The vertical extension in a light spring by a | vedclass

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Question

(C) Given mass $m = 1 \, kg$ and extension $x = 9.8 \, cm = 9.8 	imes 10^{-2} \, m$.Using Hooke's Law,$mg = kx$,where $k$ is the spring constant.Ther

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$2\pi / 10 \, s$

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(C) Given mass $m = 1 \, kg$ and extension $x = 9.8 \, cm = 9.8 	imes 10^{-2} \, m$.Using Hooke's Law,$mg = kx$,where $k$ is the spring constant.Therefore,$\frac{m}{k} = \frac{x}{g}$.The period of oscillation $T$ for a spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{k}}$.Substituting $\frac{m}{k} = \frac{x}{g}$,we get $T = 2\pi \sqrt{\frac{x}{g}}$.Substituting the values: $T = 2\pi \sqrt{\frac{9.8 	imes 10^{-2}}{9.8}} = 2\pi \sqrt{10^{-2}} = 2\pi 	imes 10^{-1} = \frac{2\pi}{10} \, s$.

The vertical extension in a light spring by a weight of $1\, kg$ suspended from it is $9.8\, cm$. What is the period of oscillation?

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