462_Physics ProblemsTechnical Physics

462_Physics ProblemsTechnical Physics - 464 P15.64...

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464 Oscillatory Motion P15.64 One can write the following equations of motion: Tk x −= 0 (describes the spring) mg T ma m dx dt −′ = = 2 2 (for the hanging object) RT T I d dt I R dt ′− = = a f 2 2 2 2 θ (for the pulley) with IM R = 1 2 2 FIG. P15.64 Combining these equations gives the equation of motion mM dt kx mg + F H G I K J += 1 2 2 2 . The solution is xt A t mg k af =+ sin ω (where mg k arises because of the extension of the spring due to the weight of the hanging object), with frequency f k M == + = + ππ π 2 1 2 1 2 100 0200 1 2 1 2 Nm kg . . (a) For M = 0 f = 356 . H z (b) For M = 0250 k g f = 279 H z (c) For M = 0750 k g f = 210 H z P15.65 Suppose a 100-kg biker compresses the suspension 2.00 cm. Then, k F x × 980 490 10 2 4 N 2.00 10 m ±Nm . If total mass of motorcycle and biker is 500 kg, the frequency of free vibration is f k m × = 1 2 1 2 500 158 4 . . kg Hz If he encounters washboard bumps at the same frequency, resonance will make the motorcycle
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