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NEET 2010 (AIPMT, Prelims) · Periodic Motion & SHM Basics

Periodic Motion & SHM Basics question from NEET 2010 (AIPMT, Prelims)

Q71NEET 2010 · AIPMT · PRELIMSEasy

Periodic Motion & SHM Basics

The displacement of a particle along the x-axis is given by x = asin2ω\omegat. The motion of the particle corresponds to

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Answer: (c)

  1. x=asin⁡2ωt=a2−a2cos⁡2ωtx = a\sin^2\omega t = \frac{a}{2} - \frac{a}{2}\cos 2\omega t.
  2. This is SHM of amplitude a/2a/2 about the shifted mean position x=a/2x = a/2, with angular frequency 2ω2\omega.
  3. Frequency f=2ω2π=ωπf = \frac{2\omega}{2\pi} = \frac{\omega}{\pi}.
  4. So the working gives (a) SHM of frequency ω/π\omega/\pi; the key's (c) treats the shifted-mean motion as non-SHM.
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