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(A) In a vertical spring-mass system,the equilibrium position is where the spring is stretched by an extension $x_0 = \frac{mg}{k}$.The period of osci

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$\frac{1}{2 \pi} \ m/s$

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(A) In a vertical spring-mass system,the equilibrium position is where the spring is stretched by an extension $x_0 = \frac{mg}{k}$.The period of oscillation is $T = 2 \pi \sqrt{\frac{m}{k}}$,which implies $\frac{k}{m} = \omega^2 = \left(\frac{2 \pi}{T}ight)^2$.Given $T = 0.1 \ s$,we have $\omega = \frac{2 \pi}{0.1} = 20 \pi \ rad/s$.The extension at equilibrium is $x_0 = \frac{g}{\omega^2} = \frac{10}{(20 \pi)^2} = \frac{10}{400 \pi^2} = \frac{1}{40 \pi^2} \ m$.Since the spring is unstretched at the highest point,the amplitude $A$ of the oscillation is equal to the equilibrium extension $x_0$,so $A = x_0 = \frac{1}{40 \pi^2} \ m$.The maximum speed $v_{max}$ in $S.H.M.$ is given by $v_{max} = A \omega$.Substituting the values: $v_{max} = \left(\frac{1}{40 \pi^2}ight) 	imes (20 \pi) = \frac{20 \pi}{40 \pi^2} = \frac{1}{2 \pi} \ m/s$.

$A$ mass suspended from a vertical spring performs $S.H.M.$ with a period of $T = 0.1 \ s$. The spring is unstretched at the highest point of its motion. What is the maximum speed of the mass? (Take gravitational acceleration $g = 10 \ m/s^2$)

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