A particle of mass m is attached to one end of a massless spring of...

Let the equation of $SHM$ be,

$x = Asin\left(\omegat\right)$

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,

$v = A\omega cos\left(\omegat\right) ⇒ v = {u}_{0}\cos \left(\omegat\right) ( sincevelocityatt = 0is{u}_{0}, so\omega A = {u}_{0} )$

Now, during collision speed of particle was $( {u}_{1} ) = 0 . 5{u}_{0}$

and time be ${t}_{1}$

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

$0 . 5{u}_{0} = {u}_{0}\cos \left(\omega{t}_{1}\right) ⇒ \cos \left(\omega{t}_{1}\right) = \dfrac{1}{2} ⇒ \omega{t}_{1} = \dfrac{\pi}{3} ⇒ {t}_{1} = \dfrac{\pi}{3\omega}$

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,

$( {T}_{1} ) = 2{t}_{1} = \dfrac{2\pi}{3\omega}$

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 $\dfrac{T}{4}$, 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 $\dfrac{T}{2}$.

Now,

$T = \dfrac{2\pi}{\omega} ⇒ \dfrac{T}{2} = \dfrac{\pi}{\omega}$

So total time required will be,

$\dfrac{\pi}{\omega} + \dfrac{2\pi}{3\omega} ⇒ \dfrac{5\pi}{3\omega}$

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

Substituting it above we get,

$TimeTaken = \dfrac{5\pi}{3}\sqrt{\dfrac{m}{k}}$