Units and Measurements Part 1

Given, accuracy of acceleration due to gravity = 4% Accuracy of time period = 3% Energy stored in pendulum at any instant is given as $E = \frac{mgL \theta^{2}}{2}$ ...(i) Time period of pendulum is given by expression $T = 2 \pi \sqrt{\frac{L}{g}}$ Rearrange the above expression for L, we get $L = g \left( \frac{T}{2 \pi} \right)^{2}$ ...(ii) Substituting the value of L in Eq. (i), we get $E = \frac{m g^{2} \theta^{2}}{2} \left( \frac{T}{2 \pi} \right)^{2} = \frac{m g^{2} \theta^{2} T^{2}}{8 \pi^{2}}$ Now, the accuracy in measurement of energy can be calculated as $\frac{\Delta E}{E} \times 100 = 2 \frac{\Delta g}{g} \times 100 + 2 \frac{\Delta T}{T} \times 100$ (∵ m,θ and π are constant) ⇒ $\frac{\Delta E}{E} \times 100 = 2 \times 4 \% + 2 \times 3 \% = 14 \%$ Thus, the accuracy to which E is known is 14%.