HW2Solutions - HW2 Solutions Notice numbers may change...

HW2Solutions
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HW2 Solutions Notice numbers may change randomly in your assignments and you may have to recalculate solutions for your specific case. Tipler 15.P.041 The wave function for a harmonic wave on a string is y ( x , t ) = ( 0.0030 m) sin(( 58.8 m -1 ) x + ( 312 s -1 ) t ). (a) In what direction does this wave travel? What is its speed? (b) Find the wavelength of this wave. Find its frequency. Find its period. (c) What is the maximum speed of any string segment? Solution: (a) The wave function is: y ( x , t ) = ( 0.0030 m) sin(( 58.8 m -1 ) x + ( 312 s -1 ) t ) = A sin(kx- ω t). So the wave is traveling in the direction of –x with speed v = " k = 312 58.8 = 5.31 m / s (b) the wavelength, frequency and period are = 2 # k =0.107m;f = $ 2 = 49.7 Hz ; T = 1 f = 0.02 s (c) the speed of any string segment is v = dy dt = A cos( kx # t ) and the maximum is for cos(kx- ω t)=±1, hence v max = A ω = 0.0030 x 312 rad/s = 0.936 m/s. Tipler 15.P.043 Waves of frequency 200 Hz and amplitude 1.2 cm move along a 20 m string that has a mass of 0.060 kg and a tension of 50 N. (a) What is the average total energy of the waves on the string? (b) Find the power transmitted past a given point on the string. Solution:
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(a) The average energy on a segment of length L is the average power P av multiplied by the time the wave takes to travel the distance L: ( " E ) av = P av " t = 1 2 μ # 2 v A 2 " t = 1 2 2 A 2 " x = 0.5 $ 0.06 20 $ (2 % $ 200) 2 $ (1.2 $ 10 2 ) 2 $ 20 = 6.82 J (b) The wave speed is: v = F T = 129.10 m / s P av = 1 2 " 2 v A 2 = 44.04 W Tipler 16.P.031 A transverse wave of frequency 32 Hz propagates down a string. Two points 8 cm apart are out of phase by π /6. (a) What is the wavelength of the wave? (b) At a given point, what is the phase difference between two displacements for times 5 ms apart? (c) What is the wave velocity? Solution: (a) The phase difference is related to the displacement between the 2 points by the relationship: = 2 $ x = 2 $ x = 2 8 10 ( 2 /6 = 0.96 m (b) v = f # $ = 2 v t = 2 f t = 2 32 5 10 ( 3 = 1.005 rad (c) the wave velocity is: v = f = 30.7 m / s Tipler 16.P.027 Two compact sources of sound oscillate in phase with a frequency of 120 Hz where the speed of sound is 340 m/s. At a point that is 3.20 meters from one source and 3.54 meters from the other, the amplitude of the sound from each source separately is A . (a) What is the phase difference in the sound waves from the two sources at that point?
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(b) What is the amplitude of the resultant wave at that point? Solution: (a) " = 2 # $ x % = 2 f $ x v = 2 120 (3.54 3.20) 340 = 0.75 rad (b) the amplitude of the resultant wave at that point is: E 1 = A sin ω t E 2 = A sin ( ω t + δ ) E P = E 1 + E 2 = A [sin ω t + sin ( ω t + δ )] E P = 2 A cos 2 sin t + 2 $ % ( ) where we used sin + sin = 2cos 1 2 ( $ )sin 1 2 ( + ) Hence, E P = 2 A cos 2 sin t + 2 $ % ( ) * A P = 2 A cos 2 $ % ( ) = 2 A cos(0.75/2) = 1.86 A Tipler 16.P.050 The normal range of hearing is about 20 to 20000 Hz. Assume
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