Question

The period of oscillation of a simple pendulum is $T=2\pi \sqrt{\frac{L}{g}}$. The measured value of $L$ is $20.0cm$ known to $1mm$ accuracy and time for $100$ oscillations of the pendulum is found to be $90s$ using a wristwatch of $1s$ resolution. What is the accuracy in the determination of $g$?

Answer 1

Given :

$T=2\pi \sqrt{\frac{L}{g}}$

Squaring both sides & rearranging

$g=4{\pi }^2\left(\frac{L}{T^2}\right)$

Now $\frac{\Delta g}{g}=\frac{\Delta L}{L}+\frac{2\Delta T}{T}$ ....(1)

$\Delta L=1mm=0.1cm$

If $t$ is time for $n$ oscillation then time taken for 1 oscillation is -

$T=\frac{t}{n}$

$\therefore \frac{\Delta T}{T}=\frac{\Delta t}{t}=\frac{1}{90}$

Substituting values in (1)

$\frac{\Delta g}{g}=\frac{0.1}{20}+2\times \frac{1}{90}$

$=0.005+0.022$

$\frac{\Delta g}{g}=0.027$

$\frac{\Delta g}{g}\times 100=0.027\times 100$

$=2.7$

$\therefore$ error is $2.7\%$