JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Elasticity & SHM

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As shown in the figure, a spring is kept in a stretched position with some extension by holding the masses 1 kg and 0.2 kg with a separation more than spring natural length and are released. Assuming the horizontal surface to be frictionless, the angular frequency (in SI unit) of the system is :

(A)30✓ Correct
(B)27
(C)20
(D)5

Answer: (A)

Solution

Two masses joined by a spring on a frictionless surface oscillate with angular frequency

ω = \frac{k}{μ}

, where

μ

is the reduced mass. With

m_{1} = 1

kg and

m_{2} = 0.2

kg,

μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}} = \frac{1 \times 0.2}{1.2} = \frac{1}{6}

kg. From the figure

k = 150

N/m, so

ω = \frac{150}{1/6} = 900 = 30 rad/s

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Single correct option

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