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NEET PHYSICSOscillationsEasy

Question

If a body is executing simple harmonic motion with frequency nn, then the frequency of its potential energy is:

A

3n3n

B

4n4n

C

nn

D

2n2n

Step-by-Step Solution

  1. Displacement Equation: For a body executing Simple Harmonic Motion (SHM) with frequency nn, the angular frequency is ω=2πn\omega = 2\pi n. The displacement xx varies with time tt as x=Asin⁡(ωt)x = A \sin(\omega t) (or a cosine function).
  2. Potential Energy Formula: The potential energy (UU) of the oscillator is given by U=12kx2U = \frac{1}{2}kx^2, where kk is the force constant.
  3. Substitution and Trigonometric Identity: Substituting the displacement equation: U=12k(Asin⁡(ωt))2=12kA2sin⁡2(ωt)U = \frac{1}{2}k (A \sin(\omega t))^2 = \frac{1}{2}kA^2 \sin^2(\omega t) Using the trigonometric identity sin⁡2θ=1−cos⁡(2θ)2\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}: U=12kA2[1−cos⁡(2ωt)2]=14kA2−14kA2cos⁡(2ωt)U = \frac{1}{2}kA^2 \left[ \frac{1 - \cos(2\omega t)}{2} \right] = \frac{1}{4}kA^2 - \frac{1}{4}kA^2 \cos(2\omega t)
  4. Frequency Analysis: The term cos⁡(2ωt)\cos(2\omega t) indicates that the potential energy oscillates with an angular frequency of 2ω2\omega. Since frequency is proportional to angular frequency (f=Ω2πf = \frac{\Omega}{2\pi}), the frequency of the potential energy is 2n2n.
  5. Conclusion: Both kinetic and potential energies in SHM vary periodically with double the frequency of the displacement (2n2n), although the total energy remains constant.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Oscillations. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSOscillationsexecutingsimpleharmonicmotionfrequency

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