A particle performs `SHM` on x- axis with amplitude `A ` and time period period `T` .The time taken by the particle to travel a distance `A//5` string from rest is

Options

A. `(T)/(20)`
B. `T/(2pi) cos^(-1)(4/5)`
C. `T/(2pi)cos^(-1)(1/5)`
D. `T/(2pi)sin^(-1)(1/5)`

Answer

B

Explanation

Particle is starting from rest, i.e. from one of its extreme position. As particle moves a distance `A/5`, we can represent it on a circle as shown.
`cos theta=(4A)/(5)=4/5 " " theta=cos^(-1)(4/5)`
`omegat=cos^(-1)(4/5) " " t=1/(omega) cos^(-1)(4/5)`
`=T/(2pi) cos^(-1)(4/5) `
As starts from rest, i.e. from extreme position `x=A sin (omegat+phi)`
At t=0 ,x=A `rArr phi=(pi)/2 " " :. A-A/5=A cos omegat`
`4/5=cos omega t " " rArr omega t=cos^(-1)(4/5) rArr " " t=T/(2pi) cos^(-1)(4/5)`