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Question: Answered & Verified by Expert
A particle of mass $m$ executes simple harmonic motion with amplitude ' $a$ ' and frequency ' $v$ '. The average kinetic energy during its motion from the position of equilibrium to the end is
PhysicsOscillationsJEE MainJEE Main 2007
Options:
  • A
    $\pi^2 \mathrm{ma}^2 v^2$
  • B
    $\frac{1}{4} \pi^2 m a^2 v^2$
  • C
    $4 \pi^2 \mathrm{ma}^2 \mathrm{v}^2$
  • D
    $2 \pi^2 m a^2 v^2$
Solution:
1098 Upvotes Verified Answer
The correct answer is:
$\pi^2 \mathrm{ma}^2 v^2$
$\frac{1}{4} m a^2 \omega^2=\pi^2 f^2 m a^2$

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