x& = aωcos( ωt + φ) .

© 2008 John Wiley & Sons, Ltd

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: η =

η=

2π 2 dt where the period is given by: T = , so we have: T ω

dx dx dx 2dt 2ω dx = = = = 2 π ω ω φ π ω φ T a cos( t + ) πa 1 − sin (ωt + φ ) π a 2 − x 2 2 a cos( t + )

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 = A ω 2 sin( ωt + φ) , where A is the amplitude, ω is the angular frequency and φ is the initial phase. So we have: Aω2 sin(ωt + φ ) = g A=

i.e.

g ω sin(ωt + φ ) 2

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

g

ω

2

=

g

4π f 2

2

=

9.8 ≈ 0.01[ m] 4 ×π 2 × 5 2

1.10

The mass of the element dy is given by: m ′ = mdy l . The velocity of an element dy of 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 © 2008 John Wiley & Sons, Ltd

of this element given by: 2

KEdy

1 1 ⎛ m ⎞⎛ y ⎞ = m ′v ′ 2 = ⎜ dy ⎟⎜ v⎟ 2 2⎝ l ⎠⎝ l ⎠

The total kinetic energy of the spring is given by: 2

mv2 1 ⎛m ⎞⎛ y ⎞ = dy v ⎜ ⎟⎜ ⎟ 0 2 2l 3 ⎝l ⎠⎝ l ⎠

l

l

KEspring = ∫ KEdy dy = ∫ 0

l

1

∫ y dy = 6 mv 2

2

0

The total kinetic energy of the system is the sum of kinetic energies of the spring and the suspended mass, and is given by: 1 1 1 KEtot = mv2 + Mv2 = (M + m 3 )v2 6 2 2 which shows the system is equivalent to a spring with zero mass with a mass of M + m 3 suspended at the end. Therefore, the frequency of the oscillation system is

given by:

ω2 =

s M+m3

1.11

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

1 2 1 2 1 2 1 mg 2 x mv + sx = mx& + 2 2 2 2 l

The equation of motion is by setting dE dt = 0 , i.e.: d ⎛ 1 2 1 mg 2 ⎞ x ⎟=0 ⎜ mx& + dt ⎝ 2 2 l ⎠ i.e.

x&& +

g x=0 l

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: 1 1 E = Iθ& 2 + Cθ 2 2 2 The equation of motion is by setting dE dt = 0 , i.e.: d ⎛ 1 &2 1 2 ⎞ ⎜ Iθ + Cθ ⎟ = 0 2 dt ⎝ 2 ⎠ © 2008 John Wiley & Sons, Ltd

C I

θ&& + θ = 0

i.e.

In Figure 1.1(c), the energy is directly given by: 1 1 E = mv 2 + sx 2 2 2 The equation of motion is by setting dE dt = 0 , i.e.: d ⎛1 2 1 2 ⎞ ⎜ mx& + sx ⎟ = 0 dt ⎝ 2 2 ⎠ x&& +

i.e.

s x =0 m

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: E=

T 1 2 1 2 1 2 1 2T 2 1 mv + sx = mx& + x = mx& 2 + x 2 l 2 2 2 2 l 2

The equation of motion is by setting dE dt = 0 , i.e.: d ⎛ 1 2 T 2⎞ ⎜ mx& + x ⎟ = 0 dt ⎝ 2 l ⎠ x&&+

i.e.

2T x=0 lm

In Figure 1.1(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 ρgAl l = 2ρgA . So the energy is given by:

1 1 1 1 1 E = mv 2 + sx 2 = ρAlx&2 + 2 ρgAx 2 = ρAlx&2 + ρgAx 2 2 2 2 2 2 The equation of motion is by setting dE dt = 0 , i.e.: d ⎛1 2 2 ⎞ ⎜ ρAlx& + ρ gAx ⎟ = 0 dt ⎝ 2 ⎠ i.e.

© 2008 John Wiley & Sons, Ltd

x&&+

2g x=0 l

In Figure 1.1(f), the gas of a mass of ρ Al 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: E=

1 2 1 2 1 1 γpA2 x2 mv + sx = ρ Alx&2 + 2 2 2 2 V

The equation of motion is by setting dE dt = 0 , i.e.: d⎛1 1 γpA 2x 2 ⎜⎜ ρAl x& 2 + dt ⎝ 2 2 V

x&& +

i.e.

⎞ ⎟⎟ = 0 ⎠

γpA x=0 lρ V

In Figure 1.1(g), the restoring force of the hydrometer is − ρgAx , then the stiffness of the system is given by s = ρ gAx x = ρgA . So the energy is given by: E=

1 2 1 2 1 2 1 mv + sx = mx& + ρ gAx2 2 2 2 2

The equation of motion is by setting dE dt = 0 , i.e.: d ⎛1 2 1 2 ⎞ ⎜ mx& + ρ gAx ⎟ = 0 dt ⎝ 2 2 ⎠ i.e.

x&& +

Aρ g x=0 m

1.12 The displacement of the simple harmonic oscillator is given by: x = a sin ωt so the velocity is given by: x& = aω cos ωt From (1.12.1) and (1.12.2), we can eliminate t and get:

x2 x& 2 + = sin 2 ωt + cos 2 ωt = 1 a 2 a2ω 2 which is an ellipse equation of points ( x, x& ) . The energy of the simple harmonic oscillator is given by: © 2008 John Wiley & Sons, Ltd

(1.12.1) (1.12.2)

(1.12.3)

E=

1 2 1 2 mx& + sx 2 2

(1.12.4)

Write (1.12.3) in form x& 2 = ω2 (a 2 − x 2 ) and substitute into (1.12.4), then we have: E=

1 1 2 1 2 1 mx& + sx = mω 2 ( a2 − x2 ) + sx2 2 2 2 2

Noting that the frequency ω is given by: ω 2 = s m , we have: E=

1 1 1 s( a2 − x 2 ) + sx2 = sa 2 2 2 2

which is a constant value. 1.13 The equations of the two simple harmonic oscillations can be written as: y1 = a sin(ωt + φ )

and

y 2 = a sin( ωt + φ + δ)

The resulting superposition amplitude is given by: R = y 1 + y 2 = a [sin(ωt + φ ) + sin(ωt + φ + δ )] = 2a sin(ωt + φ + δ 2) cos(δ 2)

and the intensity is given by: I = R2 = 4 a2 cos 2 (δ 2) sin 2 (ω t +φ + δ 2) I ∝ 4a 2 cos 2 (δ 2)

i.e.

Noting that sin 2 (ωt + φ + δ 2) varies between 0 and 1, we have: 0 ≤ I ≤ 4a 2 cos 2 (δ 2) 1.14

⎛ x ⎜⎜ sin φ2 − ⎝ a1 x2 = 2 sin2 φ2 + a1

2

2

⎞ ⎞ ⎛ y y x sin φ1 ⎟⎟ + ⎜⎜ cosφ1 − cosφ 2 ⎟⎟ a2 a1 ⎠ ⎠ ⎝ a2 2 2 y y x2 2 xy 2xy 2 2 2 − + + φ φ φ φ sin 1 sin 1 sin 2 cos 1 cos φ2 − cos φ1 cos φ2 2 2 2 a2 a1a2 a2 a1 a1 a2

2

2

=

x y 2xy (sin 2 φ2 + cos 2 φ2 ) + 2 (sin 2 φ1 + cos 2 φ1) − (sin φ1 sin φ 2 + cosφ1 cosφ 2 ) 2 a1 a2 a1a2

=

x 2 y 2 2 xy cos(φ 1 − φ 2 ) + 2− 2 a1 a2 a 1a 2

On the other hand, by substitution of : x = sin ωt cos φ1 + cos ωt sin φ1 a1

© 2008 John Wiley & Sons, Ltd

y = sin ω t cosφ2 + cosω t sinφ2 a2 2

2

⎞ ⎞ ⎛y ⎛x y x into expression ⎜⎜ sin φ2 − sin φ1 ⎟⎟ + ⎜⎜ cosφ1 − cosφ 2 ⎟⎟ , we have: a2 a1 ⎠ ⎠ ⎝ a2 ⎝ a1 2

2

⎞ ⎞ ⎛ y ⎛x y x ⎜⎜ sin φ2 − sin φ1 ⎟⎟ + ⎜⎜ cos φ1 − cos φ 2 ⎟⎟ a2 a1 ⎠ ⎠ ⎝ a2 ⎝ a1 2 2 2 = sin ωt (sin φ2 cos φ1 − sin φ1 cos φ2 ) + cos ωt (cos φ1 sin φ2 − cos φ2 sin φ1 ) 2 = (sin 2 ω t + cos2 ω t ) sin 2 (φ 2 −φ1 ) 2 = sin (φ2 − φ1 )

From the above derivation, we have: x 2 y 2 2 xy cos(φ1 − φ 2 ) = sin 2 (φ2 − φ1 ) + − a12 a22 a1a2 1.15

By elimination of t from equation x = a sin ωt and y = b cos ωt , we have: x2 y 2 + =1 a2 b2 which shows the particle follows an elliptical path. The energy at any position of x , y on the ellipse is given by: 1 1 1 1 E = mx& 2 + sx 2 + my& 2 + sy 2 2 2 2 2 1 2 2 1 2 2 2 1 2 2 2 1 2 2 2 2 = ma ω cos ωt + ma ω sin ωt + mb ω sin ωt + mb ω cos ωt 2 2 2 2 1 1 2 2 2 2 = ma ω + mb ω 2 2 1 = m ω2 (a2 + b2 ) 2 The value of the energy shows it is a constant and equal to the sum of the separate 1 mω 2 a2 and in energies of the simple harmonic vibrations in x direction given by 2 1 y direction given by m ω2 b2 . 2 At any position of x , y on the ellipse, the expression of m ( xy& − yx& ) can be written as: © 2008 John Wiley & Sons, Ltd

m( x&y − y&x) = m( − abω sin 2 ω t − abω cos 2 ω t) = − abmω (sin 2 ω t + cos2 ω t) = − abmω which is a constant. The quantity abm ω 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 2a2 high, where a1 = xmax and a 2 = ymax , see Figure 1.8.

When x = 0 in equation (1.3) the positive value of y = a2 sin(φ2 − φ1 ) . The value of ymax = a2 . So y x =0 ymax = sin( φ2 − φ1 ) which defines φ2 − φ1 . 1.17

In the range 0 ≤ φ ≤ π , the values of cosφi are −1 ≤ cos φi ≤ +1 . For n random values of φi , statistically there will be n 2 values −1 ≤ cos φi ≤ 0 and n 2 values 0 ≤ cos φi ≤ 1 . The positive and negative values will tend to cancel each other and the n

sum of the n values:

∑ cosφ

i

n

∑cos φ

→ 0 , similarly

j

→ 0 . i.e.

j=1

i=1 i≠ j

n

n

∑ cosφ ∑ cosφ i

i= 1 i≠ j

j=1

j

→0

1.18 The exponential form of the expression: a sin ωt + a sin( ωt + δ ) + a sin(ω t + 2δ ) + L + a sin[ ωt + (n −1) δ ]

is given by: aeiω t + aei(ω t+δ ) + aei (ω t+ 2δ ) + L+ aei[ω t +( n−1)δ ]

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

⎡ ⎛ n−1 ⎞ ⎤ i ⎢ω t+⎜ ⎟δ ⎥ ⎣ ⎝ 2 ⎠ ⎦

sin nδ 2 sin δ 2

with the imaginary part: ⎡ ⎛ n − 1⎞ δ ⎤ sin nδ 2 a sin ⎢ ωt + ⎜ ⎟ ⎥ ⎝ 2 ⎠ ⎦ sin δ 2 ⎣ which is the value of the original expression in sine term. © 2008 John Wiley & Sons, Ltd

1.19 From...