$\because F=-\frac{d U}{d x} \Rightarrow F=-8 \sin 2 x$
For small oscillations, $x$ will be small hence
$F=-8(2 x)=-16 x \quad \Rightarrow k=16$ and $m=1 \mathrm{kg}$
$\therefore \omega^{2}=\frac{\mathrm{k}}{\mathrm{m}}=\frac{16}{1}=16 \Rightarrow \omega=4$
$\Rightarrow \mathrm{T}=\frac{2 \pi}{\omega}=\frac{2 \pi}{4}=\frac{\pi}{2}$
$y_{2}=5(\sin 3 \pi t+\sqrt{3} \cos 3 \pi t)$
Ratio of amplitude of ${y}_{1}$ to ${y}_{2}={x}: 1$. The value of ${x}$ is ...... .
$(A)$ the speed of the particle when it returns to its equilibrium position is $u_0$.
$(B)$ the time at which the particle passes through the equilibrium position for the first time is $t=\pi \sqrt{\frac{ m }{ k }}$.
$(C)$ the time at which the maximum compression of the spring occurs is $t =\frac{4 \pi}{3} \sqrt{\frac{ m }{ k }}$.
$(D)$ the time at which the particle passes througout the equilibrium position for the second time is $t=\frac{5 \pi}{3} \sqrt{\frac{ m }{ k }}$.