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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

The equation of a wave is $$ y(x, t)=(3.5 \mathrm{cm}) \sin \left\\{\frac{\pi}{3.0 \mathrm{cm}}[x-(66 \mathrm{cm} / \mathrm{s}) t]\right\\} $$ Find (a) the amplitude and (b) the wavelength of this wave.
While testing speakers for a concert. Tomás sets up two speakers to produce sound waves at the same frequency, which is between \(100 \mathrm{Hz}\) and $150 \mathrm{Hz}$. The two speakers vibrate in phase with one another. He notices that when he listens at certain locations, the sound is very soft (a minimum intensity compared to nearby points). One such point is \(25.8 \mathrm{m}\) from one speaker and \(37.1 \mathrm{m}\) from the other. What are the possible frequencies of the sound waves coming from the speakers? (The speed of sound in air is \(343 \mathrm{m} / \mathrm{s} .\) )
A metal guitar string has a linear mass density of $\mu=3.20 \mathrm{g} / \mathrm{m} .$ What is the speed of transverse waves on this string when its tension is \(90.0 \mathrm{N} ?\)
Michelle is enjoying a picnic across the valley from a cliff. She is playing music on her radio (assume it to be an isotropic source) and notices an echo from the cliff. She claps her hands and the echo takes 1.5 s to return. (a) Given that the speed of sound in air is \(343 \mathrm{m} / \mathrm{s}\) on that day, how far away is the cliff? (b) If the intensity of the music $1.0 \mathrm{m}\( from the radio is \)1.0 \times 10^{-5} \mathrm{W} / \mathrm{m}^{2},$ what is the intensity of the music arriving at the cliff?
A standing wave has wave number $2.0 \times 10^{2} \mathrm{rad} / \mathrm{m} .$ What is the distance between two adjacent nodes?
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