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Question

(A) For a particle in $S.H.M.$ suspended from a vertical spring,the equilibrium position is where the spring force balances the gravitational force,i.

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) For a particle in $S.H.M.$ suspended from a vertical spring,the equilibrium position is where the spring force balances the gravitational force,i.e.,$kx = mg$. At the highest point of oscillation,the spring is unstretched,meaning the extension $x = 0$. Since the equilibrium position is at a distance $A$ (amplitude) below the highest point,the extension at equilibrium is $x = A$. Therefore,$kA = mg$,which gives the amplitude $A = \frac{mg}{k} = \frac{g}{\omega^2}$. Given frequency $f = 5 \ Hz$,the angular frequency is $\omega = 2 \pi f = 2 \pi 	imes 5 = 10 \pi \ rad/s$. Substituting the values,$A = \frac{10}{(10 \pi)^2} = \frac{10}{100 \pi^2} = \frac{1}{10 \pi^2} \ m$. The maximum speed is $V_{\max} = A \omega$. $V_{\max} = \left( \frac{1}{10 \pi^2} ight) 	imes (10 \pi) = \frac{1}{\pi} \ m/s$.

$A$ 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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