Observe the setup shown in the figure. The blocks are given slight...

For a two-body oscillator system, the time period of oscillation is given by $T = 2\pi \sqrt{\dfrac{\mu}{k}}$, where $\mu$ is the reduced mass of the system.

The reduced mass $\mu$ for two masses $m_1$ and $m_2$ is $\mu = \dfrac{m_1 m_2}{m_1 + m_2}$.

Given $m_1 = m$ and $m_2 = m$, we get:

$\mu = \dfrac{m \cdot m}{m + m} = \dfrac{m^2}{2m} = \dfrac{m}{2}$

Substituting the value of $\mu$ into the time period formula:

$T = 2\pi \sqrt{\dfrac{m/2}{k}} = 2\pi \sqrt{\dfrac{m}{2k}}$