Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN: 9781133104261
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 13, Problem 5P

The string shown in Figure P13.5 is driven at a frequency of 5.00 Hz. The amplitude of the motion is A = 12.0 cm, and the wave speed is v = 20.0 m/s. Furthermore, the wave is such that y = 0 at x = 0 and t = 0. Determine (a) the angular frequency and (b) the wave number for this wave. (c) Write an expression for the wave function. Calculate (d) the maximum transverse speed and (e) the maximum transverse acceleration of an element of the string.

Chapter 13, Problem 5P, The string shown in Figure P13.5 is driven at a frequency of 5.00 Hz. The amplitude of the motion is

Figure P13.5

(a)

Expert Solution
Check Mark
To determine

The angular frequency of the wave.

Answer to Problem 5P

The angular frequency of the wave is 31.4rad/s_

Explanation of Solution

Write the expression for the frequency of the string.

    f=ω2π        (I)

Here, f is the frequency of the string, ω is the angular frequency of the string.

Solve equation (I) for ω,

    ω=2πf        (II)

Conclusion:

Substitute 5.00s1 for f in equation (I) to find ω.

    ω=2π(5.00s1)=31.4rad/s

Therefore, the angular frequency of the wave is 31.4rad/s_

(b)

Expert Solution
Check Mark
To determine

The wave number of the wave.

Answer to Problem 5P

The wave number of the wave is 1.57rad/s_.

Explanation of Solution

Write the expression for the wavelength of the wave.

    λ=2πk        (III)

Here, λ is the wavelength, k is the wavenumber.

Solve equation (III) for k,

    k=2πλ        (IV)

Write the expression for the wavelength in terms of speed of the wave.

    λ=vf        (V)

Conclusion:

Substitute 20.0m/s for v, 5.00s1 for f in equation (V),

    λ=20.0m/s5.00s1=4.00m

Substitute 4.00m for λ in equation (IV) to find k.

    k=2π4.00m=1.57rad/m

Therefore, The wave number of the wave is 1.57rad/s_.

(c)

Expert Solution
Check Mark
To determine

Expression for the wave function.

Answer to Problem 5P

Expression for the wave function is y=0.120sin(1.57x31.4t)_.

Explanation of Solution

The general form of a wave function can be represented as,

    y=Asin(kxωt+ϕ)        (VI)

Here, A is the amplitude, k is the angular wave number, ω is the angular frequency, ϕ is the phase constant.

Conclusion:

Using initial conditions, to make this fit,

  y=0

  ϕ=0

In this case, taking initial conditions, substitute 12.0cm, 31.4rad/s for ω , 0 for x, 0 for t in equation (VI) to find

  y(x,t)=12.0sin(1.57x31.4t)

Therefore, Expression for the wave function is y=0.120sin(1.57x31.4t)_.

(d)

Expert Solution
Check Mark
To determine

The maximum transverse speed of the wave.

Answer to Problem 5P

The maximum transverse speed of the wave is 3.77m/s_.

Explanation of Solution

Write the expression for the transverse speed.

    v=dydt        (VII)

Differentiate equation (VI) in equation (VII),

    v=Aωcos(kxωt)        (VIII)

Conclusion:

The maximum magnitude is given by,

    vmax=Aω        (IX)

Substitute 12.0cm for A and 31.4rad/s for ω in equation (IX) to find vmax.

  vmax=(12.0cm×1m100cm)(31.4rad/s)=3.77m/s

Therefore, the maximum transverse speed of the wave is 3.77m/s_.

(e)

Expert Solution
Check Mark
To determine

The maximum transverse acceleration of an element of the string.

Answer to Problem 5P

The maximum transverse acceleration of an element of the string.is 118m/s_2.

Explanation of Solution

Write the expression for the transverse acceleration.

    ay=vyt        (VII)

 Use equation (VIII) in equation (VII),

    ay=t(Aωcos(kxωt))=Aω2sin(kxωt)

Conclusion:

The maximum magnitude is given by,

    amax=Aω2        (IX)

Substitute 12.0cm for A and 31.4rad/s for ω in equation (IX) to find vmax.

  amax=(12.0cm×1m100cm)(31.4rad/s)2=118m/s2

Therefore, The maximum transverse acceleration of an element of the string.is 118m/s_2.

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Chapter 13 Solutions

Principles of Physics: A Calculus-Based Text

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