Physics
Physics
5th Edition
ISBN: 9781260486919
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

Ch. 11.9 - Prob. 11.9CPCh. 11.10 - Prob. 11.10CPCh. 11.10 - Prob. 11.7PPCh. 11 - Prob. 1CQCh. 11 - Prob. 2CQCh. 11 - Prob. 3CQCh. 11 - Prob. 4CQCh. 11 - Prob. 5CQCh. 11 - Prob. 6CQCh. 11 - Prob. 7CQCh. 11 - Prob. 8CQCh. 11 - Prob. 9CQCh. 11 - Prob. 10CQCh. 11 - Prob. 11CQCh. 11 - Prob. 1MCQCh. 11 - Prob. 2MCQCh. 11 - Prob. 3MCQCh. 11 - Prob. 4MCQCh. 11 - Prob. 5MCQCh. 11 - Prob. 6MCQCh. 11 - Prob. 7MCQCh. 11 - Prob. 8MCQCh. 11 - Prob. 9MCQCh. 11 - Prob. 10MCQCh. 11 - Prob. 11MCQCh. 11 - Prob. 1PCh. 11 - Prob. 2PCh. 11 - Prob. 3PCh. 11 - Prob. 4PCh. 11 - 5. At what rate does the jet airplane in Problem 4...Ch. 11 - Prob. 6PCh. 11 - Prob. 7PCh. 11 - Prob. 8PCh. 11 - Prob. 9PCh. 11 - Prob. 10PCh. 11 - 11. A metal guitar string has a linear mass...Ch. 11 - Prob. 12PCh. 11 - 13. Two strings, each 15.0 m long, are stretched...Ch. 11 - Prob. 14PCh. 11 - Prob. 15PCh. 11 - Prob. 16PCh. 11 - Prob. 17PCh. 11 - 18. The speed of sound in air at room temperature...Ch. 11 - Prob. 19PCh. 11 - Prob. 20PCh. 11 - 21. A fisherman notices a buoy bobbing up and down...Ch. 11 - Prob. 22PCh. 11 - Prob. 23PCh. 11 - Prob. 24PCh. 11 - Prob. 25PCh. 11 - Prob. 26PCh. 11 - Prob. 27PCh. 11 - Prob. 28PCh. 11 - Prob. 29PCh. 11 - Problems 30–32. The graphs show displacement y as...Ch. 11 - Prob. 31PCh. 11 - 32. Rank the waves in order of maximum transverse...Ch. 11 - Prob. 33PCh. 11 - Prob. 34PCh. 11 - Prob. 35PCh. 11 - Prob. 36PCh. 11 - Prob. 37PCh. 11 - Prob. 38PCh. 11 - Prob. 39PCh. 11 - Prob. 40PCh. 11 - Prob. 41PCh. 11 - 42. A traveling sine wave is the result of the...Ch. 11 - Prob. 43PCh. 11 - Prob. 44PCh. 11 - Prob. 45PCh. 11 - Prob. 46PCh. 11 - Prob. 47PCh. 11 - Prob. 48PCh. 11 - Prob. 49PCh. 11 - Prob. 50PCh. 11 - Prob. 51PCh. 11 - Prob. 52PCh. 11 - Prob. 53PCh. 11 - Prob. 54PCh. 11 - Prob. 55PCh. 11 - Prob. 56PCh. 11 - Prob. 57PCh. 11 - Prob. 58PCh. 11 - Prob. 59PCh. 11 - Prob. 60PCh. 11 - Prob. 61PCh. 11 - Prob. 62PCh. 11 - Prob. 63PCh. 11 - Prob. 64PCh. 11 - Prob. 65PCh. 11 - Prob. 66PCh. 11 - Prob. 67PCh. 11 - Prob. 68PCh. 11 - Prob. 69PCh. 11 - Prob. 70PCh. 11 - Prob. 71PCh. 11 - Prob. 72PCh. 11 - Prob. 73PCh. 11 - Prob. 74PCh. 11 - Prob. 75PCh. 11 - Prob. 76PCh. 11 - Prob. 77PCh. 11 - Prob. 79PCh. 11 - Prob. 78PCh. 11 - Prob. 80PCh. 11 - Prob. 81PCh. 11 - Prob. 82PCh. 11 - Prob. 83PCh. 11 - Prob. 84PCh. 11 - Prob. 85PCh. 11 - Prob. 86PCh. 11 - Prob. 87PCh. 11 - Prob. 88PCh. 11 - Prob. 89PCh. 11 - Prob. 90PCh. 11 - Prob. 91PCh. 11 - Prob. 92PCh. 11 - Prob. 93PCh. 11 - Prob. 94PCh. 11 - Prob. 95PCh. 11 - Prob. 96PCh. 11 - Prob. 97PCh. 11 - Prob. 98PCh. 11 - Prob. 99PCh. 11 - Prob. 100PCh. 11 - Prob. 102PCh. 11 - 102. A harpsichord string is made of yellow brass...
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