Figure shows a spring-block system hanging in equilibrium. The block of the system is pulled down by a distance d and imparted a velocity  $v_0$  in the downward direction as shown in the figure. Find the time it will take to reach its mean position.

we know,              $v=\omega \sqrt{A^2-x^2}$
$\therefore v_0^2=\frac{k}{m}(A^2-d^2)$    $\Rightarrow A^2-d^2=\frac{m{v}_0^2}{k}$
 $\therefore A=\sqrt{\frac{m{v}_0^2}{k}+d^2}$
 $T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{k}}$
At      $t=0, y=d$    $\Rightarrow d=A\sin (0+\delta )$
 $\sin \delta =\frac{d}{A}\Rightarrow \delta ={\sin }^{-1}\left(\frac{d}{\sqrt{\frac{m{v}_0^2}{k}+d^2}}\right)$  
When it reaches mean position , y=0
$\therefore 0=A\sin (\omega t+\delta )\Rightarrow \omega t+\delta =\pi$   
$\omega t=\pi -{\sin }^{-1}\left(\frac{d}{\sqrt{\frac{m{v}_0^2}{k}+d^2}}\right)\Rightarrow t=\sqrt{\frac{m}{k}}[\pi -{\sin }^{-1}\left(\frac{d}{\sqrt{\frac{m{v}_0^2}{k}+d^2}}\right)]$