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Chapter 11: Problem 27

For a transverse wave on a string described by $y(x, t)=(0.0050 \mathrm{m}) \cos [(4.0 \pi \mathrm{rad} / \mathrm{s}) t-(1.0 \pi \mathrm{rad} / \mathrm{m}) x]$ find the maximum speed and the maximum acceleration of a point on the string. Plot graphs for one cycle of displacement \(y\) versus \(t\), velocity \(v_{y}\) versus \(t\), and acceleration \(a_{y}\) versus \(t\) at the point \(x=0.\)

Short Answer

Expert verified
Answer: The maximum speed of the transverse wave on the string is \(0.0628 \, \text{m/s}\) and the maximum acceleration is \(1.57 \, \text{m/s}^2\).

Step by step solution

01

Find the Velocity Equation

We need to differentiate the given displacement equation \(y(x, t)\) with respect to time \(t\): Let \(y(x, t) = (0.0050 \, \text{m}) \cos[(4.0 \pi \, \text{rad/s})t - (1.0 \pi \, \text{rad/m})x]\) Take the derivative of \(y(x, t)\) with respect to time, \(t\) to find the velocity function, \(v_y(x, t)\): \(v_y(x, t) = \frac{\partial y(x, t)}{\partial t} = -(0.0050 \, \text{m})(4.0 \pi \, \text{rad/s})\sin[(4.0 \pi \, \text{rad/s})t - (1.0 \pi \, \text{rad/m})x]\) To simplify the notation, let \(A = 0.0050 m\), \(B = 4.0 \pi rad/s\), and \(C = 1.0 \pi rad/m\). The velocity function now becomes: \(v_y(x, t) = -AB\sin[Bt - Cx]\)
02

Find the Maximum Speed

To find the maximum speed, we need to find the maximum value of \(v_y(x, t)\). Since \(v_y(x, t)\) is a sine function, its maximum value will be equal to its amplitude. Thus, we can find the amplitude by multiplying the coefficients of the sine function: Maximum Speed = \(|AB| = (0.0050 \, \text{m})(4.0 \pi) = 0.0628 \, \text{m/s}\)
03

Find the Acceleration Equation

Now, differentiate the velocity function \(v_y(x, t)\) with respect to time, \(t\) to find the acceleration function, \(a_y(x, t)\): \(a_y(x, t) = \frac{\partial v_y(x, t)}{\partial t} = -AB^2\cos[Bt - Cx]\)
04

Find the Maximum Acceleration

To find the maximum acceleration, we need to find the maximum value of \(a_y(x, t)\). Since \(a_y(x, t)\) is a cosine function, its maximum value will be equal to its amplitude. Thus, we can find the amplitude by multiplying the coefficients of the cosine function: Maximum Acceleration = \(|AB^2|\) = \((0.0050 \, \text{m}) (4.0 \pi \, \text{rad/s})^2\) = \(1.57 \, \text{m/s}^2\)
05

Plot the graphs for one cycle of \(y\) vs \(t\), \(v_y\) vs \(t\), and \(a_y\) vs \(t\) at \(x = 0\)

To plot the graphs, substitute \(x = 0\) in the displacement, velocity, and acceleration equations. This will give us the functions of \(y(t)\), \(v_y(t)\), and \(a_y(t)\): \(y(t) = A\cos{(Bt)}\) \(v_y(t) = -AB\sin{(Bt)}\) \(a_y(t) = -AB^2\cos{(Bt)}\) Now, plot the graphs of \(y(t)\), \(v_y(t)\), and \(a_y(t)\) for one cycle, which is the time period of the wave: Time Period, \(T = \frac{2\pi}{B} = \frac{2\pi}{4.0\pi} = 0.5 \, \text{s}\). You will see that the displacement, velocity, and acceleration functions oscillate within one cycle, with the maximum speed being \(0.0628 \, \text{m/s}\) and the maximum acceleration being \(1.57 \, \text{m/s}^2\).

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Most popular questions from this chapter

A longitudinal wave has a wavelength of \(10 \mathrm{cm}\) and an amplitude of \(5.0 \mathrm{cm}\) and travels in the \(y\) -direction. The wave speed in this medium is \(80 \mathrm{cm} / \mathrm{s},\) (a) Describe the motion of a particle in the medium as the wave travels through the medium. (b) How would your answer differ if the wave were transverse instead?
Two speakers spaced a distance \(1.5 \mathrm{m}\) apart emit coherent sound waves at a frequency of \(680 \mathrm{Hz}\) in all directions. The waves start out in phase with each other. A listener walks in a circle of radius greater than one meter centered on the midpoint of the two speakers. At how many points does the listener observe destructive interference? The listener and the speakers are all in the same horizontal plane and the speed of sound is \(340 \mathrm{m} / \mathrm{s} .\) [Hint: Start with a diagram; then determine the maximum path difference between the two waves at points on the circle. I Experiments like this must be done in a special room so that reflections are negligible.
A cord of length \(1.5 \mathrm{m}\) is fixed at both ends. Its mass per unit length is \(1.2 \mathrm{g} / \mathrm{m}\) and the tension is \(12 \mathrm{N} .\) (a) What is the frequency of the fundamental oscillation? (b) What tension is required if the \(n=3\) mode has a frequency of \(0.50 \mathrm{kHz} ?\)
A transverse wave on a string is described by $y(x, t)=(1.2 \mathrm{cm}) \sin [(0.50 \pi \mathrm{rad} / \mathrm{s}) t-(1.00 \pi \mathrm{rad} / \mathrm{m}) x]$ Find the maximum velocity and the maximum acceleration of a point on the string. Plot graphs for displacement \(y\) versus \(t\), velocity \(v_{y}\) versus \(t\), and acceleration \(a_{y}\) versus \(t\) at \(x=0.\)
What is the speed of the wave represented by \(y(x, t)=\) $A \sin (k x-\omega t),\( where \)k=6.0 \mathrm{rad} / \mathrm{cm}\( and \)\omega=5.0 \mathrm{rad} / \mathrm{s} ?$
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