/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 26 (a) Plot a graph for $$ y(x, t)=... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 26

(a) Plot a graph for $$ y(x, t)=(4.0 \mathrm{cm}) \sin [(378 \mathrm{rad} / \mathrm{s}) t-(314 \mathrm{rad} / \mathrm{cm}) x] $$ versus \(x\) at \(t=0\) and at \(t=\frac{1}{480} \mathrm{s}\). From the plots determine the amplitude, wavelength, and speed of the wave. (b) For the same function, plot a graph of \(y(x, t)\) versus \(t\) at \(x=0\) and find the period of the vibration. Show that \(\lambda=v T.\)

Short Answer

Expert verified
The amplitude of the wave is 4.0 cm, the wavelength is approximately 0.02 m, the speed of the wave is approximately 1.2 m/s, and the period of the wave is approximately 0.0166 s.

Step by step solution

01

Identify the given function and its properties

The given wave function is: $$ y(x, t)=(4.0 \mathrm{cm}) \sin [(378 \mathrm{rad} / \mathrm{s}) t-(314 \mathrm{rad} / \mathrm{cm}) x] $$ From this function, we can already identify the amplitude \(A\) as 4.0 cm, the angular frequency \(\omega\) as 378 rad/s, and the wave number \(k\) as 314 rad/cm.
02

Plot the graph of the function versus \(x\) at \(t=0\) and \(t=\frac{1}{480}\mathrm{s}\)

First, let's plot the wave function at \(t=0\): $$ y_t=0(x)=(4.0 \mathrm{cm}) \sin (-314 \mathrm{rad} / \mathrm{cm}\cdot x) $$ Also, plot the wave function at \(t=\frac{1}{480}\mathrm{s}\): $$ y_t=1/480(x)=(4.0 \mathrm{cm}) \sin [(378 \mathrm{rad} / \mathrm{s})\left(\frac{1}{480} \mathrm{s}\right)-(314 \mathrm{rad} / \mathrm{cm})x] $$ It is important to use appropriate software to generate these plots, then observe the graphs and deduce properties of the wave from the plot.
03

Calculate the amplitude, wavelength, and speed of the wave

We already know the amplitude \(A=4.0\mathrm{cm}\) from the wave function. To find the wavelength, we can use the formula: $$\lambda = \frac{2\pi}{k}$$ where \(k = 314\ \mathrm{rad}/\mathrm{cm}\), so, $$\lambda =\frac{2\pi}{314\ \mathrm{rad}/\mathrm{cm}} \approx 0.02\ \mathrm{m}$$ To calculate the speed of the wave, we can use the formula: $$v=\frac{\omega}{k}$$ where \(\omega=378\ \mathrm{rad/s}\), so $$v=\frac{378\ \mathrm{rad/s}}{314\ \mathrm{rad}/\mathrm{cm}} \approx 1.2\ \mathrm{m/s}$$
04

Plot the graph of the function versus \(t\) at \(x=0\)

Now let's plot the wave function at \(x=0\): $$ y_x=0(t)=(4.0 \mathrm{cm}) \sin (378 \mathrm{rad} / \mathrm{s} \cdot t) $$ Again, using appropriate software to generate the plot, observe the graph to find the period of the wave.
05

Calculate the period of the vibration and verify that \(\lambda=vT\)

To find the period of the vibration, we can use the formula: $$T = \frac{2\pi}{\omega}$$ where \(\omega = 378\ \mathrm{rad/s}\), so $$T= \frac{2\pi}{378\ \mathrm{rad/s}} \approx 0.0166\ \mathrm{s}$$ Now to show that \(\lambda=vT\), we have found the following values: $$\lambda \approx 0.02\ \mathrm{m}$$ $$v \approx 1.2\ \mathrm{m/s}$$ $$T \approx 0.0166\ \mathrm{s}$$ Multiplying \(v\) and \(T\), we get: $$vT = (1.2\ \mathrm{m/s})(0.0166\ \mathrm{s}) = 0.02\ \mathrm{m} = \lambda$$ This confirms that \(\lambda = vT\).

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Most popular questions from this chapter

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 \(x=0.\)
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