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Energy in SHM question from JEE Main 2007 (AIEEE)

Q73JEE Main 2007 · AIEEEMedium

Energy in SHM

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

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Answer: (b)

$KE = E\cos^2\omega t$ ; its time average over a quarter period is $\frac{E}{2}$ .
$E = \frac{1}{2}m\omega^2a^2 = \frac{1}{2}m(2\pi v)^2a^2 = 2\pi^2ma^2v^2$
Average KE $= \frac{E}{2} = \pi^2ma^2v^2$
So the answer is (b) $\pi^2ma^2v^2$ .

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