A particle executes simple harmonic motion with an amplitude of $A=4\text{ cm}$. At what displacement $x$ from the mean position is its kinetic energy equal to its potential energy?

Solution

1. The kinetic energy is $K=\frac{1}{2}k(A^2-x^2)$ and the potential energy is $U=\frac{1}{2}k{x}^2$.
2. Setting $K=U$ gives $\frac{1}{2}k(A^2-x^2)=\frac{1}{2}k{x}^2$, which simplifies to $A^2=2x^2$.
3. Solving for $x$ yields $x=\frac{A}{\sqrt{2}}$.
4. Substituting $A=4\text{ cm}$ results in $$\begin{array}{rl}x& =\frac{4}{\sqrt{2}}\\ & =2\sqrt{2}\text{ cm}\end{array}$$.
5. Hence the answer is (B).