A linear harmonic oscillator of force constant 2 × 10⁶ N/m has amplitude 0.01 m. Its (a) maximum potential energy is 100 J. (b) maximum kinetic energy is 160 J. (c) maximum potential energy is 160 J. (d) maximum potential energy is zero.

Answer: (a). Maximum PE = 1/2kA² (at the extreme). = 1/2(2 × 10⁶)(0.01)² = 100 J Maximum KE is also 100 J (at the mean position). So the answer is (a) maximum potential energy is 100 J.

A linear harmonic oscillator of force constant 2 × 10⁶ N/m has… | MHT CET 2026 (19 May, Shift 2)

Q10MHT CET 2026 · 19 MAY · SHIFT 2Medium

Energy in SHM

A linear harmonic oscillator of force constant $2 \times 10^{6}$ N/m has amplitude $0.01$ m. Its

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Answer: (a)

Maximum PE $= \frac{1}{2}kA^2$ (at the extreme).
$= \frac{1}{2}(2 \times 10^6)(0.01)^2 = 100$ J
Maximum KE is also 100 J (at the mean position).
So the answer is (a) maximum potential energy is 100 J.

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