The particle executing simple harmonic motion has a kinetic energy K0cos2 ω t. The maximum values of the potential energy and the total energy are respectively (a) K0/2 and K0 (b) K0 and 2K0 (c) K0 and K0 (d) 0 and 2K0

Answer: (c). Maximum KE = K₀ (when cos ²ω t = 1), and there PE = 0, so total energy = K₀. Maximum PE occurs when KE = 0: PEmax = K₀. So the answer is (c) K₀ and K₀.

The particle executing simple harmonic motion has a kinetic energy… | NEET 2007 (AIPMT)

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Energy in SHM

The particle executing simple harmonic motion has a kinetic energy K0cos2 $\omega$ t. The maximum values of the potential energy and the total energy are respectively

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Answer: (c)

Maximum KE $= K_0$ (when $\cos^2\omega t = 1$ ), and there PE $= 0$ , so total energy $= K_0$ .
Maximum PE occurs when KE $= 0$ : $PE_{max} = K_0$ .
So the answer is (c) $K_0$ and $K_0$ .

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