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

Options

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

Answer

A

Explanation

Tension in the string when bob passes through lowest
point `T = mg + (mv^(2))/(r ) = mg + mvomega (:' v = r omega)`
Putting `v = sqrt(2gh)` and `omega = (2pi)/(T) = (2pi)/(2) = pi`
we get `T = mg + pi sqrt(2gh))`.