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

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(B) Initially,when a $1.0\, kg$ mass is suspended,the spring stretches by $x = 5\, cm = 0.05\, m$. Using Hooke's Law,$F = kx$,where $F = mg$:$k = \fra

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(B) Initially,when a $1.0\, kg$ mass is suspended,the spring stretches by $x = 5\, cm = 0.05\, m$. Using Hooke's Law,$F = kx$,where $F = mg$:$k = \frac{mg}{x} = \frac{1.0 	imes 10}{0.05} = 200\, N/m$.Now,for a mass $M = 2.0\, kg$ suspended from the same spring,the angular frequency $\omega$ is given by:$\omega = \sqrt{\frac{k}{M}} = \sqrt{\frac{200}{2}} = \sqrt{100} = 10\, rad/s$.The block is pulled by an amplitude $A = 10\, cm = 0.1\, m$ and released. The maximum velocity $v_{\max}$ is given by:$v_{\max} = A\omega = 0.1 	imes 10 = 1\, m/s$.

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