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# A particle is executing SHM with amplitude A, time period T, maximum acceleration a0 and maximum velocity v0. It starts from mean position at t=0 and at time t it has the displacement A/2, acceleration a and velocity v, thent=T12v=v02a=a02t=T8

A
a=a02
B
t=T8
C
t=T12
D
v=v02
Solution
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#### Let the displacement at t is x=Asinωt=Asin(2πT)there,x=A2=Asin(2πT)tor, sin(2πT)t=12=sinπ6⇒t=T12dxdt=2πTAcos(2πT)td2xdt2=−(2πT)2Asin(2πT)tso, vmax=v0=(2πT)A,amax=a0=(2πT)2Anow at t=T/12, the velocity,v=dxdt=2πTAcos(2πT)T12=2πTAcosπ6=2πTA√32∴v=√3v0at t=T/12, the acceleration,a=d2xdt2=(2πT)2Asin(2πT)T12=2πTAsinπ6=2πTA12∴a=a02

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