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: Potential Energy Formulas
$P.E. = \frac{1}{2} k x^2$
$P.E._{max} = \frac{1}{2} k A^2$
Step 2: Set up Equation
$P.E. = \frac{1}{2} P.E._{max}$
$\frac{1}{2} k x^2 = \frac{1}{2} \left( \frac{1}{2} k A^2 \right)$
Step 3: Solving for x
$x^2 = \frac{A^2}{2} \Rightarrow x = \pm \frac{A}{\sqrt{2}}$.
Step 4: Conclusion
The displacement is $\pm A/\sqrt{2}$.
Final Answer:(D)