The bob of a simple pendulum is displaced from its equilibrium position 'O' to a position 'Q' which is at a height 'h' above 'O' and the bob is then released. Assuming the mass of the bob to be 'm' and time period of oscillation to be `2.0s`, the tension in the string when the bob passes through 'O' is

Options

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

Answer

A

Explanation

`T_("mean posn") = mg + (mv^(2))/(l) = mg + (2mgh)/(l)`