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(D) The reduced mass of the system is given by:$\mu = \frac{m_1 m_2}{m_1 + m_2} = \frac{m \cdot m}{m + m} = \frac{m^2}{2m} = \frac{m}{2}$The given sys

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$\pi \sqrt {\frac{{2m}}{k}} $

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(D) The reduced mass of the system is given by:$\mu = \frac{m_1 m_2}{m_1 + m_2} = \frac{m \cdot m}{m + m} = \frac{m^2}{2m} = \frac{m}{2}$The given system of two particles connected by a spring is equivalent to a single particle of mass $\mu$ connected to a fixed wall by a spring of stiffness $k$.The time period of oscillation for this equivalent system is:$T = 2\pi \sqrt{\frac{\mu}{k}}$Substituting the value of $\mu = \frac{m}{2}$:$T = 2\pi \sqrt{\frac{m/2}{k}} = 2\pi \sqrt{\frac{m}{2k}} = 2\pi \frac{1}{\sqrt{2}} \sqrt{\frac{m}{k}} = \sqrt{2} \cdot \sqrt{2} \pi \sqrt{\frac{m}{2k}} = \pi \sqrt{\frac{4m}{2k}} = \pi \sqrt{\frac{2m}{k}}$Thus,the correct option is $D$.

Two identical particles each of mass $m$ are interconnected by a light spring of stiffness $k$. The time period for small oscillations is equal to:

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