\(\Rightarrow 4 \pi^{2}=\frac{\mathrm{k}}{\mathrm{m}}\)
If block of mass \(\mathrm{m}=1 \mathrm{kg}\) is attached then, \(k=4 \pi^{2}\)
Now, identical springs are attached in parallel with mass
\(\mathrm{m}=8 \mathrm{kg}\), Hence, \(k_{e q}=2 k\)
\(f^{\prime}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{k} \times 2}{8}}=\frac{1}{2} \mathrm{Hz}\)
$(A)\;y= sin\omega t-cos\omega t$
$(B)\;y=sin^3\omega t$
$(C)\;y=5cos\left( {\frac{{3\pi }}{4} - 3\omega t} \right)$
$(D)\;y=1+\omega t+{\omega ^2}{t^2}$