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NEET2026PhysicsChapterActual

Two blocks, X and Y , are executing simple harmonic motion on smooth horizontal surfaces. Block X has mass M and is attached to a spring of force constant K . Block Y has mass 2M and is attached to a spring of force constant 8K . If both blocks cross their respective mean positions with identical speeds, what is the ratio of the amplitude of block X to the amplitude of block Y ?

Options

  1. A1 : 2
  2. B4 : 1
  3. C2 : 1
  4. D1 : 4

Correct answer

C. 2 : 1

Step-by-step solution

For block X , the angular frequency is _X = K M . For block Y , the angular frequency is _Y = 8K 2M = 4K M = 2 K M . The maximum speed of a block executing simple harmonic motion is given by v = A , which occurs at the mean position. Given that both blocks have identical speeds at their mean positions, v_X = v_Y . A_X _X = A_Y _Y A_X K M = A_Y (2 K M ) A_X A_Y = 2 1 The ratio of the amplitude of block X to the amplitude of block Y is 2 : 1 . Answer: 2 : 1

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