A simple pendulum of length 1 m is allowed to oscillate with amplitude 2^o.It collides elastically with a wall inclined at 1^o to the vertical. Its time period will be: $\text{ (use }g={\pi }^2\text{ ) }$

Time period for half part:

$T=2\pi \sqrt{\frac{l}{g}}=2\pi \sqrt{\frac{1}{g}}=\frac{2\pi }{\pi }=2\sec$

So 2^o part will be covered in a time $t=\frac{T}{2}=1\sec$

For the left 1^o part:

$\begin{array}{l}\theta ={\theta }_0\sin \left(\omega t\right)\\ 1^{\circ }=2^{\circ }\sin \left(\frac{2\pi }{T}\times t\right)\Rightarrow \frac{1}{2}=\sin \left(\frac{2\pi }{T}\times t\right)\\ \Rightarrow \frac{\pi }{6}=\pi \times t\Rightarrow t=1/6\sec \end{array}$

$\begin{array}{l}\text{ Total time }\Rightarrow \frac{T}{2}+2t\\ \Rightarrow 1+2\times \frac{1}{6}=1+\frac{1}{3}=\frac{4}{3}\sec \end{array}$