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MHT CET 2021 (20 Sep, Shift 2) · Velocity & Acceleration in SHM

Velocity & Acceleration in SHM question from MHT CET 2021 (20 Sep, Shift 2)

Q58MHT CET 2021 · 20 SEP · SHIFT 2Medium

Velocity & Acceleration in SHM

A particle executes S.H.M. of period 2π3\frac{2\pi}{\sqrt{3}} second along a straight line 4 cm4\text{ cm} long. The displacement of the particle at which the velocity is numerically equal to the acceleration is

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Answer: (b)

  1. Path length 4 cm, so A=2A = 2 cm; ω=2πT=3\omega = \frac{2\pi}{T} = \sqrt{3} rad/s.
  2. ωA2−x2=ω2x⇒4−x2=3 x\omega\sqrt{A^2 - x^2} = \omega^2 x \Rightarrow \sqrt{4 - x^2} = \sqrt{3}\,x.
  3. 4−x2=3x2⇒x=14 - x^2 = 3x^2 \Rightarrow x = 1 cm.
  4. So the answer is (b) 1 cm.
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