506_PartUniversity Physics Solution

506_PartUniversity Physics Solution - 16-14 Chapter 16...

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16-14 Chapter 16 EVALUATE: The decrease in intensity level corresponds to a decrease in intensity, and this means an increase in distance. The intensity level uses a logarithmic scale, so simple proportionality between r and β doesn&t apply. 16.57. IDENTIFY: The sound is first loud when the frequency 0 f of the speaker equals the frequency 1 f of the fundamental standing wave for the gas in the tube. The tube is a stopped pipe, and 1 . 4 v f L = . RT v M γ = The sound is next loud when the speaker frequency equals the first overtone frequency for the tube. SET UP: A stopped pipe has only odd harmonics, so the frequency of the first overtone is 31 3. f f = EXECUTE: (a) 01 1 . 44 vR T ff L LM == = This gives 22 0 16 . L Mf T R = (b) 0 f EVALUATE: (c) Measure 0 f and L . Then 0 4 v f L = gives 0 4. vL f = 16.58. IDENTIFY: beat . AB f =− 1 2 v f L = and FL v m = gives 1 1 . 2 F f mL = Apply 0 z τ = to the bar to find the tension in each wire. SET UP: For 0 z = take the pivot at wire A and let counterclockwise torques be positive. The free-body diagram for the bar is given in Figure 16.58. Let L be the length of the bar. EXECUTE: 0 z = gives lead bar (3 /4) ( / 2) 0. B FL w L w L −− = lead bar 3 / 4 /2 3(185 N)/4 (165 N)/2 221 N. B Fw w =+ = + = bar lead FFw w += + so bar lead 165 N 185 N 221 N 129 N. Fw w F = + = 1 3 1 129 N 88.4 Hz. 2 (5.50 10 kg)(0.750 m) A f == × 11 221 N 115.7 Hz. 129 N BA beat 1 1 27.3 Hz. f =−= EVALUATE: The frequency increases when the tension in the wire increases. Figure 16.58 16.59. IDENTIFY: The flute acts as a stopped pipe and its harmonic frequencies are given by Eq.(16.23). The resonant frequencies of the string are 1 , 1, 2, 3,. .. n fn f n The string resonates when the string frequency equals the flute frequency. SET UP: For the string 1s 600.0 Hz. f = For the flute, the fundamental frequency is 1f 344.0 m s 800.0 Hz. 4 4(0.1075 m) v f L = Let f n label the harmonics of the flute and let s n label the harmonics of the string. EXECUTE: For the flute and string to be in resonance, f 1f s 1s 1s , where 600.0 Hz nf f is the fundamental frequency for the string. 4 sf 1 f 1 s f 3 () . nn n s n is an integer when f 3, 1 , nN N = = 3, 5, ± (the flute has only odd harmonics). fs 3 g i v e s 4 Flute harmonic 3 N resonates with string harmonic 4 , 1, NN = 3, 5, ±. EVALUATE: We can check our results for some specific values of N . For N = f 3 n = and 3f 2400 Hz. f = For this N , s 4 n = and 4s 2400 Hz. f = For 3, N = f 9 n = and 9f 7200 Hz, f = and s 12, n = 12s 7200 Hz. f = Our general results do give equal frequencies for the two objects.
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Sound and Hearing 16-15 16.60. IDENTIFY: The harmonics of the string are 1 , 2 n v fn l ⎛⎞ == ⎜⎟ ⎝⎠ where l is the length of the string. The tube is a stopped pipe and its standing wave frequencies are given by Eq.(16.22). For the string, /, vF μ = where F is the tension in the string. SET UP: The length of the string is 10, dL = so its third harmonic has frequency string 3 1 3. 2 fF d = The stopped pipe has length L , so its first harmonic has frequency pipe s 1 .
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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506_PartUniversity Physics Solution - 16-14 Chapter 16...

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