A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is 20 m/s2 at a distance of 5 m from the mean position. The time period of oscillation is

The acceleration of particle/body executing SHM at any instant (at position x) is given as                               a = -ω2xwhere, ω is the angular frequency of the body.⇒            a=ω2x                                                                 … (i)Here,                     x=5m, a = 20 ms-2Substituting the given values in Eq. (i), we get                              20=ω2×5⇒                    ω2=205=4   or    ω=2 rad s-1As we know that, time period, T=2π/ω                                …(ii)∴Substituting the value of ro in Eq. (ii), we get                        T=2π2=π s