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 :

Solution

Physics sir · 3 steps

Both blocks move, so do not use one mass: replace them by the reduced mass and it becomes a one-block problem.

1. Why reduced mass. The surface is frictionless and the masses are released from rest, so the centre of mass stays fixed. Both blocks oscillate about it, and the two-body system behaves like a single block of reduced mass μ on the same spring.

   μ = (m₁ m₂)/(m₁ + m₂) = (1 × 0.2)/(1 + 0.2) = 0.2/1.2 = 1/6 kg

2. Put in numbers. With k = 150 N m⁻¹, ω = √(k/μ) = √(150 × 6) = √900.
3. So. ω = 30 rad s⁻¹, which is option (c).

The idea to keep. Two free blocks joined by a spring on a smooth floor oscillate with ω = √(k/μ), where μ = (m₁ m₂)/(m₁ + m₂). Remember that μ is always smaller than the lighter mass.

Answer from the NTA final answer key. Solution by PrepWiser.