# Ch15Word - Chapter 15 CHAPTER 15 Wave Motion 1 The speed of...

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Chapter 15 Page 1 CHAPTER 15 – Wave Motion 1. The speed of the wave is v = f λ = / T = (9.0 m)/(4.0 s) = 2.3 m/s . 2. For AM we find the wavelengths from AMhigher = v / f AMlower = (3.00 × 10 8 m/s)/(550 × 10 3 Hz) = 545 m ; AMlower = v / f AMhigher = (3.00 × 10 8 m/s)/(1600 × 10 3 Hz) = 188 m . For FM we have FMhigher = v / f FMlower = (3.00 × 10 8 m/s)/(88 × 10 6 Hz) = 3.4 m ; FMlower = v / f FMhigher = (3.00 × 10 8 m/s)/(108 × 10 6 Hz) = 2.78 m . 3. We find the wavelength from v = f ; 330 m/s = (262 Hz) , which gives = 1.26 m . 4. We find the speed of the longitudinal (compression) wave from v = ( B / ρ ) 1/2 for fluids and v = ( E / ) 1/2 for solids. ( a ) For water we have v = ( B / ) 1/2 = [(2.0 × 10 9 N/m 2 )/(1.00 × 10 3 kg/m 3 )] 1/2 = 1.4 × 10 3 m/s . ( b ) For granite we have v = ( E / ) 1/2 = [(45 × 10 9 N/m 2 )/(2.7 × 10 3 kg/m 3 )] 1/2 = 4.1 × 10 3 m/s . 5. The speed of the longitudinal (compression) wave is v = ( E / ) 1/2 , so the wavelength is = v / f = ( E / ) 1/2 / f = [(100 × 10 9 N/m 2 )/(7.8 × 10 3 kg/m 3 )] 1/2 /(5000 Hz) = 0.72 m . 6. We find the speed of the wave from v = [ F T /( m / L )] 1/2 = {(120 N)/[(0.65 kg)/(30 m)]} 1/2 = 74.4 m/s. We find the time from t = L / v = (30 m)/(74.4 m/s) = 0.40 s. 7. We find the tension from the speed of the wave: v = [ F T /( m / L )] 1/2 ; (4.8 m)/(0.85 s) = { F T /[(0.40 kg)/(4.8 m)]} 1/2 , which gives F T = 2.7 N . 8. The speed of the longitudinal wave is v = ( B / ) 1/2 , so the distance down and back that the wave traveled is 2 D = vt = ( B / ) 1/2 t ; 2 D = [(2.0 × 10 9 N/m 2 )/(1.00 × 10 3 kg/m 3 )] 1/2 (3.5 s), which gives D = 2.5 × 10 3 m = 2.5 km . 9. ( a ) Because the pulse travels up and back, the speed is v = 2 L / t = 2(600 m)/(16 s) = 75 m/s . ( b ) The mass density of the cable is µ = m / L = AL / L = A . We find the tension from v = ( F T / ) 1/2 = ( F T / A ) 1/2 ; 75 m/s = [ F T /(7.8 × 10 3 kg/m 3 )p(0.75 × 10 –2 m) 2 ] 1/2 , which gives F T = 7.8 × 10 3 N .

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Chapter 15 Page 2 10. ( a ) The shape is maintained and moves 1.80 m in 1.00 s. ( b ) At the instant shown, point A is moving down. We use the slope of the string to estimate the vertical speed. In the time to move vertically from 1 cm to – 1 cm, we estimate the string moves 30 cm, so the time is ? t = (0.30 m)/(1.80 m/s) = 0.17 s. The vertical velocity is – (2 cm)/(0.17 s) = – 0.1 m/s (down) . 11. ( a ) Because both waves travel the same distance, we have ? t = ( d / v S ) – ( d / v P ) = d [(1/ v S ) – (1/ v P )]; 94 s = d {[1/(5.5 km/s)] – [1/(9.0 km/s)]}, which gives d = 1.3 × 10 3 km . ( b ) The direction of the waves is not known, thus the position of the epicenter cannot be determined . Because two circles have two intersections, it would take at least two more stations. 12. Because the speed, frequency, and medium are the same for the two waves, the intensity depends on the amplitude only: I D M 2 . For the ratio of intensities we have I 2 / I 1 = ( D M2 / D M1 ) 2 ; 2 = ( D M2 / D M1 ) 2 , which gives D M2 / D M1 = 1.41 . 13. We assume that the wave spreads out uniformly in all directions. ( a ) The intensity will decrease as 1/ r 2 , so the ratio of intensities is I 2 / I 1 = ( r 1 / r 2 ) 2 = [(10 km)/(20 km)] 2 = 0.25 . ( b ) Because the intensity depends on D M 2 , the amplitude will decrease as 1/ r , so the ratio of amplitudes is D M2 / D M1 = r 1 / r 2 = (10 km)/(20 km) = 0.50 . 14. We assume that the wave spreads out uniformly in all directions.
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## This note was uploaded on 02/05/2010 for the course PHYSICS 37,38,39 taught by Professor Fielding during the Spring '09 term at Los Angeles Valley College.

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Ch15Word - Chapter 15 CHAPTER 15 Wave Motion 1 The speed of...

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