Question

The bob of a simple pendulum it displaced position `O` to a equilibrium position `Q` which is at height h above `O` and the bob to then mass released Assuming the mass of the bob is m and time period `2.0` sec of oscillation to be string when the bob passes through `O` is

A. `m(g+pisqrt(2gh))`
B. `m(g+sqrt(pi^(2)gh))`
C. `m(g+sqrt(pi^(2)/(2)gh))`
D. `m(g+sqrt(pi^(2)/(3)gh))`

Answer 1

Correct Answer - A
Tension in the string when bob passes throught lowest point `T=mg+(mv^(2))/(r)=mg+mvw`
`(` :. `v=romega)`
Putting `v=sqrt(2gh)` and `omega=(2pi)/(T)=(2T)/(2)=pi`
we get `T=m(g+pisqrt(2gh))` .