The time period of a simple harmonic oscillat | vedclass

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(D) The formula for the time period is $T = 2 \pi \sqrt{\frac{m}{k}}$.Squaring both sides, we get $T^2 = 4 \pi^2 \frac{m}{k}$, which implies $k = \fra

Vedclass · Accepted Answer

$6.76$

Vedclass · Answer

(D) The formula for the time period is $T = 2 \pi \sqrt{\frac{m}{k}}$.Squaring both sides, we get $T^2 = 4 \pi^2 \frac{m}{k}$, which implies $k = \frac{4 \pi^2 m}{T^2}$.The relative error in $k$ is given by $\frac{\Delta k}{k} = \frac{\Delta m}{m} + 2 \frac{\Delta T}{T}$.Given: $m = 10 \ g$, $\Delta m = 10 \ mg = 0.01 \ g$.Time for $50$ oscillations is $t = 60 \ s$, with resolution $\Delta t = 2 \ s$.The time period $T = \frac{t}{50} = \frac{60}{50} = 1.2 \ s$.The error in time period is $\Delta T = \frac{\Delta t}{50} = \frac{2}{50} = 0.04 \ s$.Substituting these values: $\frac{\Delta k}{k} = \frac{0.01}{10} + 2 	imes \frac{0.04}{1.2} = 0.001 + 2 	imes \frac{0.04}{1.2} = 0.001 + 0.0666... = 0.06766...$Percentage error = $0.06766 	imes 100 \% \approx 6.76 \%$.

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