Q 11-13-037JEE MainJEE Main 2026 (8 Apr, Shift 2)Easy
The frequency of oscillation of a mass $m$ suspended by a spring is $v_1$. If the length of the spring is cut to half, the same mass oscillates with frequency $v_2$. The value of $v_2/v_1$ is ______.
Answer: (C) $\sqrt{2}$
Concept. For a spring $kl$ is constant, so half the length has twice the spring constant; $f\propto\sqrt k$ for the same mass.
Step 1. New constant. $k_2=2k_1$.
Step 2. Ratio. $\dfrac{v_2}{v_1}=\sqrt{\dfrac{k_2}{k_1}}=\sqrt2$.
Answer. $\sqrt2$ (Option C).
Solution by Sreeraj P, M.Sc Physics
