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(B) The time period of a spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{k}}$,which implies $T \propto \sqrt{m}$.Initially,the total mass is $

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$2$

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(B) The time period of a spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{k}}$,which implies $T \propto \sqrt{m}$.Initially,the total mass is $M_1 = 400\,g + 500\,g + 700\,g = 1600\,g$. However,the problem states that after removing the $700\,g$ mass,the period is $3\,s$. Let the remaining mass be $m_1 = 400\,g + 500\,g = 900\,g$. So,$T_1 = 3\,s$ for $m_1 = 900\,g$.When the $500\,g$ mass is also removed,the new mass is $m_2 = 400\,g$.Using the relation $\frac{T_2}{T_1} = \sqrt{\frac{m_2}{m_1}}$:$\frac{T_2}{3} = \sqrt{\frac{400}{900}}$$\frac{T_2}{3} = \sqrt{\frac{4}{9}} = \frac{2}{3}$$T_2 = 2\,s$.

$A$ system consisting of three masses attached to a spring is in equilibrium. When the $700\,g$ mass is removed,the system oscillates with a period of $3\,s$. When the $500\,g$ mass is also removed,what will be the new time period of the system? (in $\text{seconds}$)

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