JEE Main 2026 — Jan 28, Shift 2 — Physics solution

JEE Main2026Jan 28, Shift 2Physics

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
B
27
C
20
D
5

Solution

For a two-mass spring system, oscillation frequency uses the reduced mass: $μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}} = \frac{1 \times 0.2}{1.2} = \frac{1}{6}$ kg.

$ω = \frac{k}{μ} = \frac{150}{1/6} = 900 = 30$ rad/s.

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