When a body of mass 1.0 ,kg is suspended from | vedclass

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(B) Given mass $m_1 = 1.0 \,kg$, extension $l_1 = 5 \,cm = 0.05 \,m$. Using Hooke's Law, $m_1 g = k l_1$, where $k$ is the spring constant. $k = \frac

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(B) Given mass $m_1 = 1.0 \,kg$, extension $l_1 = 5 \,cm = 0.05 \,m$. Using Hooke's Law, $m_1 g = k l_1$, where $k$ is the spring constant. $k = \frac{m_1 g}{l_1} = \frac{1.0 	imes 10}{0.05} = 200 \,N/m$. Now, for a mass $m_2 = 2.0 \,kg$, the angular frequency $\omega$ is given by $\omega = \sqrt{\frac{k}{m_2}}$. $\omega = \sqrt{\frac{200}{2.0}} = \sqrt{100} = 10 \,rad/s$. The block is pulled by $A = 10 \,cm = 0.1 \,m$, which represents the amplitude of the oscillation. The maximum velocity $v_{\max}$ is given by $v_{\max} = A \omega$. $v_{\max} = 0.1 \,m 	imes 10 \,rad/s = 1 \,m/s$.

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