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) $\sqrt{2}$
Spring constant is inversely proportional to length ($kL$ = constant), so half the spring has constant $2k$.
$v = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} \propto \sqrt{k}$, so $\dfrac{v_2}{v_1} = \sqrt{2}$.
Solution by Sreeraj P, M.Sc Physics