
Concept explainers
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.
Figure P13.5
(a)

The angular frequency of the wave.
Answer to Problem 5P
The angular frequency of the wave is
Explanation of Solution
Write the expression for the frequency of the string.
Here,
Solve equation (I) for
Conclusion:
Substitute
Therefore, the angular frequency of the wave is
(b)

The wave number of the wave.
Answer to Problem 5P
The wave number of the wave is
Explanation of Solution
Write the expression for the wavelength of the wave.
Here,
Solve equation (III) for
Write the expression for the wavelength in terms of speed of the wave.
Conclusion:
Substitute
Substitute
Therefore, The wave number of the wave is
(c)

Expression for the wave function.
Answer to Problem 5P
Expression for the wave function is
Explanation of Solution
The general form of a wave function can be represented as,
Here,
Conclusion:
Using initial conditions, to make this fit,
In this case, taking initial conditions, substitute
Therefore, Expression for the wave function is
(d)

The maximum transverse speed of the wave.
Answer to Problem 5P
The maximum transverse speed of the wave is
Explanation of Solution
Write the expression for the transverse speed.
Differentiate equation (VI) in equation (VII),
Conclusion:
The maximum magnitude is given by,
Substitute
Therefore, the maximum transverse speed of the wave is
(e)

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
Explanation of Solution
Write the expression for the transverse acceleration.
Use equation (VIII) in equation (VII),
Conclusion:
The maximum magnitude is given by,
Substitute
Therefore, The maximum transverse acceleration of an element of the string.is
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Chapter 13 Solutions
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