Q 11-13-073JEE MainJEE Main 2023 (24 Jan, Shift 1)Easy
A block of mass $2\ \text{kg}$ is attached with two identical springs of spring constant $20\ \text{N m}^{-1}$ each. The block is placed on a frictionless surface and the ends of the springs are attached to rigid supports (see figure). When the mass is displaced from its equilibrium position, it executes a simple harmonic motion. The time period of oscillation is $\dfrac{\pi}{\sqrt X}$ in SI unit. The value of $X$ is ______.
Numerical value type. Enter your answer.
Answer: 5
$k_{eff}=20+20=40\ \text{N m}^{-1}$.
$$T=2\pi\sqrt{\frac{2}{40}}=\frac{2\pi}{\sqrt{20}}=\frac{\pi}{\sqrt5}$$
So $X=5$.
Solution by Sreeraj P, M.Sc Physics