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Chapter 11: Problem 28

A transverse wave on a string is described by $y(x, t)=(1.2 \mathrm{mm}) \sin [(2.0 \pi \mathrm{rad} / \mathrm{s}) t-(0.50 \pi \mathrm{rad} / \mathrm{m}) x]$ Plot the displacement \(y\) and the velocity \(v_{y}\) versus \(t\) for one complete cycle of the point \(x=0\) on the string.

Short Answer

Expert verified
Also, calculate the period of one complete cycle and provide the expressions for the displacement and velocity of the point as a function of time.

Step by step solution

01

Identify the amplitude, frequency, and wave number

From the given wave equation, we can identify the following parameters: Amplitude: \(A = 1.2 \,\text{mm}\) Frequency: \(\omega = 2.0 \,\pi\, \text{rad/s}\) Wave number: \(k = 0.50 \,\pi\, \text{rad/m}\) Since we are asked to analyze the point \(x=0\), the wave equation simplifies to: \(y(t) = A \sin(\omega t)\)
02

Calculate the period of one complete cycle

The period \(T\) of the wave can be calculated as the inverse of its frequency: \(T = \frac{2 \pi}{\omega}\) Substitute the value of \(\omega\) and solve for \(T\): \(T = \frac{2 \pi}{2.0 \,\pi\, \text{rad/s}} = 1\, \text{s}\) So, one complete cycle takes 1 second.
03

Derive the expression for velocity \(v_y\)

To find the expression for \(v_y\), we take the time derivative of the displacement \(y(t)\): \(v_y(t) = \frac{dy(t)}{dt}\) Differentiate \(y(t) = A \sin(\omega t)\) with respect to time: \(v_y(t) = A \omega \cos(\omega t)\) Substitute the values for \(A\) and \(\omega\): \(v_y(t) = (1.2 \,\text{mm}) (2.0 \,\pi\, \text{rad/s}) \cos[(2.0 \,\pi\, \text{rad/s}) t]\)
04

Plot the displacement and velocity as a function of time

Now we can plot the displacement \(y(t)\) and the velocity \(v_y(t)\) as a function of time for one complete cycle, which takes 1 second. The displacement and velocity functions are: \(y(t) = (1.2 \,\text{mm}) \sin[(2.0 \,\pi\, \text{rad/s}) t]\) \(v_y(t) = (1.2 \,\text{mm}) (2.0 \,\pi\, \text{rad/s}) \cos[(2.0 \,\pi\, \text{rad/s}) t]\) These graphs will be sinusoidal in nature, with the displacement graph being a sine wave and the velocity graph being a cosine wave. The displacement will vary between +1.2 mm and -1.2 mm, while the velocity will vary between \((1.2 \,\text{mm}) (2.0 \,\pi\, \text{rad/s})\) and \(-(1.2 \,\text{mm}) (2.0 \,\pi\, \text{rad/s})\). The period for both graphs will be 1 second.

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Most popular questions from this chapter

A guitar's E-string has length \(65 \mathrm{cm}\) and is stretched to a tension of \(82 \mathrm{N}\). It vibrates at a fundamental frequency of $329.63 \mathrm{Hz}$ Determine the mass per unit length of the string.
A guitar string has a fundamental frequency of \(300.0 \mathrm{Hz}\) (a) What are the next three lowest standing wave frequencies? (b) If you press a finger lightly against the string at its midpoint so that both sides of the string can still vibrate, you create a node at the midpoint. What are the lowest four standing wave frequencies now? (c) If you press hard at the same point, only one side of the string can vibrate. What are the lowest four standing wave frequencies?
(a) Sketch graphs of \(y\) versus \(x\) for the function $$ y(x, t)=(0.80 \mathrm{mm}) \sin (k x-\omega t) $$ for the times \(t=0,0.96 \mathrm{s},\) and \(1.92 \mathrm{s} .\) Make all three graphs of the same axes, using a solid line for the first, a dashed line for the second, and a dotted line for the third. Use the values \(k=\pi /(5.0 \mathrm{cm})\) and $\omega=(\pi / 6.0) \mathrm{rad} / \mathrm{s}$ (b) Repeat part (a) for the function $$ y(x, t)=(0.50 \mathrm{mm}) \sin (k x+\omega t) $$ (c) Which function represents a wave traveling in the \(-x\) direction and which represents a wave traveling in the \(+x\) -direction?
Write the equation for a transverse sinusoidal wave with a maximum amplitude of \(2.50 \mathrm{cm}\) and an angular frequency of 2.90 rad/s that is moving along the positive \(x\) -direction with a wave speed that is 5.00 times as fast as the maximum speed of a point on the string. Assume that at time \(t=0,\) the point \(x=0\) is at \(y=0\) and then moves in the \(-y\) -direction in the next instant of time.
A harpsichord string of length \(1.50 \mathrm{m}\) and linear mass density $25.0 \mathrm{mg} / \mathrm{m}\( vibrates at a (fundamental) frequency of \)450.0 \mathrm{Hz}$. (a) What is the speed of the transverse string waves? (b) What is the tension? (c) What are the wavelength and frequency of the sound wave in air produced by vibration of the string? (The speed of sound in air at room temperature is \(340 \mathrm{m} / \mathrm{s} .\) )
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