468 Oscillatory MotionP15.71(a)When the mass is displaced a distance xfromequilibrium, spring 1 is stretched a distance x1andspring 2 is stretched a distance x2.By Newton’s third law, we expectkx1122=.When this is combined with the requirement thatxx x=+12,FIG. P15.71we findxkkkx12=+LNMOQPThe force on either spring is given byFxma1=+LNMOQP=where ais the acceleration of the mass m.This is in the formFkxmaeff==andTmkmkkeff+ππbg(b)In this case each spring is distorted by the distance xwhich the mass is displaced. Therefore,the restoring force isFkkx=−+andkeffso thatTm=+2π.P15.72Let Arepresent the length below water at equilibrium and
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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 .