Two simple pendulums have time period `T` and `5T//4`. They start vibrating at the same instant from the mean position in the same phase. The phase difference (in rad) between them when the smaller pendulum completes one oscillation will be

Options

A. `(pi)/(6)`
B. `(pi)/(5)`
C. `(pi)/(2)`
D. `(2pi)/(5)`

Answer

D

Explanation

The equation of motion of two pendulums are
and `theta_(10)=theta_(01)sin omega_(1)t`
and `theta_(2)=theta_(02)sin omega_(2)t`
where, `omega_(1)=(2pi)/(T)` and `omega_(2)=(2pi)/((5T//4))`
The smaller pendulum is one which is having smaller time period. Therefore, the required phase difference is
`Deltaphi=omega_(1)t-omega_(2)t=[(2pi)/(T)-(2pi)/(5T//4)]t`
At time `t=T`,
`Deltaphi=[(2pi)/(T)-(2pi)/(5T//4)]xxT=(2pi)/(5)rad`