# chapt16 - Chapter 16: Waves - I Problem 1 A wave has an...

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Chapter 16: Waves - I Problem 1 A wave has an angular frequency of 110 rad/s and a wavelength of 1.80 m. Calculate (a) the angular wave number and (b) the speed of the wave. Answer (a) 3.49 m -1 ; (b) 31.5 m/s Solution (a) The angular wave number is k = 2 π λ = 2 1.80 m = 3.49 m 1 . (b) The speed of the wave is v = ω k = 110 rad s 3.49 m 1 = 31.5 m s . Problem 8 The equation of a transverse sinusoidal wave traveling along a very long string is y = 6.0sin 0.020 x + 4.0 t ( ) , where x and y are expressed in centimeters and t is in seconds. Determine (a) the amplitude, (b) the wavelength, (c) the frequency, (d) the speed, (e) the direction of propagation of the wave, and (f) the maximum transverse speed of a particle in the string. (g) What is the transverse displacement at x = 3.5 cm when t = 0.26 s? Answer (a) 6.0 cm; (b) 100 cm; (c) 2.0 Hz; (d) 2 m/s; (e) negative x direction; (f) 75 cm/s; (g) -2.0 cm. Comparing the given equation to the general equation y = y m sin(kx ± ω t) we see that (a) The amplitude is y m = 6.0 cm. (b) We can get the wavelength from the angular wave number k = 2 or = 2 k = 2 0.020 c m 1 = 100 cm . (c) We can get the frequency from the angular frequency ω = 2 π f or f = 2 = 4 Hz 2 = 2.0 Hz . (d) The wave speed is

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16-2 Chapter 16: Waves I v = λ f = (100 cm)*(2.0 Hz) = 2 m/s. (e) Since the phase is a function of x and the two terms in the phase have the same sign the wave is traveling in the negative x direction. (f) The transverse speed is u = y t = 24 π cos(0.020 π x + 4.0 π t) . The maximum value of the transverse speed occurs when the cosine is 1. The speed is then 24 π cm/s. (g) The transverse displacement at x = 3.5 cm when t = 0.26 s is y = 6.0sin 0.020 π 3.5 cm + 4.0 0.26 s ( ) = 2.0 cm . (Don’t forget to put your calculator in radian mode for this calculation.) Problem 10 The function y(x,t) = (15.0 cm) cos( π x – 15 π t), with x in meters and t in seconds, describes a wave on a taut string. What is the transverse speed for a point on the string at an instant when that point has the displacement y = +12.0 cm? Answer 1.35 π m/s Solution The transverse velocity is v t = y ( x , t ) t = 15 15 cm sin x 15 t ( ) At the instant that y = +12.0 cm we have cos x 15 t ( ) = 12 cm 15 cm = 4 5 . When cos x 15 t ( ) = 4 5 the value of sin( π x – 15 π t) is ±3/5. Therefore the transverse speed at this instant is 15 π s -1 *15 cm*(3/5) = 1.35 π m/s. Problem 15 What is the speed of a transverse wave in a rope of length 2.0 m and mass 60 g under a tension of 500 N? Answer 130 m/s.
Chapter 16: Waves - I 16-3 Solution The linear density of the rope is μ = 60 g 2.0 m = 0.030 g m . Therefore the wave speed is v = τ μ = 500 N 0.030 kg/m = 130 m/s Problem 18 The speed of a wave on a string is 170 m/s when the tension is 120 N. To what value must the tension be increased in order to raise the wave speed to 180 m/s? Answer 135 N. The linear density of the string is related to the tension, τ , and the wave speed. μ = τ /v 2 . If the linear density of the string does not change when the tension is increased then τ 1 v 1 2 = 2 v 2 2 2 = 1 v 2 2 v 1 2 = 120 N 180 m s 2 170 m s 2 = 135 N Problem 19 The linear density of a vibrating string is 1.6 × 10 -4 kg/m. A transverse wave is propagating

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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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chapt16 - Chapter 16: Waves - I Problem 1 A wave has an...

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