Question

The period of oscillation of a simple pendulum of length l is given by $T=2\pi \sqrt{l/g}$. The length l is about 10 cm and is known to 1 mm accuracy. The period of oscillation is about 0.5 s. The time of 100 oscillations has been measured with a stop watch of 1 s resolution. Find the percentage error in the determination of g.

Answer 1

Formula for the period of oscillation of a simple pendulum is

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

Squaring both sides

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

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

The error in the determination of g

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

here $L=10cm$

$\Delta L=0.1cm$

$T=0.5\times 100=50s$

$\Delta T=1s$

here $T=\frac{t}{n}$ and $\Delta T=\frac{\Delta t}{n}$

So,

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

As errors in both L and t are least count errors

hence from (1)

$\frac{\Delta g}{g}=\frac{0.1}{10}+2\left(\frac{1}{50}\right)$

$=0.005\%$