Question

In an experiment of simple pendulum, the errors in the measurement of length of the pendulum (L) and time period (T) are 3% and 2% respectively. The maximum percentage error in the value of $\frac{L}{T^{2}}$ is _____ %

Answer 1

Time period of simple pendulum is
$T = 2\pi \sqrt{\frac{L}{g}}$
Squaring both sides, we get
$\therefore \quad8 = 4\pi^{2} \frac{L}{g}$
or $\quad 8 = 4\pi ^{2} \frac{L}{T^{2}}$
The maximum percentage error in g is
$\frac{\Delta g}{g}\times100 = \frac{\Delta L}{L}\times 100+2\left(\frac{\Delta T}{T}\right)\times 100 = 3\% + 2 × 2\% = 7\%$
From $\left(i\right)$, we get $\frac{L}{T^{2}} = \frac{g}{4\pi ^{2}}$
The maximum percentage error in $\frac{L}{T^{2}}$ is
$\frac{\Delta\left(\frac{L}{T^{2}}\right)}{\frac{L}{T^{2}}}\times 100$ $ = \frac{\Delta g}{g}\times 100 = 7\%$