Question

A particle executes SHM, at what value of displacement are the kinetic and potential energies equal ?

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

Answer 1

The correct option is B $\frac{A}{\sqrt{2}}$

We know that,
For a particle executing SHM, about mean position $x=0$.
The total energy of the particle
$E=\frac{1}{2}m{\omega }^2{A}^2$
Kinetic energy of the particle
$K=\frac{1}{2}m{\omega }^2(A^2-x^2)$
Potential energy of the particle
$U=\frac{1}{2}m{\omega }^2{x}^2$
If, $K=U$
$\frac{1}{2}m{\omega }^2(A^2-x^2)=\frac{1}{2}m{\omega }^2{x}^2$
$\Rightarrow 2x^2=A^2$
Then, $x=\frac{A}{\sqrt{2}}=0.707A$
Thus, option (b) is the correct answer.