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 :

Question

G1 Physics · Accepted Answer

For two masses connected by a spring on a frictionless surface, the relative oscillation is governed by the reduced mass μ = m₁m₂/(m₁+m₂). Here, m₁ = 1 kg, m₂ = 0.2 kg. Therefore, μ = (1×0.2)/(1+0.2) = 0.2/1.2 = 1/6 kg. The spring constant is k = 150 N/m. Hence the angular frequency is ω = √(k/μ) = √[150/(1/6)] = √900 = 30 rad/s. Correct option: A.