# Sol-Ch39 - CHAPTER 39 E xercises 1 = 5/4 so v = 0.6c 2(a 1...

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Exercises 1. γ = 5/4, so v = 0.6c 2. (a) γ ≈ 1 + v 2 /2c 2 ; (b) 1/ γ = 1 – v 2 /2c 2 . 3. L o = γ L = (5/4)(1.2) = 1.5m 4. γ = 10/3, so 1 – v 2 /c 2 = 9/100, thus v = 0.954c. 5. (a) ∆T/T o = γ – 1 ≈ +v 2 /2c; (b) ∆L/L o = 1 – 1/ γ ≈ –v 2 /2c 6. (a) γ = 1.005, so T – T o = T o ( γ – 1) = 43.9 h (b) γ = 15.8, so T – T o = T o ( γ – 1) = 14.8 h 7. Use Exercise 2: T – T o = ( γ – 1)T o ≈ (v 2 /2c 2 )T o Thus, v/c = (2∆T/T o ) 1/2 = 2.5 × 10 –4 , so v = 75 km/s 8. T o = 2.2 μs, L o = 400 m. In frame S, v = L o / γ T o , so v 2 γ 2 = (L o /T o ) 2 . Solve to find v = 1.56 ×  10 8 m/s. 9. (a) γ = 5/3, T = γ T o = 8.33 μs; (b) L o = (0.8c)T = 2 km; (c) L = L o / γ = 1.2 km 10. T – T o = T(1 – 1/ γ ) = (500 s)(–v 2 /2c 2 ) = –4.44 ×  10 –10 s 11. T = γ T o and γ = 2. Find v = 0.866c 12. (a) T o = L o / γ v = (4.2y ×  c)/5(0.98c) = 0.857 y (b) T = γ T o = 4.29 y; (c) L = L o / γ = 0.84 1y. 13. (a) L o = γ L = (1.02)(150) = 153 m; (b) T o = 150 m/v = 2.5 µs; (c) T = γ T o = 2.55 µs 14. T o = L/v = L o / γ v, thus γ v = 80c/70 = 3.43 ×  108 m/s. Solve to find v = 2.26 ×  108 m/s CHAPTER 39

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15. (a) 320/0.6c = 1.75 µs (b) T = γ T o = (5/4)(1.78 µs) = 2.22 µs 16. (a) L o = γ L = 1.5 km; (b) 1.5 km/0.6c = 8.33 µs (c) T o = T/ γ = 4T/5 = 6.67 µs 17. (a) T o = 120/0.98c = 0.408 µs (b) T = γ T o = 5.03T o = 2.05 µs 18. (a) γ = L o /L = 4.2/3.6 = 1.166; and (1 – v 2 /c 2 ) = 1/ γ 2 Find v = 0.52c (b) T o = L o / γ v = 24 y; thus γ v = 0.175c. Solve to find v = 1.72c = 5.25 ×  10 7 m/s (c) T = 4.2c/0.172c = 24.4 y 19. (a) 500 km/ γ v = 8.17 ms; (b) 500 km/ γ = 490 km 20. (a) γ = 5/3, so T = (5/3)T o = 43.3 ns; (b) vT = 10.4 m; (c) vT o = 6.24 m 21. (a) γ = 10, so T = 10T o = 22 µs (b) T = 10 km/v = 33.5 µs; (c) T o = 3.35 µ 22. (a) f’ = (1.6/0.4) 1/2 f o = 2f o = 144 beats/min (b) f’ = (0.4/1.6) 1/2 f o = 0.5f o = 36 beats/min 23. f o = 10 10 Hz, and v = 30 m/s. Received by car f 1 = [(c + v)/(c – v)] 1/2 f o Received by police f 2 = [(c + v)/(c – v)] 1/2 f 1 ∆f = f 2 – f o = 2vf o /(c – v) = 2 kHz 24. λ = [(c – v)/(c + v)] 1/2 λ o , thus (c – v)/(c + v) = 25/49 Find v = 0.324c 25. λ = [(c + v)/(c – v)] 1/2 λ o = (1.2/0.8) 1/2 λ o , so λ o = 490 nm 26. f” = [(c + v)/(c – v)] f o = 1.22 ×  10 9 Hz
27. (a) f” = [(c + v)/(c – v)]f o . Note (1 – v/c) –1 ≈ 1 + v/c. Find f” – f

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## This note was uploaded on 11/09/2011 for the course EE EE taught by Professor Ee during the Spring '11 term at National Chiao Tung University.

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Sol-Ch39 - CHAPTER 39 E xercises 1 = 5/4 so v = 0.6c 2(a 1...

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