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A simple harmonic oscillator at the point x = 0 generates

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 21E Chapter 15

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 21E

A simple harmonic oscillator at the point x = 0 generates a wave on a rope. The oscillator operates at a frequency of 40.0 Hz and with an amplitude of 3.00 cm. The rope has a linear mass density of 50.0 g/m and is stretched with a tension of 5.00 N. (a) Determine the speed of the wave. (b) Find the wavelength. (c) Write the wave function y(x, t) for the wave. Assume that the oscillator has its maximum upward displacement at time t = 0. (d) Find the maximum transverse acceleration of points on the rope. (e) In the discussion of transverse waves in this chapter, the force of gravity was ignored. Is that a reasonable approximation for this wave? Explain.

Step-by-Step Solution:

Solution 21E Introduction The frequency of the wave will be equal to the frequency of the oscillator. Now knowing the tension of the string and density we can calculate the speed of the wave on the rope. Now from the speed of the wave and frequency we can calculate the wavelength and hence the wavenumber. Then, using the initial condition we can write the wave equation. From the wave equation we can calculate maximum transverse acceleration. Then we have to discuss if it is okay to ignore the gravitational acceleration. Step 1 The linear mass density of the rope is = 50.0 g/m = 50.0 kg/m And the tension on the string is = 5.00 N Hence the speed of the wave is given by Step 2 The frequency of the wave will be equal to the frequency of the oscillator, hence the frequency of the wave is f = 40.0 Hz . Hence we the wavelength is given by Step 3 The wave function of the wave can be written as Now we have And Also as given at t = 0, and x = 0 we have Hence we have Hence the wave equation becomes

Step 4 of 5

Chapter 15, Problem 21E is Solved
Step 5 of 5

Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

The full step-by-step solution to problem: 21E from chapter: 15 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. Since the solution to 21E from 15 chapter was answered, more than 597 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: University Physics, edition: 13. This full solution covers the following key subjects: wave, rope, Oscillator, Find, transverse. This expansive textbook survival guide covers 26 chapters, and 2929 solutions. The answer to “A simple harmonic oscillator at the point x = 0 generates a wave on a rope. The oscillator operates at a frequency of 40.0 Hz and with an amplitude of 3.00 cm. The rope has a linear mass density of 50.0 g/m and is stretched with a tension of 5.00 N. (a) Determine the speed of the wave. (b) Find the wavelength. (c) Write the wave function y(x, t) for the wave. Assume that the oscillator has its maximum upward displacement at time t = 0. (d) Find the maximum transverse acceleration of points on the rope. (e) In the discussion of transverse waves in this chapter, the force of gravity was ignored. Is that a reasonable approximation for this wave? Explain.” is broken down into a number of easy to follow steps, and 122 words. University Physics was written by and is associated to the ISBN: 9780321675460.

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