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JEE Main 2021 (27 Jul, Shift 1) · Energy in SHM

Energy in SHM question from JEE Main 2021 (27 Jul, Shift 1)

Q55JEE Main 2021 · 27 JUL · SHIFT 1Easy

Energy in SHM

A particle starts executing simple harmonic motion (SHM) of amplitude 'a' and total energy E. At any instant, its kinetic energy is 3E4\frac{3E}{4} then its displacement 'y' is given by:

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Answer: (d)

  1. KE=3E4⇒PE=E4KE = \frac{3E}{4} \Rightarrow PE = \frac{E}{4}.
  2. y2a2=14⇒y=a2\frac{y^2}{a^2} = \frac{1}{4} \Rightarrow y = \frac{a}{2}
  3. So the answer is (d) y=a2y = \frac{a}{2}.
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