# Chapter 13 - 13.1: a) b) T 1 4 ( 220 Hz) 1 f 4 . 55 1 . 14...

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13.1 : a) s. rad 10 38 . 1 2 s, 10 55 . 4 3 2 3 1 × = = = × = = - πf ω T T π f b) s. rad 10 53 . 5 2 s, 10 14 . 1 3 3 Hz) 220 ( 4 1 × = = × = - πf ω 13.2: a) Since the glider is released form rest, its initial displacement (0.120 m) is the amplitude. b) The glider will return to its original position after another 0.80 s, so the period is 1.60 s. c) The frequency is the reciprocal of the period (Eq. (13.2)), = = s 60 . 1 1 f Hz. 625 . 0 13.3: The period is s 10 14 . 1 3 440 s 50 . 0 - × = and the angular frequency is = = T π ω 2 s. rad 10 53 . 5 3 × 13.4: (a) From the graph of its motion, the object completes one full cycle in 2.0 s; its period is thus 2.0 s and its frequency . s 5 . 0 period 1 1 - = = (b) The displacement varies from m, 20 . 0 to m 20 . 0 + - so the amplitude is 0.20 m. (c) 2.0 s (see part a) 13.5: This displacement is 4 1 of a period. s. 0500 . 0 so s, 200 . 0 1 = = = t f T 13.6: The period will be twice the time given as being between the times at which the glider is at the equilibrium position (see Fig. (13.8)); m. N 292 . 0 kg) 200 . 0 ( s) 60 . 2 ( 2 2 2 2 2 2 = = = = π m T π m ω k 13.7: a) kg. 084 . 0 c) s. rad 7 . 37 2 b) s. 167 . 0 2 1 = = = = = = ω k f m πf ω T 13.8: Solving Eq. (13.12) for k , m. N 10 05 . 1 s 150 . 0 2 kg) 600 . 0 ( 2 3 2 2 × = = = π T π m k 13.9: From Eq. (13.12) and Eq. (13.10), Hz, 66 . 2 s, 375 . 0 2 1 m N 140 kg 500 . 0 = = = = T f π T s. rad 7 . 16 2 = = πf ω 13.10: a) ) ( so , ) sin( 2 2 2 2 t x x ω β ωt A ω a dt x d x - = + - = = is a solution to Eq. (13.4) if ω A a ω m k 2 b) . 2 = = a constant, so Eq. (13.4) is not satisfied. c) , ) ( β ωt i dt dx x v + = = = - = = = + m k ω t x x ω Ae a β ωt i dt dv x x 2 2 ) ( 2 if (13.4) Eq. o solution t a is ) ( so , ) (

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