PHYSICS
PHYSICS
5th Edition
ISBN: 2818440038631
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 11, Problem 25P

(a)

To determine

The amplitude of the transverse wave on a string described by y(x,t)=(0.35mm){(π3.0radm)[x(66ms)]t}.

(a)

Expert Solution
Check Mark

Answer to Problem 25P

The amplitude of the transverse wave on the string is 0.35mm.

Explanation of Solution

Write the standard expression of a transverse wave travelling along x direction.

y(x,t)=Asin(kxωt) (I)

Here, A is the amplitude of wave, k is the wave vector, ω is the angular velocity, x is position of wave at any time t and y(x,t) is the displacement of wave at position x and time t.

Write the equation of wave given in question.

y(x,t)=(0.35mm){(π3.0radm)[x(66ms)]t} (II)

Conclusion:

Compare equation (II) with (I) to get A.

A=0.35mm

Therefore, the amplitude of the transverse wave on the string is 0.35mm.

(b)

To determine

The wavelength of the transverse wave on a string described by y(x,t)=(0.35mm){(π3.0radm)[x(66ms)]t}.

(b)

Expert Solution
Check Mark

Answer to Problem 25P

The wavelength of the transverse wave on a string described by y(x,t)=(0.35mm){(π3.0radm)[x(66ms)]t} is 6.0m.

Explanation of Solution

Write the standard expression of a transverse wave travelling along x direction.

y(x,t)=Asin(kxωt) (I)

Here, A is the amplitude of wave, k is the wave vector, ω is the angular velocity, x is position of wave at any time t and y(x,t) is the displacement of wave at position x and time t.

Write the equation of wave given in question.

y(x,t)=(0.35mm){(π3.0radm)[x(66ms)]t} (II)

Expand above equation to compare with standard form given in equation (I).

y(x,t)=(0.35mm)[(π3.0radm)x(π3.0radm)(66ms)t] (III)

Write the expression for wavelength of wave.

λ=2πk (IV)

Conclusion:

Compare equation (III) with (I) to get k.

k=π3.0radm

Substitute π3.0radm for k in equation (IV) to get λ.

λ=2ππ3.0radm=6.0m

Therefore, the wavelength of the transverse wave on a string is 6.0m.

(c)

To determine

The frequency of the transverse wave on a string described by y(x,t)=(0.35mm){(π3.0radm)[x(66ms)]t}.

(c)

Expert Solution
Check Mark

Answer to Problem 25P

The frequency of the transverse wave on a string described by y(x,t)=(0.35mm){(π3.0radm)[x(66ms)]t} is 11Hz.

Explanation of Solution

Write the standard expression of a transverse wave travelling along x direction.

y(x,t)=Asin(kxωt) (I)

Here, A is the amplitude of wave, k is the wave vector, ω is the angular velocity, x is position of wave at any time t and y(x,t) is the displacement of wave at position x and time t.

Write the equation of wave given in question.

y(x,t)=(0.35mm){(π3.0radm)[x(66ms)]t} (II)

Expand above equation to compare with standard form given in equation (I).

y(x,t)=(0.35mm)[(π3.0radm)x(π3.0radm)(66ms)t] (III)

Write the expression for frequency of wave.

f=ω2π (V)

Conclusion:

Compare equation (III) with (I) to get ω.

ω=((π3.0radm)66ms)

Substitute ((π3.0radm)66ms) for ω in equation (V) to get f.

f=π3.0radm×66ms2π=11Hz

Therefore, the frequency of the transverse wave on a string described by y(x,t)=(0.35mm){(π3.0radm)[x(66ms)]t} is 11Hz.

(d)

To determine

The direction of the transverse wave on a string described by y(x,t)=(0.35mm){(π3.0radm)[x(66ms)]t}.

(d)

Expert Solution
Check Mark

Answer to Problem 25P

The transverse wave on a string described by y(x,t)=(0.35mm){(π3.0radm)[x(66ms)]t} travels in the +x direction.

Explanation of Solution

Write the standard expression of a transverse wave travelling along x direction.

y(x,t)=Asin(kxωt) (I)

Here, A is the amplitude of wave, k is the wave vector, ω is the angular velocity, x is position of wave at any time t and y(x,t) is the displacement of wave at position x and time t.

Write the equation of wave given in question.

y(x,t)=(0.35mm){(π3.0radm)[x(66ms)]t} (II)

Equation (I) represents a transverse traveling along +x direction. This is possible if sign of term ωt is negative and sign of kx is positive.

Conclusion:

The wave in equation (II) represents same type of wave that travels in +x direction.

Therefore, the transverse wave on a string described by y(x,t)=(0.35mm){(π3.0radm)[x(66ms)]t} travels in the +x direction.

(e)

To determine

The maximum transverse speed of a point on the string.

(e)

Expert Solution
Check Mark

Answer to Problem 25P

The maximum transverse speed of a point on the. string described by y(x,t)=(0.35mm){(π3.0radm)[x(66ms)]t} is 24mm/s.

Explanation of Solution

Write the standard expression of a transverse wave travelling along x direction.

y(x,t)=Asin(kxωt) (I)

Here, A is the amplitude of wave, k is the wave vector, ω is the angular velocity, x is position of wave at any time t and y(x,t) is the displacement of wave at position x and time t.

Write the equation of wave given in question.

y(x,t)=(0.35mm){(π3.0radm)[x(66ms)]t} (II)

Expand above equation to compare with standard form given in equation (I).

y(x,t)=(0.35mm)[(π3.0radm)x(π3.0radm)(66ms)t] (III)

Write the expression for maximum transverse speed of a point on the string.

vmax=ωA (VI)

Conclusion:

Substitute π3.0m(66ms) for ω and 0.35mm for A in above equation to get vmax.

vmax=(π3.0m(66ms))(0.35mm×1m1000mm)=24mm/s

Therefore, the maximum transverse speed of a point on the. string described by y(x,t)=(0.35mm){(π3.0radm)[x(66ms)]t} is 24mm/s.

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Chapter 11 Solutions

PHYSICS

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