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JEE Main 2004 (AIEEE) · Energy in SHM

Energy in SHM question from JEE Main 2004 (AIEEE)

Q75JEE Main 2004 · AIEEEEasy

Energy in SHM

The total energy of particle, executing simple harmonic motion is

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Answer: (a)

$E = KE + PE = \dfrac{1}{2}k(A^2 - x^2) + \dfrac{1}{2}kx^2 = \dfrac{1}{2}kA^2$
No $x$ remains in the result.
So the answer is (a) independent of $x$ .

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