Physics for Scientists and Engineers, Technology Update (No access codes included)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN: 9781305116399
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 18, Problem 18.18P

A standing wave is described by the wave function

y = 6 sin ( π 2 x ) cos ( 100 π t )

where x and y are in meters and t is in seconds. (a) Prepare graphs showing y as a function of x for five instants: t = 0, 5 ms, 10 ms, 15 ms, and 20 ms. (b) From the graph, identify the wavelength of the wave and explain how to do so. (c) From the graph, identify the frequency of the wave and explain how to do so. (d) From the equation, directly identify the wavelength of the wave and explain bow to do so. (e) From the equation, directly identify the frequency and explain how to do so.

(a)

Expert Solution
Check Mark
To determine

To draw: The graphs showing y as a function of x for five instants t=0 , t=5ms , t=10ms , t=15ms and t=20ms .

Answer to Problem 18.18P

The graph of y as a function of x at an instant t=0 is shown below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 18, Problem 18.18P , additional homework tip  1

Figure (1)

The graph of y as a function of x at an instant t=5ms is shown below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 18, Problem 18.18P , additional homework tip  2

Figure (2)

The graph of y as a function of x at an instant t=10ms is shown below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 18, Problem 18.18P , additional homework tip  3

Figure (3)

The graph of y as a function of x at an instant t=15ms is shown below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 18, Problem 18.18P , additional homework tip  4

Figure (4)

The graph of y as a function of x at an instant t=20ms is shown below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 18, Problem 18.18P , additional homework tip  5

Figure (5)

Explanation of Solution

Introduction:

The values of y varies in sinusoidal form. Initially it increases from zero to its maximum value and then decreases from maximum value to zero. This phenomenon gets repeated in periodic form.

Explanation:

Given info: The sinusoidal waves function is,

y=6sin(π2x)cos(100πt) . (1)

For t=0 :

Substitute 0 for t in the equation (1).

y=6sin(π2x)cos(100π×0)=6sin(π2x)cos(0)=6sin(π2x)×1=6sin(π2x)

The graph of y as a function of x at an instant t=0 is shown below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 18, Problem 18.18P , additional homework tip  6

Figure (1)

For t=5ms :

Substitute 5ms for t in the equation (1).

y=6sin(π2x)cos(100π×(5ms))=6sin(π2x)cos(100π×(5×103s))=6sin(π2x)×cos(0.5π)=0

The graph of y as a function of x at an instant t=5ms is shown below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 18, Problem 18.18P , additional homework tip  7

Figure (2)

For t=10ms :

Substitute 10ms for t in the equation (1).

y=6sin(π2x)cos(100π×10ms)=6sin(π2x)cos(100π×10×103s)=6sin(π2x)×cos(π)=6sin(π2x)

The graph of y as a function of x at an instant t=10ms is shown below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 18, Problem 18.18P , additional homework tip  8

Figure (3)

For t=15ms :

Substitute 15ms for t in the equation (1).

y=6sin(π2x)cos(100π×15ms)=6sin(π2x)cos(100π×15×103s)=6sin(π2x)×cos(1.5π)=0

The graph of y as a function of x at an instant t=15ms is shown below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 18, Problem 18.18P , additional homework tip  9

Figure (4)

For t=20ms :

Substitute 20ms for t in the equation (1).

y=6sin(π2x)cos(100π×20ms)=6sin(π2x)cos(100π×20×103s)=6sin(π2x)×cos(2π)=6sin(π2x)

The graph of y as a function of x at an instant t=20ms is shown below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 18, Problem 18.18P , additional homework tip  10

Figure (5)

(b)

Expert Solution
Check Mark
To determine
The wavelength of the wave using the graph of y as a function of x .

Answer to Problem 18.18P

The wavelength of the wave is 4m .

Explanation of Solution

Given info: The sinusoidal waves function is y=6sin(π2x)cos(100πt) .

The distance between the two-crest point or two trough point is called the wavelength of the wave.

From the figure (1), the distance between the two crest point is 4m . Therefore, the wavelength of the wave is,

λ=4m

Conclusion:

Therefore, the wavelength of the wave is 4m .

(c)

Expert Solution
Check Mark
To determine
The frequency of the wave using the graph of y as a function of x .

Answer to Problem 18.18P

The frequency of the wave is 50Hz .

Explanation of Solution

Given info: The sinusoidal waves function is y=6sin(π2x)cos(100πt) .

Compare the equation (1) with y=Asinkxcosωt .

ω=100π

The frequency of the wave is,

f=ω2π

Substitute 100π for ω in the above equation.

f=100π2π=50Hz

Conclusion:

Therefore, the frequency of the wave is 50Hz .

(d)

Expert Solution
Check Mark
To determine
The wavelength of the wave using the equation of the wave.

Answer to Problem 18.18P

The wavelength of the wave is 4m .

Explanation of Solution

Given info: The sinusoidal waves function is y=6sin(π2x)cos(100πt) .

Compare the equation (1) with y=Asinkxcosωt .

k=π2

The frequency of the wave is,

λ=2πk

Substitute π2 for k in the above equation.

λ=2π(π2)=4m

Conclusion:

Therefore, the wavelength of the wave is 4m .

(e)

Expert Solution
Check Mark
To determine
The frequency of the wave using the equation of the wave.

Answer to Problem 18.18P

The frequency of the wave is 50Hz .

Explanation of Solution

Given info: The sinusoidal waves function is y=6sin(π2x)cos(100πt) .

Compare the equation (1) with y=Asinkxcosωt .

ω=100π

The frequency of the wave is,

f=ω2π

Substitute 100π for ω in the above equation.

f=100π2π=50Hz

Conclusion:

Therefore, the frequency of the wave is 50Hz .

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A standing wave is described by the wave function                                    y = 6 sin (π/2 x) cos (100πt)where x and y are in meters and t is in seconds. (a) Prepare graphs showing y as a function of x for five instants: t = 0, 5 ms, 10 ms, 15 ms, and 20 ms. (b) From the graph, identify the wavelength of the wave and explain how to do so. (c) From the graph, identify the frequency of the wave and explain how to do so. (d) From the equation, directly identify the wavelength of the wave and explain how to do so. (e) From the equation, directly identify the frequency and explain how to do so.
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Chapter 18 Solutions

Physics for Scientists and Engineers, Technology Update (No access codes included)

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