A mass 0.9 kg, attached to a horizontal spring, executes SHM with an amplitude A₁. When this mass passes through its mean position, then a smaller mass of 124 g is placed over it and both masses move together with amplitude A₂. If the ratio A₁/A₂ is α/(α-1), then the value of α will be ___________.

Solution

Physics sir · 4 steps

The mass is added at the mean position, so conserve momentum, then write each top speed as Aω.

1. Momentum at mean position. At the mean position the spring is unstretched, so it exerts no horizontal force while the small mass is placed on. Horizontal momentum is conserved: M v₁ = (M+m) v₂. Kinetic energy is not conserved, because the two masses come to a common speed.
2. Speed from amplitude. The speed at the mean position is the maximum speed, v = Aω. Here ω₁ = √(k/M) before and ω₂ = √(k/(M+m)) after, with the same spring constant k.

   M A₁√(k/M) = (M+m) A₂√(k/(M+m)) ⇒ A₁/A₂ = √((M+m)/M)

3. Put in numbers. M = 0.9 kg and m = 0.124 kg, so (M+m)/M = 1.024/0.9 = 1024/900. Its square root is 32/30 = 16/15.
4. So. A₁/A₂ = 16/15 = α/(α - 1), which gives α = 16.

The idea to keep. When a mass is added at the mean position, momentum is conserved and energy is not. Write each maximum speed as Aω with ω = √(k/m), and the amplitude ratio follows.

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