Two masses m_1 = 1 , kg and m_2 = 0.5 , kg ar | vedclass

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(C) Initially,the spring is stretched by both masses $m_1$ and $m_2$ in equilibrium.The initial extension $x_i$ is given by $k x_i = (m_1 + m_2) g$,so

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$80$

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(C) Initially,the spring is stretched by both masses $m_1$ and $m_2$ in equilibrium.The initial extension $x_i$ is given by $k x_i = (m_1 + m_2) g$,so $x_i = \frac{(m_1 + m_2) g}{k}$.When $m_1$ is removed,the new equilibrium position (mean position) for the remaining mass $m_2$ is $x_f = \frac{m_2 g}{k}$.The amplitude of oscillation $A$ is the difference between the initial position and the new equilibrium position: $A = x_i - x_f$.$A = \frac{(m_1 + m_2) g}{k} - \frac{m_2 g}{k} = \frac{m_1 g}{k}$.Substituting the values $m_1 = 1 \, kg$,$g = 10 \, m/s^2$,and $k = 12.5 \, N/m$:$A = \frac{1 	imes 10}{12.5} = \frac{10}{12.5} = 0.8 \, m$.Converting to centimeters: $A = 0.8 	imes 100 = 80 \, cm$.

Two masses $m_1 = 1 \, kg$ and $m_2 = 0.5 \, kg$ are suspended together by a massless spring of spring constant $k = 12.5 \, N/m$. When the masses are in equilibrium,$m_1$ is removed without disturbing the system. The new amplitude of oscillation will be .......... $cm$.

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