Energy of a simple pendulum in terms of $m, g, T$

For a simple pendulum, $T = 2\pi \sqrt{\frac{\ell}{g}}$ so, $\ell = \frac{gT^2}{4\pi^2}$

The energy supplied to raise the bob through height comparable to the pendulum length is proportional to gravitational potential energy: $E \propto mg\ell$

Substituting $\ell$, $E \propto mg\left(\frac{gT^2}{4\pi^2}\right)$ $E \propto m g^2 T^2$

Since $m$ is known exactly, only $g$ and $T$ contribute to error.

Use percentage error propagation

If $E \propto g^2 T^2$ then maximum percentage error is $\frac{\Delta E}{E}\times 100 = 2\left(\frac{\Delta g}{g}\times 100\right) + 2\left(\frac{\Delta T}{T}\times 100\right)$

Given: $\frac{\Delta g}{g}\times 100 = 4\%$ $\frac{\Delta T}{T}\times 100 = 3\%$

Therefore, $\frac{\Delta E}{E}\times 100 = 2(4) + 2(3) = 8 + 6 = 14\%$

Final answer

The accuracy to which $E$ is known corresponds to an error of $14\%$