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Chapter 11: Problem 22

A transverse wave on a string is described by the equation $y(x, t)=(2.20 \mathrm{cm}) \sin [(130 \mathrm{rad} / \mathrm{s}) t+(15 \mathrm{rad} / \mathrm{m}) x].$ (a) What is the maximum transverse speed of a point on the string? (b) What is the maximum transverse acceleration of a point on the string? (c) How fast does the wave move along the string? (d) Why is your answer to (c) different from the answer to (a)?

Short Answer

Expert verified
(a) The maximum transverse speed of a point on the string is 286 cm/s. (b) The maximum transverse acceleration of a point on the string is 37180 cm/s虏. (c) The wave moves along the string with a speed of 8.67 m/s. (d) The maximum transverse speed (a) refers to the speed at which the string oscillates up and down in the vertical direction, while the speed of wave propagation along the string (c) relates to how fast the wave moves horizontally along the string. They are two different aspects of the wave motion, and thus their speeds are not directly related.

Step by step solution

01

Differentiate y(x, t) with respect to time

To find the transverse speed and acceleration, we first need to find the rate of change of \(y(x, t)\) with respect to time \(t\). So, let's differentiate it w.r.t \(t\): \(\frac{\partial y}{\partial t} = \frac{\partial}{\partial t}[(2.20 \mathrm{cm}) \sin((130\mathrm{rad/s})t+(15\mathrm{rad/m})x)]\) Using the chain rule, the derivative becomes: \(\frac{\partial y}{\partial t} = (2.20 \mathrm{cm})(130 \mathrm{rad/s}) \cos((130 \mathrm{rad/s})t+(15 \mathrm{rad/m})x)\)
02

Find the maximum transverse speed

The maximum transverse speed occurs when the \(\cos\) term equals 1. So, the maximum value for the derivative of \(y(x, t)\) is: \(v_{max} = (2.20 \mathrm{cm})(130 \mathrm{rad/s})\) Calculating the maximum transverse speed: \(v_{max} = 286 \space \mathrm{cm/s}\) Therefore, the maximum transverse speed of a point on the string is 286 cm/s.
03

Differentiate y'(x, t) with respect to time

To find the maximum transverse acceleration, we need to find the second derivative of \(y\) with respect to time \(t\). So, let's differentiate \(\frac{\partial y}{\partial t}\) w.r.t \(t\): \(\frac{\partial^2 y}{\partial t^2} = \frac{\partial^2}{\partial t^2}[(2.20 \mathrm{cm})(130 \mathrm{rad/s}) \cos((130 \mathrm{rad/s})t+(15 \mathrm{rad/m})x)]\) Using the chain rule again, the second derivative becomes: \(\frac{\partial^2 y}{\partial t^2} = -(2.20 \mathrm{cm})(130\mathrm{rad/s})^2 \sin((130 \mathrm{rad/s})t+(15 \mathrm{rad/m})x)\)
04

Find the maximum transverse acceleration

The maximum transverse acceleration occurs when the \(\sin\) term equals 1. So, the maximum value for the second derivative of \(y(x, t)\) is: \(a_{max} = -(2.20 \mathrm{cm})(130 \mathrm{rad/s})^2\) Calculating the maximum transverse acceleration: \(a_{max} = -37180 \space \mathrm{cm/s^2}\) Therefore, the maximum transverse acceleration of a point on the string is \(37180\, \mathrm{cm/s^2}\)
05

Find the speed of wave propagation along the string

To find the speed of wave propagation \(v\), we need to find the ratio of the angular frequency \(\omega\) to the wave number \(k\): \(v = \frac{\omega}{k}\) In our case, the angular frequency \(\omega = 130 \space\mathrm{rad/s}\) and wave number \(k = 15 \space\mathrm{rad/m}\). So, the wave speed will be: \(v = \frac{130\, \mathrm{rad/s}}{15\, \mathrm{rad/m}}\) Calculating the wave speed: \(v = 8.67 \space \mathrm{m/s}\) Therefore, the wave moves along the string with a speed of \(8.67\, \mathrm{m/s}\).
06

Discuss the difference between (a) and (c)

The maximum transverse speed of a point on the string (a) refers to the speed at which the string oscillates up and down in the vertical direction. On the other hand, the speed of wave propagation along the string (c) relates to how fast the wave moves horizontally along the string. In this exercise, the maximum transverse speed (286 cm/s) is larger than the wave speed (8.67 m/s). This difference is because the maximum transverse speed describes the motion of the individual particles of the string, while the wave speed describes the overall motion of the wave pattern. They are two different aspects of the wave motion, and thus their speeds are not directly related.

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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 \(x=0.\)

Comprehensive Problems The speed of waves on a lake depends on frequency. For waves of frequency \(1.0 \mathrm{Hz}\), the wave speed is \(1.56 \mathrm{m} / \mathrm{s} ;\) for \(2.0-\mathrm{Hz}\) waves, the speed is \(0.78 \mathrm{m} / \mathrm{s} .\) The 2.0-Hz waves from a speedboat's wake reach you \(120 \mathrm{s}\) after the 1.0 -Hz waves generated by the same boat. How far away is the boat?
A uniform string of length \(10.0 \mathrm{m}\) and weight \(0.25 \mathrm{N}\) is attached to the ceiling. A weight of \(1.00 \mathrm{kN}\) hangs from its lower end. The lower end of the string is suddenly displaced horizontally. How long does it take the resulting wave pulse to travel to the upper end? [Hint: Is the weight of the string negligible in comparison with that of the hanging mass?]
What is the wavelength of the radio waves transmitted by an FM station at 90 MHz? (Radio waves travel at \(3.0 \times 10^{8} \mathrm{m} / \mathrm{s} .\)
What is the frequency of a wave whose speed and wavelength are $120 \mathrm{m} / \mathrm{s}\( and \)30.0 \mathrm{cm},$ respectively?
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