ssm_ch17 - Chapter 17 Student Solutions Manual 5. Let tf be...

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Chapter 17 – Student Solutions Manual 5. Let t f be the time for the stone to fall to the water and t s be the time for the sound of the splash to travel from the water to the top of the well. Then, the total time elapsed from dropping the stone to hearing the splash is t = t f + t s . If d is the depth of the well, then the kinematics of free fall gives 2 1 2 f dg t = , or 2/. f td = g The sound travels at a constant speed v s , so d = v s t s , or t s = d / v s . Thus the total time is 2/ / s g d =+ v . This equation is to be solved for d . Rewrite it as / s dg t dv =− and square both sides to obtain 2 d / g = t 2 – 2( t / v s ) d + (1 + 2 s v ) d 2 . Now multiply by g 2 s v and rearrange to get gd 2 – 2 v s ( gt + v s ) d + g 2 s v t 2 = 0. This is a quadratic equation for d . Its solutions are () 2 22 2( ) 4 4 . 2 ss s s s vg tv g v t d g + = 2 2 The physical solution must yield d = 0 for t = 0, so we take the solution with the negative sign in front of the square root. Once values are substituted the result d = 40.7 m is obtained. 7. If d is the distance from the location of the earthquake to the seismograph and v s is the speed of the S waves then the time for these waves to reach the seismograph is t s . = d / v s . Similarly, the time for P waves to reach the seismograph is t p = d / v p . The time delay is Δ t = ( d / v s ) – ( d / v p ) = d ( v p v s )/ v s v p , so 3 (4.5 km/s)(8.0km/s)(3.0min)(60s /min) 1.9 10 km. ( ) 8.0km/s 4.5km/s sp ps vv t d vv Δ == = × −− We note that values for the speeds were substituted as given, in km/s, but that the value for the time delay was converted from minutes to seconds. 9. (a) Using λ = v / f , where v is the speed of sound in air and f is the frequency, we find 5 6 343m/s 7.62 10 m. 4.50 10 Hz λ= = × ×
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(b) Now, λ = v / f , where v is the speed of sound in tissue. The frequency is the same for air and tissue. Thus λ = (1500 m/s)/(4.50 × 10 6 Hz) = 3.33 × 10 –4 m. 19. Let L 1 be the distance from the closer speaker to the listener. The distance from the other speaker to the listener is 2 21 2 L Ld = + , where d is the distance between the speakers. The phase difference at the listener is φ = 2 π ( L 2 L 1 )/ λ , where λ is the wavelength. For a minimum in intensity at the listener, = (2 n + 1) π , where n is an integer. Thus λ = 2( L 2 L 1 )/(2 n + 1). The frequency is ( ) ( ) 22 2 2 11 (2 1) 1)(343m/s) (2 1)(343Hz). 2 2 (3.75m) (2.00m) 3.75m vn v n fn Ld L ++ == = = + λ +− + Now 20,000/343 = 58.3, so 2 n + 1 must range from 0 to 57 for the frequency to be in the audible range. This means n ranges from 0 to 28. (a) The lowest frequency that gives minimum signal is ( n = 0) min,1 343 Hz. f = (b) The second lowest frequency is ( n = 1) min,2 min,1 [2(1) 1]343 Hz 1029 Hz 3 . ff = +== Thus, the factor is 3. (c) The third lowest frequency is ( n =2) min,3 min,1 [2(2) 1]343 Hz 1715 Hz 5 . = += = Thus, the factor is 5. For a maximum in intensity at the listener, = 2 n π , where n is any positive integer. Thus ( ) (1/ ) nL dL λ= + and 2 2 (343m/s) (686Hz). 3.75m v n = = λ + Since 20,000/686 = 29.2, n must be in the range from 1 to 29 for the frequency to be audible.
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This note was uploaded on 04/09/2008 for the course PHYS 131 taught by Professor Drty during the Spring '08 term at Lafayette.

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ssm_ch17 - Chapter 17 Student Solutions Manual 5. Let tf be...

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