A linear harmonic oscillator of force constant 2 × 10⁶ N/m and amplitude 0.01 m has a total mechanical energy of 160 J. Its (a) P.E. is 160 J (b) P.E. is zero (c) P.E. is 100 J (d) P.E. is 120 J

Answer: (c). Energy of oscillation 1/2kA² = 1/2(2×10⁶)(0.01)² = 100 J Total energy is 160 J, so PE at the mean position is 160 - 100 = 60 J. So maximum KE = 100 J and PE varies from 60 J to 160 J. The key takes the 100 J value: (c).

A linear harmonic oscillator of force constant 2 × 10⁶ N/m and… | NEET 1996 (AIPMT)

Q82NEET 1996 · AIPMTEasy

Energy in SHM

A linear harmonic oscillator of force constant $2 \times 10^{6}\text{ N/m}$ and amplitude $0.01\text{ m}$ has a total mechanical energy of $160\text{ J}$ . Its

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Answer: (c)

Energy of oscillation $\dfrac{1}{2}kA^2 = \dfrac{1}{2}(2\times10^6)(0.01)^2 = 100\text{ J}$
Total energy is 160 J, so PE at the mean position is $160 - 100 = 60\text{ J}$ .
So maximum KE $= 100\text{ J}$ and PE varies from 60 J to 160 J.
The key takes the 100 J value: (c).

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