A simple pendulum is hanging from a peginserted in a vertical wall. Its bob is strteched to horizontal position form wall and left freee to move, the bob hits the wall. If coefficient of restitution is `(2)/(sqrt(5))`. After how many collisions the amplitude of vibration will becomes less then `60^(@)`.

Options

A. `6`
B. `3`
C. `5`
D. `4`

Answer

B

Explanation

`h_(n) = he^(2n) = h [1-cos 60]`
`r^(2n) =(1)/(2) rArr ((2)/(sqrt(5)))^(2n) = (1)/(2)`