A particle is executing SHM on a straight line. A and B are two points at which its  velocity is zero. It passes through a certain point P (AP < BP) at successive  intervals of 0.5 s and 1.5 s with a speed of  3ms−1. Then the

It is clear that time period, T = 0.5 sec + 1.5 sec  = 2 secLet O be the mean position. tPAP=0.5  sec.  But tPA=tAP   So,  tPA=tAP=0.25  secSimilarly, tPBP=1.5  sec  But    tPB=tBPSo, tPB=tBP=0.75  sec  tPB=tPO+tOB⇒0.75=tPO+T4=tPO+0.5⇒tPO=0.25  secSo, all segments AP, PO, OB, BO, OP, PA take same amount of time = 0.25 seconds.Let equation of motion be  x=−RsinωtSubstituting the values,  x=−OP=−Rsin2π20.25⇒OP=R2mSo, v=−Rωcosωt Substituting the values, −3=−R2π2cos2π2tPO⇒−3=−R2π2cos2π20.25 ⇒R=32πmMaximum speed =  Rω=32π2π2=32ms−1 APBP=AO−POBO+OP=R−R2R+R2=2−12+1.