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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 F m f k m F mA F = = = = = = = = F H G I K J × = 0 2 0 2 2 1 0 2 40 0 9 80 2 0 2 0 2 0 2 2 20 0 200 49 0 40 0 9 80 2 00 10 3 950 49 0 318 ω ω ω π π ω ω ω e j e j c h e j e j b g . . . . . . . . s s N P15.48 A F m b m = + ext ω ω ω 2 0 2 2 2 e j b g With b = 0, A F m F m F m = = ± = ± ext ext ext ω ω ω ω ω ω 2 0 2 2 2 0 2 2 0 2 e j e j Thus, ω ω 2 0 2 6 30 0 150 1 70 0 440 = ± = ± = ± F m A k m F mA ext ext N m kg N 0.150 kg m . . . . b ga f This yields ω = 8 23 . rad s or ω = 4 03 . rad s Then, f = ω π 2 gives either f = 1 31 . Hz or f = 0 641 . Hz 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 9 80 0 082 1 1 74 π
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