Physics Solution Manual for 1100 and 2101

Chapter 17 problems 911 12 reasoning equation 171

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Unformatted text preview: R SUPERPOSITION AND INTERFERENCE PHENOMENA ANSWERS TO FOCUS ON CONCEPTS QUESTIONS ____________________________________________________________________________________________ 1. (d) If we add pulses 1 and 4 as per the principle of linear superposition, the resultant is a straight horizontal line that extends across the entire graph. 2. (a) These two pulses combine to produce a peak that is 4 units high and a valley that is 2 units deep. No other combination gives greater values. 3. (c) The smallest difference in path lengths for destructive interference to occur is one-half a wavelength ( 1 λ ) . As the frequency goes up, the wavelength goes down, so the separation 2 between the cellists decreases. 4. Smallest separation = 1.56 m 5. (b) According to Equation 17.2, the diffraction angle θ is related to the wavelength and diameter by sin θ = 1.22 ( λ / D ) and is determined by the ratio λ/D. Here the ratio is 2λ0 / D0 and is the largest of any of the choices, so it yields the largest diffraction angle. 6. θ = 30.0 degrees 7. (e) Since the wavelength is directly proportional to the speed of the sound wave (see Section 16.2), the wavelength is greatest in the helium-filled room. The greater the wavelength, the greater the diffraction angle θ (see Section 17.3). Thus, the greatest diffraction occurs in the helium-filled room. 8. (d) The trombones produce 6 beats every 2 seconds, so the beat frequency is 3 Hz. The second trombone can be producing a sound whose frequency is either 438 Hz − 3 Hz = 435 Hz or 438 Hz + 3 Hz = 441 Hz. 9. Beat frequency = 3.0 Hz 10. (d) According to the discussion in Section 17.5, one loop of a transverse standing wave corresponds to one-half a wavelength. The two loops in the top picture mean that the wavelength of 1.2 m is also the distance L between the walls, so L = 1.2 m. The bottom picture contains three loops in a distance of 1.2 m, so its wavelength is 2 (1.2 m ) = 0.8 m. 3 Chapter 17 Answers to Focus on Concepts Questions 901 11. (b) The frequency of a standing wave is directly proportional to the speed of the traveling waves that form it (see Equation 17.3). The speed of the waves, on the other hand, depends on the mass m of the string through the relation v = F / ( m / L ) , so the smaller the mass, the greater is the speed and, hence, the greater the frequency of the standing wave. 12. (c) For a string with a fixed length, tension, and linear density, the frequency increases when the harmonic number n increases from 4 to 5 (see Equation 17.3). According to λ = v/f (Equation 16.1), the wavelength decreases when the frequency increases. 13. Fundamental frequency = 2.50 × 102 Hz 14. (c) One loop of a longitudinal standing wave corresponds to one-half a wavelength. Since this standing wave has two loops, its wavelength is equal to the length of the tube, or 0.80 m. 15. (b) There are two loops in this longitudinal standing wave. This means that the 2nd harmonic is being generated. According to Equation 17.4, the nth harmonic frequency is v v is the fundamental frequency. Since f2 = 440 Hz and n = 2, we fn = n , where 2L 2L v 440 Hz have = = 220 Hz . 2L 2 16. (c) The standing wave pattern in the drawing corresponds to n = 3 (the 3rd harmonic) for a tube open at only one end. Using Equation 17.5, the length of the tube is n v ( 3) ( 343 m/s ) L= = = 0.39 m . 4 fn 4 ( 660 Hz ) 17. Frequency of 3rd harmonic = 9.90 × 102 Hz 902 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA CHAPTER 17 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA PROBLEMS 1. REASONING For destructive interference to occur, the difference in travel distances for the sound waves must be an integer number of half wavelengths. For larger and larger distances between speaker B and the observer at C, the difference in travel distances becomes smaller and smaller. Thus, the largest possible distance between speaker B and the observer at C occurs when the difference in travel distances is just one half wavelength. SOLUTION Since the triangle ABC in Figure 17.7 is a right triangle, we can apply the 2 ( 5.00 m )2 + d BC . Pythagorean theorem to obtain the distance dAC as Therefore, the difference in travel distances is 2 ( 5.00 m )2 + d BC − d BC = λ 2 = v 2f where we have used Equation 16.1 to express the wavelength λ as λ = v/f. Solving for the distance dBC gives 2 ( 5.00 m ) + d BC = d BC + 2 2 2 ( 5.00 m )2 + d BC = d BC + ( 5.00 m ) d BC = 2. 2 v f v2 −2 4f v 2f d BC v f or + v2 4f2 ( 5.00 m ) − 2 = v 2 ( 5.00 m )2 + d BC = d BC + 2f or ( 5.00 m )2 = ( 343 m/s )2 2 4 (125 Hz ) 343 m/s 125 Hz d BC v f + 2 v2 4f2 = 8.42 m REASONING When the difference in path lengths traveled by the two sound waves is a half-integer number ( 1 , 1 1 , 2 1 , K) 2 2 2 of wavelengths, the waves are out of phase and destructive interference occurs at the listener. The smallest separation d between the Chapter 17 Problems 903 speakers is when the difference in path lengths is 1 of a wavelength, so d = 1 λ...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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