Question

The period of oscillation of a simple pendulum is given by $T=2\pi \sqrt{\frac{l}{g}}$ where $l$ is about $100cm$ and known to have $1mm$ accuracy. The period is about $2s$. The time of $100$ oscillations is measured by a stopwatch of least count $0.1s$. The percentage error in $g$ is

Answer 1

Find percentage error in $l$.

Given, $l=100cm$

Accuracy,$\Delta l=1mm=0.1cm$

So, percentage error in $l$,

$\frac{\Delta l}{l}\times 100=\frac{0.1}{100}\times 100$

$\frac{\Delta l}{l}\times 100=0.1$

Given, $T=2s$

And $100$ oscillations using a stopwatch of least count $0.1s$,

Accuracy, $\Delta T=\frac{0.1}{100}$

So, percentage error in $T$,

$\frac{\Delta T}{T}\times 100=\frac{0.1}{100\times 2}\times 100$

$\frac{\Delta T}{T}\times 100=0.05$

Find percentage error in value of $g$,

Given ,time period of simple pendulum
$T=2\pi \sqrt{\frac{l}{g}}$

$g=\frac{4{\pi }^{2}l}{T^2}$

Percentage error in value of $g$.

$\frac{\Delta g}{g}\times 100=\%errorinvalueofl$ + 2 $\%errorinvalueofT)$

$\frac{\Delta g}{g}\times 100=0.1+2\left(0.05\right)$

$\frac{\Delta g}{g}\times 100=0.2$

Final answer : $0.20$