464 Oscillatory MotionP15.64One can write the following equations of motion:Tkx−=0(describes the spring)mg Tmamdxdt−′= =22(for the hanging object)RT TIddtIRdt′−==af2222θ(for the pulley)with IMR=122FIG. P15.64Combining these equations gives the equation of motionmMdtkxmg+FHGIKJ+=1222.The solution is xtAtmgkaf=+sinω(where mgkarises because of the extension of the spring due tothe weight of the hanging object), with frequencyfkM==+=+πππ2121210002001212Nmkg..(a)For M=0f=356. Hz(b)For M=0250kgf=279Hz(c)For M=0750kgf=210HzP15.65Suppose a 100-kg biker compresses the suspension 2.00 cm.Then,kFx×=×−980490 1024N2.00 10m±Nm.If total mass of motorcycle and biker is 500 kg, the frequency of free vibration isfkm×=12125001584..kgHzIf he encounters washboard bumps at the same frequency, resonance will make the motorcycle
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