JEE Main 2026 Apr 8 Shift 2, Physics Q12: SHM in Spring-Mass System

JEE Main2026Apr 8, Shift 2Physics

The frequency of oscillation of a mass $m$ suspended by 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 $\frac{v _{2}}{v _{1}}$ is ________.

A
$1$
B
$2$
C
$2$
D
$3$

Solution

For a uniform spring the product $k ℓ$ is constant (stiffness is inversely proportional to length). Halving the length doubles the spring constant: $k_{2} = 2 k_{1}$ .

Frequency $f = \frac{1}{2 π} \frac{k}{m} \propto k$ for fixed $m$ .

$\frac{v _{2}}{v _{1}} = \frac{k _{2}}{k _{1}} = 2$ .

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JEE Main 2026 Apr 8 Shift 2, Physics Q12: SHM in Spring-Mass System

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