A particle undergoes simple harmonic motion with amplitude $A$. At what displacement from the equilibrium position is the kinetic energy equal to the potential energy?

Solution

1. The total mechanical energy of a particle in simple harmonic motion is $E=\frac{1}{2}k{A}^2$, and the potential energy at displacement $x$ is $U=\frac{1}{2}k{x}^2$.
2. Given that kinetic energy equals potential energy, the total energy is the sum of two equal parts, meaning $U=\frac{1}{2}E$.
3. Substituting the expressions for $U$ and $E$, we have $\frac{1}{2}k{x}^2=\frac{1}{2}(\frac{1}{2}k{A}^2)$.
4. Simplifying the equation gives $x^2=\frac{A^2}{2}$, which results in $x=\frac{A}{\sqrt{2}}$.
5. Hence the answer is (B).