A mass of 0.2,kg is attached to the lower end | vedclass

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(D) $1$. Equilibrium stretching: At equilibrium,the weight of the mass is balanced by the spring force. $mg = kx \implies 0.2 	imes 10 = 200 	imes x

Vedclass · Accepted Answer

All of the above.

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(D) $1$. Equilibrium stretching: At equilibrium,the weight of the mass is balanced by the spring force. $mg = kx \implies 0.2 	imes 10 = 200 	imes x \implies x = 0.01\,m = 1\,cm.$ Thus,statement $A$ is true.$2$. Motion from unstretched state: When the mass is raised to the unstretched position and released,the amplitude of oscillation $A$ is equal to the equilibrium extension $x = 1\,cm.$ The mass will oscillate between the unstretched position $(0\,cm)$ and the lowest point $(2A = 2\,cm)$. Thus,it will go down by $2\,cm$ before moving upwards. Statement $B$ is true.$3$. Frequency of oscillation: The frequency $f$ is given by $f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} = \frac{1}{2\pi} \sqrt{\frac{200}{0.2}} = \frac{1}{2\pi} \sqrt{1000} \approx \frac{31.62}{6.28} \approx 5.03\,Hz.$ This is nearly $5\,Hz.$ Statement $C$ is true.Since all statements are true,the correct option is $D$.

$A$ mass of $0.2\,kg$ is attached to the lower end of a massless spring of force-constant $200\,N/m,$ the upper end of which is fixed to a rigid support. Which of the following statements is/are true?

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