Question

A particle executes Simple Harmonic Motion (SHM) with amplitude \( A \). The distance from the mean position when its kinetic energy becomes equal to its potential energy is:

• A. \( \sqrt{2} A \)
• B. \( 2A \)
• C. \( \frac{A}{\sqrt{2}} \)
• D. \( \frac{A}{2} \)

Hint:

In SHM, when kinetic energy = potential energy, the displacement is: \[ x = \frac{A}{\sqrt{2}} \] This occurs because total energy is equally shared between kinetic and potential forms.

Answer 1

The Correct Option is C

Step 1: Understanding the energy distribution in SHM. - The total energy in SHM is given by: \[ E = \frac{1}{2} k A^2 \] where: - \( k \) is the force constant (spring constant), - \( A \) is the amplitude. - The kinetic energy (KE) at displacement \( x \) is: \[ KE = \frac{1}{2} k (A^2 - x^2) \] - The potential energy (PE) at displacement \( x \) is: \[ PE = \frac{1}{2} k x^2 \] Step 2: Equating kinetic and potential energy. \[ KE = PE \] \[ \frac{1}{2} k (A^2 - x^2) = \frac{1}{2} k x^2 \] Canceling \( \frac{1}{2} k \): \[ A^2 - x^2 = x^2 \] \[ A^2 = 2x^2 \] \[ x^2 = \frac{A^2}{2} \] \[ x = \frac{A}{\sqrt{2}} \] Final Answer: \[ \boxed{\frac{A}{\sqrt{2}}} \]