\(4 \frac{d^2 x}{d t^2}+320 x=0\)

\(4 a=-320 x\)

\(a=-80 x\)

Since \(a=-\omega^2 x\) in \(S.H.M.\)

\(80=\omega^2\)

\(\sqrt{16 \times 5}=\omega\)

or \(\omega=4 \sqrt{5}\)

\(T=\frac{2 \pi}{\omega}=\frac{2 \pi}{4 \sqrt{3}}=\frac{\pi}{2 \sqrt{5}} s\)