454_Physics ProblemsTechnical Physics

454_Physics ProblemsTechnical Physics - 456 P15.47...

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456 Oscillatory Motion P15.47 From the equation for the amplitude of a driven oscillator with no damping, A Fm f k m A F = == = = = =− = F H G I K J ×− = −− 0 2 0 2 2 1 0 2 40 0 980 2 0 2 0 2 0 2 22 0 0 200 49 0 40 0 200 10 3950 490 318 ωω ωπ π ω ej ch bg .. . . . . s s N P15.48 A bm = −+ ext 2 0 2 2 2 With b = 0, A = = ±− ext ext ext 2 0 2 2 2 0 2 2 0 2 Thus, 2 0 2 630 0150 170 0440 = ± = ± A k m F mA ext ext Nm kg N 0.150 kg m . . . . a f This yields = 823 . rad s or = 403 . r a d s Then, f = 2 gives either f = 131 H z o r f = 0641 H z P15.49 The beeper must resonate at the frequency of a simple pendulum of length 8.21 cm: f g L = 1 2 1 2 00821 174 ππ . . . ±ms m Hz 2 . *P15.50 For the resonance vibration with the occupants in the car, we have for the spring constant of the
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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