In the given figure, a mass M is attached to a horizontal spring which is fixed on one side to a rigid support. The spring constant of the spring is k. The mass oscillates on a frictionless surface with time period T and amplitude A. When the mass is in equilibrium position, as shown in the figure, another mass m is gently fixed upon it. The new amplitude of oscillations will be :

Solution

Physics sir · 4 steps

At equilibrium M has its top speed, and sticking m on is a collision, so conserve momentum, not energy.

1. Speed at equilibrium. At the equilibrium position the spring force is zero and the speed is maximum: v = Aω = A√(k/M).
2. Momentum conserved. When m is placed gently, it has no horizontal velocity and the spring exerts no force at that instant. So horizontal momentum is conserved: Mv = (M+m)v', which gives v' = M/(M+m)v. Kinetic energy is not conserved, because the two masses stick like an inelastic collision.
3. New amplitude. The system is still at equilibrium, so v' is the new maximum speed. The new angular frequency is ω' = √(k/(M+m)), and the amplitude is A' = v'/ω'.

   A' = M/(M+m) A√(k/M) × √((M+m)/k) = A√(M/(M+m))

4. So. The amplitude decreases to A√(M/(M+m)), which is option (d).

The idea to keep. When a mass is dropped on or stuck to an oscillating body, momentum is conserved at that instant. Find the new maximum speed, then divide by the new ω to get the new amplitude.

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