A particle performs simple harmonic motion about `O` with amolitude `A` and time period `T`. The magnitude of its acceleration at `t=(T)/(8)`s after the particle reaches the extreme position would be

Options

A. `(4pi^2A)/(sqrt(2)T^2)`
B. `(4pi^2A)/(T^2)`
C. `(2pi^2A)/(sqrt(2T^2))`
D. none of these

Answer

A

Explanation

Let at `t=0` the particle is at extreme position, then the equation of SHM can be written as
`x=Acos(omegat)=Acos((2pi)/(T)t)`
At `x=(T)/(8)`
`x=Acos(A)/(4)=(A)/(sqrt2)`
Acceleration `=-omega^2x=-((2pi)/(T))^2xx(A)/(sqrt2)`
Magnitude of acceleration`=(4pi^2A)/(sqrt(2T^2))`