506_PartUniversity Physics Solution

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

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16-14 Chapter 16 E VALUATE : 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. I DENTIFY : 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. S ET U P : A stopped pipe has only odd harmonics, so the frequency of the first overtone is 3 1 3 . f f = E XECUTE : (a) 0 1 1 . 4 4 v RT f f L L M γ = = = This gives 2 2 0 16 . L Mf T R γ = (b) 0 3 . f E VALUATE : (c) Measure 0 f and L . Then 0 4 v f L = gives 0 4 . v Lf = 16.58. I DENTIFY : beat . A B f f 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. S ET U P : For 0 z τ = 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. E XECUTE : 0 z τ = gives lead bar (3 /4) ( / 2) 0. B F L w L w L = lead bar 3 / 4 /2 3(185 N)/4 (165 N)/2 221 N. B F w w = + = + = bar lead A B F F w w + = + so bar lead 165 N 185 N 221 N 129 N. A B F w w F = + = + = 1 3 1 129 N 88.4 Hz. 2 (5.50 10 kg)(0.750 m) A f = = × 1 1 221 N 115.7 Hz. 129 N B A f f = = beat 1 1 27.3 Hz. B A f f f = = E VALUATE : The frequency increases when the tension in the wire increases. Figure 16.58 16.59. I DENTIFY : 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 f nf n = = The string resonates when the string frequency equals the flute frequency. S ET U P : 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. E XECUTE : For the flute and string to be in resonance, f 1f s 1s 1s , where 600.0 Hz n f n f f = = is the fundamental frequency for the string. 4 s f 1f 1s f 3 ( ) . n n f f n = = s n is an integer when f 3 , 1, n N N = = 3, 5, ± (the flute has only odd harmonics). f s 3 gives 4 n N n N = = Flute harmonic 3 N resonates with string harmonic 4 , 1, N N = 3, 5, ±. E VALUATE : We can check our results for some specific values of N . For 1, 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. I DENTIFY : The harmonics of the string are 1 , 2 n v f nf n 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, / , v F μ = where F is the tension in the string. S ET U P : The length of the string is 10, d L = so its third harmonic has frequency string 3 1 3 .
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