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Chapter 21 - PHYS 132 WEEK#3 21.2 Model The principle of...

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PHYS 132, WEEK #3 21.2. Model: The principle of superposition comes into play whenever the waves overlap. Visualize: The snapshot graph at t = 1 s differs from the graph t = 0 s in that the left wave has moved to the right by 1 m and the right wave has moved to the left by 1 m. This is because the distance covered by each wave in 1 s is 1 m. The snapshot graphs at t = 2 s, 3 s, and 4 s are a superposition of the left and the right moving waves. The overlapping parts of the two waves are shown by the dotted lines.
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21.8. Model: Reflections at the string boundaries cause a standing wave on the string. Solve: Figure Ex21.8 indicates one and a half full wavelengths on the string. Hence l = 2 3 (60 cm) = 40 cm. Thus v = l f = 0.40 m ( ) 100 Hz ( ) = 40 m / s 21.10. Model: A string fixed at both ends supports standing waves. Solve: (a) A standing wave can exist on the string only if its wavelength is l m = 2 L m m = 1, 2, 3, K The three longest wavelengths for standing waves will therefore correspond to m = 1, 2, and 3. Thus, l 1 = 2 2.40 m ( ) 1 = 4.80 m l 2 = 2 2.40 m ( ) 2 = 2.40 m l 3 = 2 2.40 m ( ) 3 = 1.60 m (b) Because the wave speed on the string is unchanged from one m value to the other, f 2 l 2 = f 3 l 3 f 3 = f 2 l 2 l 3 = 50 Hz ( ) 2.40 m ( ) 1.60 m = 75 Hz 21.13. Model: A string fixed at both ends forms standing waves. Solve: (a) Three antinodes means the string is vibrating as the m = 3 standing wave. The frequency is f 3 = 3 f 1 , so the fundamental frequency is f 1 = 1 3 (420 Hz) = 140 Hz. The fifth harmonic will have the frequency f 5 = 5 f 1 = 700 Hz. (b) The wavelength of the fundamental mode is l 1 = 2 L = 1.20 m. The wave speed on the string is v = l 1 f 1 = (1.20 m)(140 Hz) = 168 m/s. Alternatively, the wavelength of the n = 3 mode is l 3 = 1 3 (2 L ) = 0.40 m, from which v = l 3 f 3 = (0.40 m)(420 Hz) = 168 m/s. The wave speed on the string is given by v = T S m T S = m v 2 = 0.0020 kg / m ( ) 168 m / s ( ) 2 = 56.4 N Assess: You must remember to use the linear density in SI units of kg/m. Also, the speed is the same for all modes, but you must use a matching l and f to calculate the speed. 21.14. Model: The laser light forms a standing wave inside the cavity. Solve: The wavelength of the laser beam is l m = 2 L m l 100,000 = 2 0.5300 m ( ) 100,000 = 10.6 m m The frequency is f 100 ,000 = c l 100 ,000 = 3.0 10 8 m / s 10.6 10 - 6 m = 2.83 10 13 Hz 21.15. Model: We have an open-open tube that forms standing sound waves. Visualize: Please refer to Figure Ex21.15. Solve: The gas molecules at the ends of the tube exhibit maximum displacement, making antinodes at the ends. There is another antinode in the middle of the tube. Thus, this is the m = 2 mode and the wavelength of the standing wave is equal to the length of the tube, that is, l = 0.80 m. Since the frequency f = 500 Hz, the speed of sound in this case is v = f l = (500 Hz)(0.80 m) = 400 m/s. Assess: The experiment yields a reasonable value for the speed of sound.
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21.18. Solve: For the open-open tube, the fundamental frequency of the standing wave is f 1 = 1500 Hz when the tube is filled with helium gas at 0 ° C. Using l m = 2 L m , f 1 helium = v helium l 1 = 970 m / s 2 L Similarly, when the tube is filled with air, f 1 air = v air l 1 = 331 m / s 2 L f 1 air f 1 helium = 331 m / s 970 m / s f 1 air = 331 m / s 970 m / s 1500 Hz ( ) = 512 Hz Assess: Note that the length of the tube is one-half the wavelength whether the tube is filled with helium or air.
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