JEE Main 2026 — Oscillations Question with Solution

At equilibrium, the restoring force of the spring balances the weight of the mass:

$mg=k{x}_0$

$k=\frac{mg}{x_0}=\frac{0.2\times 10}{2\times 10^{-3}}=1000$ N/m

The frequency of oscillation is given by:

$f=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$

$f=\frac{1}{2\pi }\sqrt{\frac{1000}{0.2}}=\frac{1}{2\pi }\sqrt{5000}=\frac{10\sqrt{50}}{2\pi }=\frac{5\sqrt{50}}{\pi }$ Hz

The maximum energy in the spring corresponds to its maximum elastic potential energy, which occurs at the maximum extension.

Maximum extension $x_{max}=x_0+A=2\text{ mm}+2\text{ mm}=4\text{ mm}=4\times 10^{-3}$ m

Maximum energy in the spring $U_{max}=\frac{1}{2}k{x}_{max}^2$

$U_{max}=\frac{1}{2}\times 1000\times (4\times 10^{-3}{)}^2$

$U_{max}=500\times 16\times 10^{-6}=8000\times 10^{-6}=8\times 10^{-3}$ J

Answer: $\frac{5\sqrt{50}}{\pi }$ and $8\times 10^{-3}$