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Question

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, then
  1. t=T12
  2. v=v02
  3. a=a02
  4. t=T8

A
a=a02
B
t=T8
C
t=T12
D
v=v02
Solution
Verified by Toppr

Let the displacement at t is x=Asinωt=Asin(2πT)t
here,x=A2=Asin(2πT)t
or, sin(2πT)t=12=sinπ6t=T12
dxdt=2πTAcos(2πT)t
d2xdt2=(2πT)2Asin(2πT)t
so, vmax=v0=(2πT)A,amax=a0=(2πT)2A
now at t=T/12, the velocity,v=dxdt=2πTAcos(2πT)T12
=2πTAcosπ6=2πTA32
v=3v0
at t=T/12, the acceleration,a=d2xdt2=(2πT)2Asin(2πT)T12
=2πTAsinπ6=2πTA12
a=a02

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