Question

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.

Answer 1

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`