Q 11-13-052JEE MainJEE Main 2025 (29 Jan, Shift 2)Easy
Two bodies A and B of equal mass are suspended from two massless springs of spring constants $k_1$ and $k_2$, respectively. If the bodies oscillate vertically such that their amplitudes are equal, the ratio of the maximum velocity of A to the maximum velocity of B is
Answer: (B) $\sqrt{\dfrac{k_1}{k_2}}$
$v_{max} = A\omega = A\sqrt{\dfrac{k}{m}}$. With equal $A$ and $m$:
$$\frac{v_A}{v_B} = \sqrt{\frac{k_1}{k_2}}$$
Solution by Sreeraj P, M.Sc Physics