given: \(m=326 g , T=0.250 s\), Total Energy \(=5.83 J\) required \(V_{\max }=\) ?
recall, \(V_{\max }=\omega A, A=\frac{V_{\max }}{\omega}\)
where \(A=\) maximum displacement or amplitude
Since, we are given the total energy \(E_{T}\), we also recall that the total energy of a simple harmonic motion is
$$
\begin{aligned}_{T} &=\frac{1}{2} k A^{2} \\
&=\frac{1}{2} m V^{2}+\frac{1}{2} k x^{2}
\end{aligned}
$$
using \(E_{T}=\frac{1}{2} k A^{2}\)
substitute \(A=\frac{V_{\max }}{\omega}\)
$$_{T}=\frac{1}{2} k\left(\frac{V_{\max }^{2}}{\omega^{2}}\right) \ldots \ldots \ldots
$$
next, we find \(k\) from
$$
\begin{aligned}
\omega &=\sqrt{\frac{k}{m}} \\
\omega^{2} &=\frac{k}{m} ; k=m \omega^{2}
\end{aligned}
$$
Substitute \(k\) into the energy equation above \(\left( eq ^{*}\right)\)
$$
\begin{aligned}
&E_{T}=\frac{1}{2} \times m \omega^{2} \frac{V_{\max }^{2}}{\omega^{2}} \\
&E_{T}=\frac{1}{2} m V_{\max }^{2} \\
&\Rightarrow 2 \times E_{T}=m V_{\max }^{2} \\
&V_{\max }^{2}=\frac{2 \times E_{T}}{m} \\
&V_{\max }=\sqrt{\frac{2 \times E_{T}}{m}} \\
&V_{\max }=\sqrt{\frac{2 \times 5.83}{0.326}} \\
&V_{\max }=\sqrt{35.767} \\
&V_{\max }=5.98 m / s
\end{aligned}
$$