1. Write the SHM equation

For a mass-spring system released from an extreme position, we can take $x(t)=A\cos(\omega t)$ where $A$ is amplitude and $\omega=\dfrac{2\pi}{T}$.

2. Write kinetic and potential energies

In SHM,

• Potential energy: $U=\frac{1}{2}kx^2=\frac{1}{2}kA^2\cos^2(\omega t)$
• Total energy: $E=\frac{1}{2}kA^2$
• Kinetic energy: $K=E-U=\frac{1}{2}kA^2\sin^2(\omega t)$

3. Condition for kinetic energy = potential energy

We need $K=U$ So, $\frac{1}{2}kA^2\sin^2(\omega t)=\frac{1}{2}kA^2\cos^2(\omega t)$ which gives $\sin^2(\omega t)=\cos^2(\omega t)$ Therefore, $\tan^2(\omega t)=1$ So the first time this happens is $\omega t=\frac{\pi}{4}$

4. Find the corresponding time

Since $\omega=\frac{2\pi}{T}$ we get $t=\frac{\pi/4}{\omega}=\frac{\pi/4}{2\pi/T}=\frac{T}{8}$

Thus, $t=\frac{T}{x}=\frac{T}{8}$ So, $x=8$

5. Comparison with stored answer

Stored correct answer = $8$.

Our derived answer also is $8$, so they agree.