A particle executes a simple harmonic motion of time period T. Find the time taken by the particle to go directly from its mean position to half the amplitude.

Explanation

Let the equatio of motion be `x=Asinomegat`.
At t=0, x=0 and hence tehparticle is at its mean position. Its velocity is
`v=Aomegacosomegat=Aomega`
which is positive. So it is going towards `x=A/2`
The particle will be at `x=A/2`, at time t where `A/2=A sinomegat`
or `sinomegat=1/2`
or` omegat=pi/6`
here minimum positive value of `omegat` is chosen because we are intgerested in finding teh time taken by the particle to directly go from `x=0 to x=A/2`.
Thus `t=pi/(6omega)=(pi)/(6(2pi/T))=T/12`