Question

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

• A. $\frac{a}{2}$
• B. $\frac{a}{4}$
• C. $\frac{2a}{3}$
• D. none of these

Answer 1

Answer: (A)

$\frac{a}{2}$

We have,

$PE$ may $=\frac{1}{2}k{a}^2$

$PE=\frac{1}{2}k{x}^2$

According to question,

$\begin{array}{c}\frac{1}{2}k{x}^2=\frac{1}{4}.\frac{1}{2}k{A}^2\\ x=\frac{a}{2}\end{array}$

Then,

Option $A$ is correct answer.