Question

A particle is executing Simple Harmonic Motion (SHM). The ratio of potential energy and kinetic energy of the particle when its displacement is half of its amplitude will be

• A. 1:4
• B. 1:3
• C. 2:1
• D. 1:1

Answer 1

The Correct Option is B

To solve this problem, we need to determine the ratio of potential energy to kinetic energy for a particle executing Simple Harmonic Motion (SHM) when its displacement is half of its amplitude. Let's denote the amplitude as A, and the displacement as x = \frac{A}{2}.

1. In SHM, the total energy E of the system is given by: E = \frac{1}{2} m \omega^2 A^2, where m is the mass of the particle and \omega is the angular frequency.
2. The potential energy U at displacement x is: U = \frac{1}{2} m \omega^2 x^2. Substituting x = \frac{A}{2} gives: U = \frac{1}{2} m \omega^2 \left( \frac{A}{2} \right)^2 = \frac{1}{2} m \omega^2 \frac{A^2}{4} = \frac{1}{8} m \omega^2 A^2.
3. The total energy is constant and is equal to the sum of kinetic energy K and potential energy U. Thus: E = U + K = \frac{1}{2} m \omega^2 A^2. So, the kinetic energy K is: K = E - U = \frac{1}{2} m \omega^2 A^2 - \frac{1}{8} m \omega^2 A^2 = \frac{4}{8} m \omega^2 A^2 - \frac{1}{8} m \omega^2 A^2 = \frac{3}{8} m \omega^2 A^2.
4. The ratio of potential energy to kinetic energy is then: \frac{U}{K} = \frac{\frac{1}{8} m \omega^2 A^2}{\frac{3}{8} m \omega^2 A^2} = \frac{1}{3}.

Thus, the ratio of potential energy to kinetic energy when the displacement is half of the amplitude is 1:3.