Question

Two identical simple pendulums A and B have the same point of suspension, having length $\ell$ each. They are displaced by an angle ($\alpha$ and $\beta$ are very small and $\beta >\alpha$) and released from rest. Find the time after which B reaches its initial position for the first time. Assume collision to be elastic and both pendulums move in the same plane.

• A. $\pi \sqrt{\frac{\ell }{g}}$
• B. $2\pi \sqrt{\frac{\ell }{g}}$
• C. $\frac{\pi \beta }{\alpha }\sqrt{\frac{\ell }{g}}$
• D. $\frac{2\pi \beta }{\alpha }\sqrt{\frac{\ell }{g}}$

Answer 1

The correct option is B $2\pi \sqrt{\frac{\ell }{g}}$

Time taken by B for first collision (i.e from angular diplacement $\beta$ to mean position) $=\frac{T}{4}$
Since the collision is elastic, the maximum velocities of A and B are exchanged at the mean position.

$v_{max}=A\omega$ where $A$ is the amplitude and $\omega$ is the angular frequency which does not change.

Therefore, after first collision, angular displacement amplitude of B will be $\alpha$.

Time taken for B to go to angular displacement $\alpha$ and back to mean position $=\frac{T}{4}+\frac{T}{4}=\frac{T}{2}$
During second collision, again the maximum velocities of A and B are exchanged, so B will reach back to its initial angular displacement $\beta$ in time $\frac{T}{4}+\frac{T}{4}=\frac{T}{2}$.

So, total time taken is $t=\frac{T}{4}+\frac{T}{4}+\frac{T}{4}+\frac{T}{4}=T$