A particle of mass m is attached to one end of a massless spring of force constant k , lying on a…

Let the equation of $SHM$ be,

$x=A\sin \left(\omega t\right)$

Here, $x$ is the position of the particle at any time $t$

$A$ is the amplitude of the $SHM$

$\omega$ is the Angular frequency of $SHM$

So, equation for the velocity will be,

$$\begin{gathered} v=A\omega \cos \left(\omega t\right) \\ \Rightarrow v=u_0\cos \left(\omega t\right) (\mathrm{since} \mathrm{velocity} \mathrm{at} \mathrm{t}=0 \mathrm{is} {\mathrm{u}}_0, \mathrm{so} \mathrm{\omega A}={\mathrm{u}}_0) \end{gathered}$$

Now, during collision speed of particle was $\left(u_1\right)=0.5u_0$

and time be $t_1$

So, Substituting the values in equation of velocity we get,

$$\begin{gathered} 0.5u_0=u_0\cos \left(\omega t_1\right) \\ \Rightarrow \cos \left(\omega t_1\right)=\frac{1}{2} \\ \Rightarrow \omega t_1=\frac{\mathrm{\pi }}{3} \\ \Rightarrow t_1=\frac{\mathrm{\pi }}{3\omega } \end{gathered}$$

As the energy will be conserve, so it will take same time to reach Equlibrium position first time,$i.e.$,
Time at which particle crosses equlibrium first time, 

$\left(T_1\right)=2t_1=\frac{2\mathrm{\pi }}{3\omega }$

Now, for the second half of the Oscillation, particle dosen't undergo collision that's why it will have normal Time period,$i.e.$

Time taken for the particle to go to the left will be $\frac{T}{4}$, as it was in the general case.

Here, $T$ is time period of $SHM$.

So to come back to Equlibrium position for second time time required will be $\frac{T}{2}$.

Now,

 $$\begin{gathered} T=\frac{2\mathrm{\pi }}{\omega } \\ \Rightarrow \frac{T}{2}=\frac{\mathrm{\pi }}{\omega } \end{gathered}$$

So total time required will be,

$$\begin{gathered} \frac{\mathrm{\pi }}{\omega }+\frac{2\mathrm{\pi }}{3\omega } \\ \Rightarrow \frac{5\mathrm{\pi }}{3\omega } \end{gathered}$$

Here, $\omega$ is the angular frequency,$i.e.$, $\omega =\sqrt{\frac{k}{m}}$

Substituting it above we get,

$Time Taken=\frac{5\mathrm{\pi }}{3}\sqrt{\frac{m}{k}}$