A particle executes SHM along a straight line. Its amplitude is A. The potential energy of the particle is equal to the kinetic energy when the displacement of the particle from the mean position is:
Correct answer: C. ±A/√2
- A. Zero
- B. ±A/2
- C. ±A/√2
- D. 2A
Explanation
When the potential energy is equal to the kinetic energy, we have:(1/2)kx^2 = (1/2)mv^2x^2 = (mv^2)/kv^2 = kx^2/mSubstituting v^2 into the expression for K, we have:K = (1/2)mv^2 = (1/2)m(kx^2/m) = (1/2)kx^2Thus, when the potential energy is equal to the kinetic energy, we have:2K = 2U2(1/2)kx^2 = (1/2)kA^2x^2 = A^2/2x = ±A/√2Therefore, the displacement of the particle from the mean position when the potential energy is equal to the kinetic energy is ±A/√2.lAs it is numerical, it can have only one possible answer.
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About Simple Harmonic Motion
Simple harmonic motion is oscillation in which acceleration is directly proportional to displacement and directed toward the equilibrium position. Work includes displacement, velocity, acceleration, phase, period, frequency, amplitude and energy, with applications to springs and simple pendulums. The restoring force and the conditions for SHM distinguish it from general periodic motion.
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