A particle performs SHM about `x=0` such that at `t=0` it is at `x=0` and moving towards positive extreme. The time taken by it to go from `x=0` to `x=(A)/(2)` is time the time taken to go from `x=(A)/(2)` to a. The most suitable option for the blank space is

Options

A. 2
B. `(1)/(2)`
C. `(3A)/(T)`
D. `(A)/(2T)`

Answer

B

Explanation

From circular motion representation we can represent SHM by uniform circular motion.
Let `t_1` is the time taken by paritcle to go from `x=0` to `x=(A)/(2)`
`sintheta_1=((A)/(2))/(A)=(1)/(2)`
`theta_1=(pi)/(6)impliesomegat_1=(pi)/(6)omega_1=(pixxT)/(6xx2pi)=(T)/(12)`
So, time taken to go from `x=(A)/(2)` to `A` is
`t_2=(T)/(4)-t_1=(T)/(4)-(T)/(12)=(T)/(6)`
Hence `(t_1)/(t_2)=((T)/(12))/((T)/(6))=(1)/(2)`