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ExamsJEE MainPhysics

A block of mass M is attached to a horizontal spring and oscillates in simple harmonic motion with amplitude A1. When the block crosses the equilibrium position, a smaller mass m is gently placed on it, and thereafter the two move together. The new amplitude is A2. What is the value of the ratio A1/A2?

  1. (M + m)/M
  2. (M/(M + m))^(1/2)
  3. ((M + m)/M)^(1/2)
  4. M/(M + m)

Correct answer: ((M + m)/M)^(1/2)

Solution

The correct option is derived from the conservation of energy principle in simple harmonic motion. When the smaller mass m is added to the block of mass M at the equilibrium position, the total mass increases, which reduces the amplitude of oscillation. The new amplitude A2 is related to the original amplitude A1 by the ratio of the square root of the total mass to the original mass, leading to the expression A1/A2 = (M + m)/M.

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