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

Method 1: 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.
`costheta=((4A)/(5))/(A)=(4)/(5)impliestheta=cos^-1((4)/(5))`
`omegat=cos^-1((4)/(5))impliest=(1)/(omega)cos^-1((4)/(5))=(T)/(2pi)cos^-1((4)/(5))`
Method 2: As the particle starts from rest i.e., from extreme position `x=Asin(omegat-phi)`.
At `t=0`,`x=Aimpliesphi=(pi)/(2)`
`A-(A)/(5)=Acosomegat`
`(4)/(5)=cosomegatimpliesomegat=cos^-1(4)/(5)`
`t=(T)/(2pi)cos^-1((4)/(5))`