Ch15Word - Chapter 15 CHAPTER 15 – Wave Motion 1 The...

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Unformatted text preview: Chapter 15 CHAPTER 15 – Wave Motion 1. The speed of the wave is v = f λ = λ / T = (9.0 m)/(4.0 s) = 2.3 m/s . 2. For AM we find the wavelengths from λ AMhigher = v / f AMlower = (3.00 × 10 8 m/s)/(550 × 10 3 Hz) = 545 m ; λ AMlower = v / f AMhigher = (3.00 × 10 8 m/s)/(1600 × 10 3 Hz) = 188 m . For FM we have λ FMhigher = v / f FMlower = (3.00 × 10 8 m/s)/(88 × 10 6 Hz) = 3.4 m ; λ 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 λ ; 330 m/s = (262 Hz) λ , which gives λ = 1.26 m . 4. We find the speed of the longitudinal (compression) wave from v = ( B / ρ ) 1/2 for fluids and v = ( E / ρ ) 1/2 for solids. ( a ) For water we have v = ( B / ρ ) 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 / ρ ) 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 / ρ ) 1/2 , so the wavelength is λ = v / f = ( E / ρ ) 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 / ρ ) 1/2 , so the distance down and back that the wave traveled is 2 D = vt = ( B / ρ ) 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 = ρ AL / L = ρ A . We find the tension from v = ( F T / μ ) 1/2 = ( F T / ρ 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 . Page 1 Chapter 15 10. ( a ) The shape is maintained and moves 1.80 m in 1.00 s....
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This homework help was uploaded on 03/26/2008 for the course PHYS 208 taught by Professor Ross during the Spring '08 term at Texas A&M.

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Ch15Word - Chapter 15 CHAPTER 15 – Wave Motion 1 The...

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