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(B) The time period of a mass-spring system is given by $T = 2\pi \sqrt{\frac{m}{k}}$.Given $T_1 = 2\pi \sqrt{\frac{m}{k}} = 0.6 \ s$ ... $(i)$And $T_

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$3.5$

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(B) The time period of a mass-spring system is given by $T = 2\pi \sqrt{\frac{m}{k}}$.Given $T_1 = 2\pi \sqrt{\frac{m}{k}} = 0.6 \ s$ ... $(i)$And $T_2 = 2\pi \sqrt{\frac{m + m'}{k}} = 0.7 \ s$ ... $(ii)$Dividing $(ii)$ by $(i)$,we get $\frac{T_2}{T_1} = \sqrt{\frac{m + m'}{m}} = \frac{0.7}{0.6} = \frac{7}{6}$.Squaring both sides,$\frac{m + m'}{m} = \frac{49}{36}$,which implies $1 + \frac{m'}{m} = \frac{49}{36}$.Thus,$\frac{m'}{m} = \frac{49}{36} - 1 = \frac{13}{36}$.The extension $x$ caused by additional mass $m'$ is $x = \frac{m'g}{k}$.From $(i)$,$\frac{m}{k} = \frac{(0.6)^2}{4\pi^2} = \frac{0.36}{4\pi^2}$.Substituting $m' = \frac{13}{36}m$ into the extension formula: $x = \frac{13}{36} \cdot \frac{mg}{k} = \frac{13}{36} \cdot g \cdot \frac{m}{k}$.Using $g \approx 10 \ m/s^2$ and $\pi^2 \approx 10$,we get $x = \frac{13}{36} \cdot 10 \cdot \frac{0.36}{4 \cdot 10} = \frac{13}{36} \cdot \frac{0.36}{4} = \frac{13}{36} \cdot 0.09 = 13 \cdot 0.0025 = 0.0325 \ m = 3.25 \ cm$.Rounding to the nearest provided option,the extension is approximately $3.5 \ cm$.

Similar Questions

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