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(C) The angular frequency $\omega$ of a two-body spring-mass system is given by $\omega = \sqrt{\frac{k_{	ext{eq}}}{\mu}}$,where $k_{	ext{eq}}$ is t

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$10$

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(C) The angular frequency $\omega$ of a two-body spring-mass system is given by $\omega = \sqrt{\frac{k_{	ext{eq}}}{\mu}}$,where $k_{	ext{eq}}$ is the equivalent spring constant and $\mu$ is the reduced mass.$1$. Calculate the equivalent spring constant $k_{	ext{eq}}$:The springs are connected in series. Therefore,$k_{	ext{eq}} = \frac{k_1 k_2}{k_1 + k_2}$.Given $k_1 = k$ and $k_2 = 4k$,we have:$k_{	ext{eq}} = \frac{k 	imes 4k}{k + 4k} = \frac{4k^2}{5k} = \frac{4k}{5}$.Substituting $k = 20 \, N/m$:$k_{	ext{eq}} = \frac{4 	imes 20}{5} = 16 \, N/m$.$2$. Calculate the reduced mass $\mu$:$\mu = \frac{m_1 m_2}{m_1 + m_2}$.Given $m_1 = 200 \, g = 0.2 \, kg$ and $m_2 = 800 \, g = 0.8 \, kg$:$\mu = \frac{0.2 	imes 0.8}{0.2 + 0.8} = \frac{0.16}{1.0} = 0.16 \, kg$.$3$. Calculate the angular frequency $\omega$:$\omega = \sqrt{\frac{16}{0.16}} = \sqrt{100} = 10 \, rad/s$.

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