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MHT CET 2026 (15 Apr, Shift 1) · Velocity & Acceleration in SHM

Velocity & Acceleration in SHM question from MHT CET 2026 (15 Apr, Shift 1)

Q10MHT CET 2026 · 15 APR · SHIFT 1Medium

Velocity & Acceleration in SHM

A particle executes linear S.H.M. with amplitude 44 cm. The magnitude of velocity and acceleration is equal when it is at 33 cm from mean position. Time period is

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Answer: (b)

  1. ∣v∣=ωA2−x2|v| = \omega\sqrt{A^2 - x^2}, ∣a∣=ω2x|a| = \omega^2 x.
  2. ω16−9=ω2(3)⇒ω=73\omega\sqrt{16 - 9} = \omega^2(3) \Rightarrow \omega = \frac{\sqrt{7}}{3} rad/s.
  3. T=2πω=6π7T = \frac{2\pi}{\omega} = \frac{6\pi}{\sqrt{7}} s.
  4. So the answer is (b) 6π7\frac{6\pi}{\sqrt{7}} s.
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