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
- A1 : 2
- B4 : 1
- C2 : 1
- 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