Question

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

• A. \( 2\pi^2 m a^2 \nu^2 \)
• B. \( \pi^2 m a^2 \nu^2 \)
• C. \( \dfrac{1}{4} m a^2 \nu^2 \)
• D. 4π² m a² ν²

Hint:

Average kinetic and potential energies are equal over symmetric SHM intervals.

Answer 1

The Correct Option is B

Step 1: Total energy in SHM: E = (1)/(2) m ω² a²
Step 2: Average KE from mean to extreme is half of total energy: ⟨ K ⟩ = (E)/(2)
Step 3: ω = 2πν ⟹ ⟨ K ⟩ = π² m a² ν²