Assume all surfaces depicted in the diagram lack friction. The car...

Let $x_1$ and $x_2$ be the displacements of the block of mass $m$ and the car of mass $M$ from their respective equilibrium positions.

Since no external horizontal force acts on the system, the center of mass remains stationary. Therefore:

$m x_1 + M x_2 = 0 \Rightarrow x_2 = -\dfrac{m}{M} x_1$

The net extension or compression in the spring is the relative displacement $x = x_1 - x_2$.

Substituting the value of $x_2$, we get:

$x = x_1 - \left(-\dfrac{m}{M} x_1\right) = x_1 \left(1 + \dfrac{m}{M}\right) = x_1 \left(\dfrac{M + m}{M}\right)$

The restoring force acting on the block of mass $m$ is $F = -kx$.

$F = -k \left(\dfrac{M + m}{M}\right) x_1$

The acceleration of the block is $a_1 = \dfrac{F}{m} = -\dfrac{k(M + m)}{mM} x_1$.

Comparing this with the standard equation of simple harmonic motion $a = -\omega^2 x$, we find the angular frequency $\omega$:

$\omega^2 = \dfrac{k(M + m)}{mM} \Rightarrow \omega = \sqrt{\dfrac{k(M + m)}{mM}}$

The time period of oscillation is given by $T = \dfrac{2\pi}{\omega}$:

$T = 2\pi \sqrt{\dfrac{mM}{k(M + m)}}$

Alternatively, this is a standard two-body oscillator problem where the time period is given by $T = 2\pi \sqrt{\dfrac{\mu}{k}}$, with the reduced mass $\mu = \dfrac{mM}{M + m}$.