Question

The time $t$, of a complete oscillation of a simple pendulum of length $l$ is given by $t=2\pi \sqrt{\frac{l}{g}}$ where $g$ is gravitational constant. Find the approximate percentage of errors in $t$ when the percentage of error $l$ is $1$%.

Answer 1

$\because T=2\pi \sqrt{\frac{l}{g}}$

$\because \frac{\Delta T}{T}=\frac{1}{2}\frac{\Delta l}{l}$

$\because \frac{\Delta T}{T}\%=\frac{1}{2}\times 1=\frac{1}{2}\%$