Question

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

• A. 2/3 sec
• B. 4/3 sec
• C. 2 sec
• D. none of these

Answer 1

Answer: (B)

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 $2^{\circ}$ part will be covered in a time $t=\frac{T}{2}=1\, sec$
For the left $1^{\circ}$ part:
$\theta=\theta_{0} \sin (\omega t) $
$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 $
Total time $\Rightarrow \frac{T}{2}+2 t $
$ \Rightarrow 1+2 \times \frac{1}{6}=1+\frac{1}{3}=\frac{4}{3} \,sec$