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CALC Figure P16.75 shows the pressure fluctuation p of a

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

Solution for problem 83CP Chapter 16

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 83CP

CALC Figure P16.75 shows the pressure fluctuation p of a non sinusoidal sound wave as a function of x for t = 0. The wave is traveling in the +x-direction. (a) Graph the pressure fluctuation p as a function of t for x = 0. Show at least two cycles of oscillation. (b) Graph the displacement y in this sound wave as a function of x at t = 0. At x = 0, the displacement at t = 0 is zero. Show at least two wavelengths of the wave. (c) Graph the displacement y as a function of t for x = 0. Show at least two cycles of oscillation. (d) Calculate the maximum velocity and the maximum acceleration of an element of the air through which this sound wave is traveling. (e) Describe how the cone of a loudspeaker must move as a function of time to produce the sound wave in this problem.

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Solution 83CP (a) Since the pressure as a function is a triangular wave, the pressure as a function of time will also be triangular wave. And since the as given the pressure at x=0, and t=0 is zero, the pressure as a function of time at x=0 will also start from zero. Now the speed of the sound is c = 344 m/s. And the wavelength here is = 0.200 m. Hence, the time period of the oscillation is 0.200 m 3 T = v = 344 m/s 58 × 10 s = 58 ms The graph as a function of pressure is shown below. (b) We know that the pressure in a wave is given by P(x) = B y(x) x 5 Where B is the bulk modulus and for air B = 1.42 × 10 Pa. So the displacement y(x) can be written as y(x) = 1 P(x)dx B Now suppose the equation of P(x) for the first half of the pressure vs x graph at t=0, is given by P(x) = mx + C Now from the figure we can see that p(x) = 40 for x=0 and p(x)=-40 for x=0.100 m. Hence we have C = 40 And 40 = m(0.100) + 40 m = 80 = 800 0.100 Therefore the equation of the pressure is P(x) = 800x + 40 So the displacement for this half is given by y(x) = 1 ( 800x + 40)dx = 2.81 × 10 x 2.81 × 10 x 4 1.24×10 This is a parabola. Since the wave is symmetric, the wave will look same but opposite in sign for the next half of the triangle (that is in the part where the pressure is increasing). The graph is shown...

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Chapter 16, Problem 83CP is Solved
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Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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CALC Figure P16.75 shows the pressure fluctuation p of a

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