A particle executes SHM with a time period of `4 s`. Find the time taken by the particle to go directly from its mean position to half of its amplitude.

Options

A. 1s
B. `(1)/(2)s`
C. `(1)/(3)s`
D. `(1)/(4)s`

Answer

C

Explanation

`x=Asin(omegat+phi_(@))`
at t=0, `x=0impliesAsinphi_(@)=0` or `phi_(@)=0`
Hence `x=Asin(omegat)`
or `A//2=Asin(omegat)`
or 1/2`=sin(omegat)`
`omegat=sin^(-1)((1)/(2))=(pi)/(6)`
`t=(pi)/(6omega)=(pi.T)/(6(2pi))`
as `omega=2pi//Timpliest T//12=1//3s`