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

In Simple Harmonic Motion, the displacement of the particle can be represented as: \[ x = \frac{A}{2} \] Where \( A \) is the amplitude of the motion. The potential energy \( P.E. \) and kinetic energy \( K.E. \) in SHM are given by the equations: \[ P.E. = \frac{1}{2}kx^2 \] \[ K.E. = \frac{1}{2}kA^2 - \frac{1}{2}kx^2 \] Substitute \( x = \frac{A}{2} \) into the equations: \[ P.E. = \frac{1}{2} k \left( \frac{A}{2} \right)^2 = \frac{A^2 k}{8} \] \[ K.E. = \frac{1}{2}kA^2 - \frac{A^2 k}{8} = \frac{3A^2 k}{8} \] Now, find the ratio of \( P.E. \) to \( K.E. \): \[ \frac{P.E.}{K.E.} = \frac{\frac{A^2 k}{8}}{\frac{3A^2 k}{8}} = \frac{1}{3} \] Thus, the ratio of potential energy to kinetic energy is \( \boxed{1:3} \).