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(B) The frequency of oscillation is $n = 10 \ Hz$. The angular frequency is $\omega = 2 \pi n = 20 \pi \ rad/s$.For a vertical spring-mass system,the

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

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(B) The frequency of oscillation is $n = 10 \ Hz$. The angular frequency is $\omega = 2 \pi n = 20 \pi \ rad/s$.For a vertical spring-mass system,the equilibrium position is at a distance $x_0 = \frac{mg}{k}$ below the unstretched position.Since the spring is unstretched at the highest point,the amplitude $A$ of the oscillation is equal to the displacement of the equilibrium position from the unstretched position,so $A = x_0 = \frac{mg}{k}$.We know that $\omega^2 = \frac{k}{m}$,so $k = m \omega^2$.Substituting $k$ into the expression for $A$: $A = \frac{mg}{m \omega^2} = \frac{g}{\omega^2}$.The maximum speed is $v_{\max} = \omega A = \omega \left( \frac{g}{\omega^2} \right) = \frac{g}{\omega}$.Substituting the values: $v_{\max} = \frac{10}{20 \pi} = \frac{1}{2 \pi} \ m/s$.

Frequency of a particle performing $S.H.M.$ is $10 \ Hz$. The particle is suspended from a vertical spring. At the highest point of its oscillation,the spring is unstretched. The maximum speed of the particle is $(g = 10 \ m/s^2)$.

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