Question

A particle of mass \( m \) executes SHM with amplitude \( a \) and frequency \( v \). The average kinetic energy during its motion from the position of equilibrium to the end is

• A. \( 2 \pi^2 ma^2 v^2 \)
• B. \( 4 \pi^2 ma^2 v^2 \)
• C. \( \frac{1}{2} \pi^2 ma^2 v^2 \)
• D. \( \pi^2 ma^2 v^2 \)

Hint:

In SHM, the average kinetic energy can be derived using the angular frequency and amplitude of the motion.

Answer 1

The Correct Option is D

The average kinetic energy of a particle in SHM is given by: \[ K_{{avg}} = \frac{1}{2} m \omega^2 a^2 \] where \( \omega = 2 \pi v \) is the angular frequency. Substituting the expression for \( \omega \), the average kinetic energy becomes: \[ K_{{avg}} = \pi^2 ma^2 v^2 \] Hence, the correct answer is (d).