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MHT CET 2025 (22 Apr, Shift 2) · Velocity & Acceleration in SHM

Velocity & Acceleration in SHM question from MHT CET 2025 (22 Apr, Shift 2)

Q15MHT CET 2025 · 22 APR · SHIFT 2Easy

Velocity & Acceleration in SHM

A particle executes S.H.M. starting from the mean position. Its amplitude is ' a ' and its periodic time is ' TT '. At a certain instant, its speed ' uu ' is half that of maximum speed VVmax. The displacement of the particle at that instant is

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Answer: (d)

  1. v=ωa2−x2v = \omega\sqrt{a^2 - x^2}, Vmax=ωaV_{max} = \omega a
  2. ωa2−x2=ωa2⇒a2−x2=a24\omega\sqrt{a^2 - x^2} = \frac{\omega a}{2} \Rightarrow a^2 - x^2 = \frac{a^2}{4}
  3. x=3a2x = \frac{\sqrt{3}a}{2}
  4. So the answer is (d) 3a2\frac{\sqrt{3}a}{2}.
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