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(D) The frequency of $S.H.M.$ is given by $n = 5 Hz$. The angular frequency is $\omega = 2 \pi n = 2 \pi 	imes 5 = 10 \pi rad s^{-1}$.At the highest

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$\frac{1}{\pi} m s^{-1}$

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(D) The frequency of $S.H.M.$ is given by $n = 5 Hz$. The angular frequency is $\omega = 2 \pi n = 2 \pi 	imes 5 = 10 \pi rad s^{-1}$.At the highest point of oscillation,the spring is unstretched,meaning the extension $x = 0$. In $S.H.M.$,the equilibrium position is where the spring force balances gravity,$k x_0 = mg$,where $x_0$ is the static extension.The amplitude $A$ of the oscillation is equal to this static extension $x_0$,because the particle reaches the unstretched position $(x=0)$ at the top of its path. Thus,$A = x_0 = \frac{mg}{k}$.We know that $\omega^2 = \frac{k}{m}$,so $k = m \omega^2 = m(10 \pi)^2 = 100 \pi^2 m$.Substituting $k$ into the amplitude equation: $A = \frac{mg}{100 \pi^2 m} = \frac{g}{100 \pi^2} = \frac{10}{100 \pi^2} = \frac{1}{10 \pi} m$.The maximum speed is $V_{max} = \omega A = (10 \pi) 	imes (\frac{1}{10 \pi}) = 1 m s^{-1}$.Wait,re-evaluating the calculation: $V_{max} = \omega A = (10 \pi) 	imes \frac{g}{\omega^2} = \frac{g}{\omega} = \frac{10}{10 \pi} = \frac{1}{\pi} m s^{-1}$.

$A$ mass is suspended from a vertical spring which is executing $S.H.M.$ of frequency $5 Hz$. The spring is unstretched at the highest point of oscillation. The maximum speed of the mass is [acceleration due to gravity $g=10 m s^{-2}$].

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