Question

A simple pendulum of length $1m$ 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:
$T=2\pi \sqrt{\frac{l}{g}}=2\pi \sqrt{\frac{1}{g}}=\frac{2\pi }{\pi }=2sec$
So $2^{\circ }$ part will be covered in a time $t=\frac{T}{2}=1sec$
For the left $1^{\circ }$ part:
$\theta ={\theta }_0\sin (\omega t)$
$1^{\circ }=2^{\circ }\sin \left(\frac{2\pi }{T}\times t\right)\Rightarrow \frac{1}{2}=\sin \left(\frac{2\pi }{T}\times t\right)$
$\Rightarrow \frac{\pi }{6}=\pi \times t\Rightarrow t=1/6sec$
Total time $\Rightarrow \frac{T}{2}+2t$
$\Rightarrow 1+2\times \frac{1}{6}=1+\frac{1}{3}=\frac{4}{3}sec$