Question
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 : (use $g = \pi ^2$ )
✓
Answer
Time period for half part $T=2 \pi \sqrt{\frac{1}{g}}=2 \pi \sqrt{\frac{1}{g}}=2 \pi \sqrt{\frac{1}{g}}=\frac{2 \pi}{g}=2 \mathrm{sec}$
So $2^{\circ}$ part will be covered in a time $t=\frac{T}{2}=1$ sec
For the half $1^{\circ}$ part
$\theta=\theta_{2} \sin (\omega t)$
$1^{\circ}=2^{\circ} \sin \left(\frac{2 \pi}{T} \times t\right) \Rightarrow \frac{t}{2}=\sin \left(\frac{2 \pi}{T} \times t\right)$
$\Rightarrow \frac{\pi}{6}=\pi \times 1 \Rightarrow t=1 / 6 \mathrm{sec}$
Total time $\Rightarrow \frac{T}{2}+2 t$
$\Rightarrow 1+2 \times \frac{1}{6}=1+\frac{1}{3}=\frac{4}{3} \mathrm{sec}$
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