JEE Main 2025 Jan 29 Shift 2, Physics Q10: SHM in Spring-Mass System

JEE Main2025Jan 29, Shift 2Physics

Two bodies $A$ and $B$ of equal mass are suspended from two massless springs of spring constant $k_{1}$ 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

A
$\frac{k _{1}}{k _{2}}$
B
$\frac{k _{1}}{k _{2}}$
C
$\frac{k _{2}}{k _{1}}$
D
$\frac{k _{2}}{k _{1}}$

Solution

For a spring-mass oscillator, $v_{m a x} = A ω$ with $ω = k / m$ .

With equal amplitudes $A_{1} = A_{2}$ and equal masses,

$\frac{v _{A, m a x}}{v _{B, m a x}} = \frac{ω _{1}}{ω _{2}} = \frac{k _{1} / m}{k _{2} / m} = \frac{k _{1}}{k _{2}}$ .

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JEE Main 2025 Jan 29 Shift 2, Physics Q10: SHM in Spring-Mass System

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