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{m_1 g}

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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 $m_2 = 2.0 ~kg$ suspended from the same spring,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 simple harmonic motion.The maximum velocity $v_{\max}$ is given by $v_{\max} = A \omega$.$v_{\max} = 0.1 ~m 	imes 10 ~rad/s = 1 ~m/s$.

When a body of mass $1.0 ~kg$ is suspended from a certain light spring hanging vertically,its length increases by $5 ~cm$. By suspending a $2.0 ~kg$ block to the spring,if the block is pulled through $10 ~cm$ and released,the maximum velocity in $m/s$ is: (Acceleration due to gravity $= 10 ~m/s^2$)

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