Chapter 15455Section 15.7Forced OscillationsP15.44(a)For resonance, her frequency must matchfkm00321212430 1012 5295===×=ωπππ...NmkgHz .(b)From xAt=cos, vdxdtAt−ωωsin, and advdt−2cos, the maximum accelerationis A2. When this becomes equal to the acceleration due to gravity, the normal forceexerted on her by the mattress will drop to zero at one point in the cycle:Ag2=orAgggmkkm=2A=×=9801252853....mskgcm2ejbgP15.45Ft=3002.cosNandk=20 0. Nm(a)22Trad ssoT=100. s(b)In this case, 020 0200316=km..radsThe equation for the amplitude of a driven oscillator,with b=0, givesAFm=FHGIKJ−=−−−02021221324316af.ThusA00509509m cm.P15.46kxmdxdt022cos0=km(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 .