JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Elasticity & SHM

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 ____.

Answer: (C)

Solution

The stiffness of a spring is inversely proportional to its length, so cutting it to half its length doubles the force constant:

k_{2} = 2 k_{1}

. The frequency of a spring-mass oscillator is

v = \frac{1}{2 π} \frac{k}{m}

, hence

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

. The ratio

\frac{v _{2}}{v _{1}}

2

JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Single correct option