Question

A particle is executing linear simple harmonic motion of amplitude A. At what displacement is the energy of the particle half potential and half kinetic?

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

Answer 1

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

The total energy of a particle executing SHM $=\frac{1}{2}m{\omega }^2{A}^2$

The PE of the particle at a distance x from the equilibrium position

$=\frac{1}{2}m{\omega }^2{x}^2$

From the question, $\frac{1}{2}m{\omega }^2{x}^2=\frac{1}{2}\left(\begin{array}{c}\frac{1}{2}m{\omega }^2{A}^2\end{array}\right)$

$\Rightarrow x^2=\frac{A^2}{2}\Rightarrow x=\frac{A}{\sqrt{2}}$