# Sol-Ch17 - CHAPTER 17 E xercises 1 = v/f = 3.4 mm 2 = v/f =...

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Exercises 1. λ = v/f = 3.4 mm 2. λ = v/f = 9.7 mm 3. λ = (1500 m/s)/(4 ×   10 6 Hz) = 0.375 mm 4. D(1/340 –1/1500) = 3.2 s, thus D = 1.41 km 5. (a) v = (B/ ρ ) 1/2 = 1.43 km/s ; (b) λ = v/f = 1.43 m 6. v = (B/ ρ ) 1/2 : (a) 314 m/s; (b) 972 m/s 7. v = (Y/ ρ ) 1/2 = 5.06 km/s 8. Δt = (100 m)(1/340 – 1/1320) = 218 ms 9. v = (S/ ρ ) 1/2 = 3.04 km/s 10. v = (15.4 × 10 10 / ρ ) 1/2 = 416 km/s 11. ∂s/∂t = – ω s o cos(kx – ω t). Using v = ω /k and P o = Bks o ’ , we see that (B/v) ∂s/∂t = P (See Eq. 17.14) 12. P = 2 sin(5.3x + 1800t) 13. λ = v/f =77.3 cm. L 1 = λ /4 = 19.3 cm; L 2 = 3 λ /4 = 58.0 cm 14. Plot f vs 1/2 L, then slope, v = 350 m/s 15. Pipe : λ = 4L/5 = 0.8 m = 340/f, thus f = 425 Hz. String: λ = L = 0.6 m = v/f; so v = 255 m/s. F = µv 2 = 78 N 16. (a) F = (40)(1200/60) = 800 Hz; (b) No, there are many harmonics. 17. f = nv/2L : 8.5 Hz, 17 Hz, 25.5 Hz 18. f 2 /f 1 = (T 2 /T 1 ) 1/2 , f 2 =983 Hz CHAPTER 17

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19. (a) f = v/2L = 283 Hz; (b) L = v/2f = 51.5 cm 20. (a) L = v/4f = 3.4 m; (b) L = v/2L = 34 cm. 21. f 20 = v/2L = 573.3 Hz; f 5 = 558.3 Hz, thus 22. (a) (340/310)(1200) = 1320 Hz; (b) (340/370)(1200) = 1100 Hz 23. (a) f’ = 400(340/315) = 431.7 Hz, λ ’ = v/f’ = 340/431.7 = 78.8 cm (b) f’ = 400(340/365) = 372.6 Hz, λ ’ = v/f’ = 340/372.6 = 91.3 cm 24. λ = v/f o : (a) 85 cm; (b) 85 cm 25. (a) f’ = 200(340/300) = 227 Hz, λ ’ = v/f’ = 340/227 = 1.50 m (b) f’ = 200(380/340) =224 Hz, λ ’ = v/f’ = 380/224 = 1.70 m (c) f’ = 200(360/320) =225 Hz, λ ’ = v/f’ = 360/225 = 1.60 m 26. v = 18.1 m/s. f’ = 1800(340)/(340 + 18.1) = 1710 Hz, 1900 Hz 27. (a) f 1 ’ = 400(325/300) = 433.3 Hz, f 2 ’ = 400(355/380) = 373.7 Hz. Thus Δf’ = 59.6 Hz; (b) f 1 ’ = 400(355/300) = 473.3 Hz, f 2 ’ = 400(325/380) = 342.1 Hz. Thus Δf’ = 131 Hz 28. f’ = 600(320/300) = 640 Hz, received by truck
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