Question

The acceleration due to gravity is found upto an accuracy of $4\%$ on a planet. The energy supplied to a simple pendulum to known mass ' $m$ ' to undertake oscillations of time period $T$ is being estimated. If time period is measured to an accuracy of $3\%$, the accuracy to which $E$ is known as ...........$\%$

Answer 1

Answer: 14

$T=2\pi \sqrt{\frac{\ell }{g}}$
$\Rightarrow \ell =\frac{T^{2}g}{4{\pi }^2}$
$E=mg\ell \frac{{\theta }^2}{2}$
$=m{g}^2\frac{T^2{\theta }^2}{8{\pi }^2}$
$\frac{dE}{E}=2\left(\frac{dg}{g}+\frac{dT}{T}\right)$
$=(4+3)=14\%$