Question

A particle is executing S.H.M. of amplitude 'A'. When the potential energy of the particle is half of its maximum value during the oscillation, its displacement from the equilibrium position is

• A. $\pm \frac{A}{4}$
• B. $\pm \frac{A}{2}$
• C. $\pm \frac{A}{\sqrt{3}}$
• D. $\pm \frac{A}{\sqrt{2}}$

Hint:

At $x = A/\sqrt{2}$, Potential Energy = Kinetic Energy = half of Total Energy.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The potential energy in SHM varies with displacement $x$, reaching its maximum at the extreme positions ($x = \pm A$).
Step 2: Key Formula or Approach:
1. Potential Energy $PE = \frac{1}{2} m \omega^2 x^2$.
2. Maximum Potential Energy $PE_{max} = \frac{1}{2} m \omega^2 A^2$.
Step 3: Detailed Explanation:
Given: $PE = \frac{1}{2} PE_{max}$.
\[ \frac{1}{2} m \omega^2 x^2 = \frac{1}{2} \cdot \left(\frac{1}{2} m \omega^2 A^2\right) \]
Divide both sides by $\frac{1}{2} m \omega^2$:
\[ x^2 = \frac{1}{2} A^2 \]
Taking the square root on both sides:
\[ x = \pm \frac{A}{\sqrt{2}} \]
Step 4: Final Answer:
The displacement is $\pm \frac{A}{\sqrt{2}}$.