A block of mass m attached to a massless spring is performing oscillatory motion of amplitude ‘A’ on a frictionless horizontal plane. If half of the mass of the block breaks off when it is passing through its equilibrium point, the amplitude of oscillation for the remaining system become fA. The value of f is :

Solution

Physics sir · 4 steps

At the mean position the block has all the energy as speed; the broken piece carries away half of it.

1. Speed at equilibrium. At the mean position the block moves at its maximum speed v = Aω = A√(k/m), and the spring is unstretched, so all the energy is kinetic. The piece that breaks off does not jerk the rest, so the remaining mass m/2 keeps the same speed v.
2. Energy that stays. Only the remaining block stays on the spring. Its kinetic energy is 1/2(m/2)v², which is half of the original energy 1/2kA². At the new extreme position this energy is all stored in the spring as 1/2kA'².

   1/2kA'² = 1/2(1/2kA²) ⇒ A' = A/√2

3. Check with ω. The new angular frequency is ω' = √(k/(m/2)) = √2 ω. Then A' = v/ω' = Aω/(√2 ω) = A/√2, the same result.
4. So. The new amplitude is A/√2, so f = 1/√2, option (d).

The idea to keep. When mass changes suddenly at the mean position, speed stays the same, not amplitude. Use A' = v/ω' with the new ω', or match the kinetic energy that remains to 1/2kA'².

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