Question

When the potential energy of a particle executing simple harmonic motion is one-fourth of its maximum value during the oscillation, its displacement from the equilibrium position in terms of its amplitude $a$ is

• A. a/4
• B. a/3
• C. a/2
• D. 2a/3

Answer 1

Answer: (C)

Potential energy of particle executing SHM is given by
$PE=\frac{1}{2}m{\omega }^2{y}^2$
maximum $PE\left(U_{\max }\right)=\frac{1}{2}m{\omega }^2{a}^2$
Given $PE(U)=\frac{1}{4}{U}_{\max }$
$\therefore \frac{1}{2}m{\omega }^2{y}^2=\frac{1}{4}\frac{1}{2}m{\omega }^2{a}^2$
or $y^2=a^2/4$
$\Rightarrow y=\pm \frac{a}{2}$