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Question: Answered & Verified by Expert
A particle executes simple harmonic motion along a straight line with an amplitude $A$. The potential energy is maximum when the displacement is
PhysicsOscillationsJEE Main
Options:
  • A $\pm A$
  • B Zero
  • C $+\frac{A}{2}$
  • D $+\frac{A}{\sqrt{2}}$
Solution:
1706 Upvotes Verified Answer
The correct answer is: $\pm A$
P.E. $=\frac{1}{2} m \omega^2 x^2$
It is clear P.E. will be maximum when $x$ will be maximum i.e., at $x= \pm A$

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