An oscillator of mass M is at rest in its equilibrium position in a...

In the first collision& $\text{mu}$ momentum will be imparted to the system. In the second collision when the momentum of $( M + m )$ is in the opposite direction mu momentum of the particle will make its momentum zero. On ${13}^{th}$ collision,~


Applying momentum conservation
$mu = ( M + 13m ) v$

$⇒ v = \dfrac{mu}{M + 13m} = \dfrac{u}{15}$~$⇒ M = 2m$
$⇒ v = \omegaA$

$⇒ \dfrac{u}{15} = \sqrt{\dfrac{K}{M + 13m}} \times A$

$⇒ A = \dfrac{1}{15}\sqrt{\dfrac{75}{1}} = \dfrac{1}{3}$