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

                               $\mathrm{a} = -{\mathrm{\omega }}^2\mathrm{x}$

where, $\mathrm{\omega }$ is the angular frequency of the body.

$\Rightarrow \left|\mathrm{a}\right|={\mathrm{\omega }}^2\mathrm{x}$                                                                 … (i)

Here,                     $\mathrm{x}=5\mathrm{m}, \left|\mathrm{a}\right| = 20 {\mathrm{ms}}^{-2}$

Substituting the given values in Eq. (i), we get

                              $20={\mathrm{\omega }}^2\times 5$

$\Rightarrow {\mathrm{\omega }}^2=\frac{20}{5}=4 \mathrm{or} \mathrm{\omega }=2 \mathrm{rad} {\mathrm{s}}^{-1}$

As we know that, time period, $\mathrm{T}=2\mathrm{\pi }/\mathrm{\omega }$                                …(ii)

$\therefore$Substituting the value of ro in Eq. (ii), we get

                        $\mathrm{T}=\frac{2\mathrm{\pi }}{2}=\mathrm{\pi } \mathrm{s}$