The acceleration due to gravity is found up to an accuracy of 4 %...

$T = 2\pi\sqrt{\dfrac{l}{g}}$

$\dfrac{\Deltal}{l} \times 100 = \left(\dfrac{\Deltag}{g} \times 100\right) + \left(\dfrac{2\DeltaT}{T} \times 100\right)$

$⇒ \dfrac{\Deltal}{l} \times 100 = 4 + \left(2 \times 3\right) = 10$

$E = \dfrac{1}{2}m{\omega}^{2}{A}^{2} = \dfrac{1}{2}m\dfrac{g}{l}{A}^{2}\text{Therefore}, \dfrac{\DeltaE}{E} \times 100 = \left(\dfrac{\Deltag}{g} \times 100\right) + \left(\dfrac{\Deltal}{l} \times 100\right) = 4 + 10 = 14$