453_Physics ProblemsTechnical Physics

453_Physics ProblemsTechnical Physics - Chapter 15 Section...

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Chapter 15 455 Section 15.7 Forced Oscillations P15.44 (a) For resonance, her frequency must match f k m 0 0 3 2 1 2 1 2 430 10 12 5 295 == = × = ω ππ π . . . Nm kg Hz . (b) From xA t = cos , v dx dt At ωω sin , and a dv dt 2 cos , the maximum acceleration is A 2 . When this becomes equal to the acceleration due to gravity, the normal force exerted on her by the mattress will drop to zero at one point in the cycle: Ag 2 = or A gg g m k k m = 2 A = × = 980 125 285 3 .. . . ms kg cm 2 ej bg P15.45 Ft = 300 2 .c o s N and k = 20 0 . Nm (a) 2 2 T rad s so T = 100 . s (b) In this case, 0 20 0 200 316 = k m . . r a d s The equation for the amplitude of a driven oscillator, with b = 0, gives A F m = F H G I K J −= 0 2 0 2 1 2 2 1 3 2 43 1 6 af . Thus A 00509 509 m c m . P15.46 k x m dx dt 0 2 2 cos 0 = k m (1) t =+ cos
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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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