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NEET 1996 (AIPMT) · Velocity & Acceleration in SHM

Velocity & Acceleration in SHM question from NEET 1996 (AIPMT)

Q85NEET 1996 · AIPMTEasy

Velocity & Acceleration in SHM

A particle starts with S.H.M. from the mean position as shown in the figure. Its amplitude is AA and its time period is TT. At one time, its speed is half that of the maximum speed. What is this displacement?

Figure

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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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