Two simple pendulum of length `5m` and `20m` respectively are given small displacement in one direction at the same time. They will again be in the same phase when the pendulum of shorter length has completed oscillation.

Options

A. `5`
B. `1`
C. `2`
D. `3`

Answer

C

Explanation

`T_1=2pisqrt(5/g)=T`, `T_2=2pisqrt((20)/(g))=2*2pisqrt(5/g)=2T`
The pendulum will be in phase if
`(omega_1t-omega_2t)=2npi`, `n=0,1,2,…..`
`n=1`
`((2pi)/(T)-(2pi)/(2T))t=2pi`
`t=2T`
Number of oscillations completed by shorter pendulum
`=(2T)/(T)=2`
Alternatively:
The pendulums will be in phase if
`(n+1)T_1=nT_2`
n: number of oscillations completed by large pendulum
`(n+1)T=2*2Timpliesn=1`
Number of oscillations by shorter pendulum `=n+1=1+1=2`