Q 11-13-066JEE MainJEE Main 2024 (31 Jan, Shift 2)Medium
The time period of simple harmonic motion of mass $M$ in the given figure is $\pi\sqrt{\dfrac{\alpha M}{5K}}$, where the value of $\alpha$ is ______.
Numerical value type. Enter your answer.
Answer: 12
Right side: two springs $k$ in parallel ($2k$) in series with $k$:
$$k_r = \frac{2k\cdot k}{3k} = \frac{2k}{3}$$
This acts in parallel with the left spring: $k_{eq} = k + \dfrac{2k}{3} = \dfrac{5k}{3}$.
$$T = 2\pi\sqrt{\frac{3M}{5k}} = \pi\sqrt{\frac{12M}{5k}}$$
$\alpha = 12$.
Solution by Sreeraj P, M.Sc Physics