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(D) The time period of a vertical spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{k}}$.Given $T = 6 \ s$,we have $6 = 2\pi \sqrt{\frac{m}{k}}$

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$9$

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(D) The time period of a vertical spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{k}}$.Given $T = 6 \ s$,we have $6 = 2\pi \sqrt{\frac{m}{k}}$,which simplifies to $3 = \pi \sqrt{\frac{m}{k}}$.Squaring both sides,we get $9 = \pi^2 \frac{m}{k}$.Since $g = \pi^2$,we can write $9 = g \frac{m}{k}$,or $\frac{m}{k} = \frac{9}{g}$.When the mass is at rest,the spring is stretched by a distance $x$ such that the spring force equals the gravitational force: $kx = mg$.Therefore,the extension $x = \frac{mg}{k} = m \cdot \frac{g}{k}$.Substituting $\frac{m}{k} = \frac{9}{g}$,we get $x = \frac{g}{k} \cdot m = \frac{9}{g} \cdot g = 9 \ m$.

$A$ vertical spring oscillates with a period of $6 \ s$ when a mass $m$ is suspended from it. When the mass is at rest,the spring is stretched through a distance of (Take acceleration due to gravity $g = \pi^2 = 10 \ m/s^2$): (in $m$)

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