JEE Main 2026 — Oscillations Question with Solution

The frequency of oscillation of a spring-mass system is given by $v=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$, where $k$ is the spring constant and $m$ is the mass.

The spring constant $k$ is inversely proportional to the length of the spring $L$, so $k\propto \frac{1}{L}$.

When the length of the spring is cut to half, the new length is $L^{\prime }=\frac{L}{2}$. The new spring constant becomes $k^{\prime }=2k$.

The new frequency of oscillation is $v_2=\frac{1}{2\pi }\sqrt{\frac{k^{\prime }}{m}}=\frac{1}{2\pi }\sqrt{\frac{2k}{m}}=\sqrt{2}{v}_1$.

Therefore, the ratio $\frac{v_2}{v_1}=\sqrt{2}$.

Answer: $\sqrt{2}$