At any position, let the velocities be $$v_{1}$$ and $$v_{2}$$ respectively.
Here, $$v_{1}=$$ velocity of $$m$$ with respect to $$M$$.
By energy method
Total Energy = Constant
$$(1/2)Mv^{2}+(1/2)m(v_{1}-v_{2})^{2}+(1/2)k(x_{1}+x_{2})^{2}=$$ Constant ..(i)
[$$v_{1}-v_{2}=$$ Absolute velocity of mass $$m$$ as seen from the road.]
Again, from law of conservation of momentum
$$mx_{2}=mx_{1}\Rightarrow x_{1}=\dfrac{M}{m}x_{2}$$ $$....(1)$$
$$mv_{2}=m(v_{1}-v_{2})\Rightarrow (v_{1}-v_{2})=\dfrac{M}{m}v_{2}$$ $$.....(2)$$
Putting the above values in equation $$(1)$$, we get
$$\dfrac{1}{2}Mv_{2}^{2}+\dfrac{1}{2}m\dfrac{M^{2}}{m^{2}}v_{2}^{2}+\dfrac{1}{2}kx_{2}^{2}\left(1+\dfrac{M}{m}\right)^{2}=$$ constant
$$\therefore M\left(1+\dfrac{M}{m}\right)v_{2}+k\left(1+\dfrac{M}{m}\right)^{2}x_{2}^{2}=$$ Constant.
$$mv_{2}^{2}+l\left(1+\dfrac{M}{m}\right)x_{2}^{2}=$$ constant
Taking derivative of both sides,
$$M\times 2v_{2}\dfrac{dv_{2}}{dt}+k\dfrac{(M+m)}{m}--ex_{2}^{2}\dfrac{dx_{2}}{dt}=0$$
$$\Rightarrow ma_{2}+k\left(\dfrac{M+m}{m}\right)x_{2}=0$$ [because, $$c_{2}=\dfrac{dx_{2}}{dt}$$]
Taking derivative of both sides,
$$M\times 2v_{2}\dfrac{dv_{2}}{}+k\dfrac{(M+m)}{m}-ex_{2}^{2}\dfrac{dx_{2}}{dt}=0$$
$$\Rightarrow ma_{2}+k\left(\dfrac{M+m}{m}\right)x_{2}=0$$ [because, $$v_{2}=\dfrac{dx_{2}}{dt}$$]
$$\Rightarrow \dfrac{a_{2}}{x_{2}}=-\dfrac{k(M+m)}{Mm}=\omega^{2}$$
$$\therefore \omega=\sqrt{\dfrac{k(M+m)}{Mm}}$$
So, Time period, $$T=2\pi \sqrt{\dfrac{Mm}{k(M+m)}}$$