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(A) The frequency of the $S.H.M.$ is $f = 5 \ Hz$.Angular frequency $\omega = 2\pi f = 2\pi 	imes 5 = 10\pi \ rad/s$.At the highest point of oscillat

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

Vedclass · Answer

(A) The frequency of the $S.H.M.$ is $f = 5 \ Hz$.Angular frequency $\omega = 2\pi f = 2\pi 	imes 5 = 10\pi \ rad/s$.At the highest point of oscillation,the spring is unstretched,which means the extension at the equilibrium position is equal to the amplitude $A$ of the oscillation.Thus,$A = \Delta l = \frac{mg}{k}$.Since $\omega^2 = \frac{k}{m}$,we have $\frac{g}{\omega^2} = \frac{mg}{k} = A$.Substituting the values: $A = \frac{10}{(10\pi)^2} = \frac{10}{100\pi^2} = \frac{1}{10\pi^2} \ m$.The maximum speed is given by $V_{\max} = A\omega$.$V_{\max} = \left(\frac{1}{10\pi^2}ight) 	imes (10\pi) = \frac{1}{\pi} \ m/s$.

$A$ particle 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. What is the maximum speed of the particle? (Take $g = 10 \ m/s^2$)

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