Two simple pendulums of length 0.4 m and 1.6 m respectively are given small linear displacements in one direction at the same ins"tan"t. After how many complete oscillations of the pendulum of the shorter length, they will again be in phase ?

Options

A. 1
B. 2
C. 4
D. 6

Answer

B

Explanation

When the shorter pendulum completes n oscillations, the longer pendulum will complete (n-1) oscillation.
`therefore nT_(1)=(n-1)T_(2)`
`nxx2pi sqrt((l_(1))/(g))=(n-1)2pi sqrt((l_(2))/(g))`
`therefore (n)/(n-1)= sqrt((l_(2))/(l_(1)))=sqrt((1.6)/(0.4))=2`
`therefore n=2n-2 therefore n=2`