Q 11-13-118JEE MainJEE Main 2021 (26 Feb, Shift 2)Medium
Time period of a simple pendulum is $T$. The time taken to complete $\frac{5}{8}$ oscillations starting from mean position is $\frac{\alpha}{12}T$. The value of $\alpha$ is ______.
Numerical value type. Enter your answer.
Answer: 7
In one oscillation the bob covers a path of $4A$, so $\frac{5}{8}$ oscillation means a path of $\frac{5}{8}\times4A = \frac{5}{2}A$.
Starting at the mean position: mean $\to$ extreme $\to$ mean covers $2A$ in $\dfrac{T}{2}$.
The remaining $\dfrac{A}{2}$ is from the mean position to $x = \dfrac{A}{2}$: $\dfrac{A}{2} = A\sin\omega t \Rightarrow \omega t = \dfrac{\pi}{6} \Rightarrow t = \dfrac{T}{12}$.
$$t_{total} = \frac{T}{2} + \frac{T}{12} = \frac{7T}{12} \Rightarrow \alpha = 7$$
Solution by Sreeraj P, M.Sc Physics