A particle executing `SHM` while moving from executy it found at distance `x_(1) x_(2)` and `x_(2)` from comes at the and of three successive second The period of oscilation is
where `theta = cos^(-1)((x_1 + x_2)/(2x_(2)))`

Options

A. `2 pi//theta`
B. `pi//theta`
C. `theta`
D. `pi//2theta`

Answer

A

Explanation

Displacement time equation of the particle will be
`X = A cos omega t`
Given that `x_(1) = A cos omega`
`and x_(2) = A cos 2omega`
`Now (x_(1) + x_(3))/(2x_(2)) = (A (cos theta + cos 3 omega))/(2A cos 2omega)`
`= (2A cos 2 omega cos omega)/(2A cos 2 omega)`
`= cos omega`
`omega = cos^(-1)((x_(1) +x_(2))/(2x_(2))) = (2pi)/(T)`
`or T = (2pi)/(theta) where theta = cos^(-1)((x_(1) +x_(2))/(2x_(2)))`