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

Options

A. `2pi//theta`
B. `2pitheta`
C. `theta//2pi`
D. `((2pi)/(theta))^(2)`

Answer

A

Explanation

`x_(1)=rcos(omegaxx1),x_(2)=rcos(omegaxx2)`
and `x_(3)=rcos(omegaxx3)`
`costheta=[(x_(1)+x_(3))/(2x_(2))]=(rcosomega+rcos3omega)/(2rcos2omega)`
`=(2rcos2omegacosomega)/(2rcos2omega)=cosomega`
`:. theta=omega=(2pi)/(T) or T=(2pi)/(theta)`