chapt17 - Chapter 17: Waves - II Problem 1 When the door of...

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Chapter 17: Waves - II Problem 1 When the door of the Chapel of the Mausoleum in Hamilton, Scotland, is slammed shut, the last echo heard by someone standing just inside the door reportedly comes 15 s later. (a) If that echo were due to a single reflection off a wall opposite the door, how far from the door would that wall be? (b) If, instead, the wall is 25.7 m away, how many reflections (back and forth) correspond to the last echo? Answer (a) 2.6 km; (b) 50 Solution (a) The distance would be equal to the product of the wave speed and the time. The time in this case is half of 15 s since the sound has to travel to the wall and back. d = vt = 343 m/s * 7.5 s = 2.6 km (b) The back and forth distance is 2 * 25.7 m = 51.4 m and the number of back and forth transits is N = 2.6 km 51.4 m = 50 Problem 4 What is the bulk modulus of oxygen if 32.0 g of oxygen occupies 22.4 L and the speed of sound in the oxygen is 317 m/s? Answer 1.44 × 10 5 Pa Solution The bulk modulus is related to the density of the gas and the speed of sound in the gas. B = ρ v 2 = 32 g 22.4 L 317 m s 2 = 1.44 × 10 5 Pa Problem 8 A man strikes one end of a thin rod with a hammer The speed of sound in the rod is 15 times the sped of sound in air. . A woman, at the other end with her ear close to the rod, hears the sound of the blow twice with a 0.120 s interval between; one sound comes
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17-2 Chapter 17: Waves II through the rod and the other comes through the air alongside the rod. If the speed of sound in air is 343 m/s, what is the length of the rod? Answer 44.1 m. Solution The difference in the times which it takes the sound to travel through the air and through the rod of length L is Δ t = L v A L v R where v A and v R are the speeds of the sound in air and the rod respectively. Since v R = 15v a , the length of the rod is L = v A v R Δ t v R v A = 15 v A 2 Δ t 15 v A v A = 15 14 v A Δ t = 15 14 343 m s 0.120 s = 44.1 m Problem 10 The pressure in a traveling sound wave is given by the equation Δ p = (1.50 Pa) sin π [(0.900 m -1 )x - (315 s -1 )t]. Find the (a) pressure amplitude, (b) frequency, (c) wavelength, and (d) speed of the wave. Answer (a) 1.50 Pa; (b) 158 Hz; (c) 2.22 m; (d) 350 m/s. Solution (a) The pressure amplitude is 1.50 Pa. (b) The angular frequency is π (315 s -1 ). Therefore the frequency is f = ω /2 π = π (315 s -1 )/2 π = 158 Hz. (c) The angular wave number is π (0.900 m -1 ). Therefore the wavelength is λ = 2 π /k = 2 π / π (0.900 m -1 ) = 2.22 m. (d) The wave speed is v = λ f = (2.22 m)*(158 Hz) = 350 m/s.
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Chapter 17: Waves - II 17-3 Problem 11 If the form of a sound wave traveling through air is s(x,t) =(6.0 nm) cos[kx + (3000 rad/s)t + φ ] how much time does any given air molecule along the path take to move between displacements s = +2.0 nm and s = - 2.0 nm? Answer 0.23 ms Solution Solving the wave equation for the time t yields t = 1 3000 rad s cos 1 s x , t ( ) 6.0 nm kx φ , The time difference when the air molecule moves between the two displacements is Δ t = 1 3000 rad s cos 1 2.0 nm 6.0 nm kx 1 3000 rad s cos 1 2.0 nm 6.0 nm kx Δ t = 1 3000 rad s cos 1 2.0 nm 6.0
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This note was uploaded on 08/03/2009 for the course PHYSICS 2049 taught by Professor C during the Spring '09 term at Vanderbilt.

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chapt17 - Chapter 17: Waves - II Problem 1 When the door of...

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