The period of oscillation of a simple pendulum is given by $\mathrm{T}=2 \pi \sqrt{(} \ell / \mathrm{g})$ what is the equation of the relative error \(\Delta \mathrm{T} / \mathrm{T}\) in measurement of period \(\mathrm{T}\) ? (a) \((1 / 2)(\Delta \ell / \ell)\) (b) \(2(\Delta \ell / \ell)\) (c) \((1 / 4)(\Delta \ell / \ell)\) (d) \(4(\Delta \ell / \ell)\)

The formula for the period of oscillation of a simple pendulum is given by: \[T = 2\pi\sqrt{\frac{\ell}{g}}\]

To find the relative error, we need the derivative of T with respect to the length, \(\frac{dT}{d\ell}\). To do this, we can first rewrite the equation above as: \[T = 2\pi\left(\frac{\ell}{g}\right)^{\frac{1}{2}}\] Now, applying the chain rule, we get: \[\frac{dT}{d\ell} = 2\pi\frac{1}{2}\left(\frac{\ell}{g}\right)^{-\frac{1}{2}}\frac{d(\frac{\ell}{g})}{d\ell}\] Next, find the derivative of \(\frac{\ell}{g}\) with respect to \(\ell\): \[\frac{d(\frac{\ell}{g})}{d\ell} = \frac{1}{g}\] Finally, substitute the result into the equation: \[\frac{dT}{d\ell} = \frac{2\pi}{2}\left(\frac{\ell}{g}\right)^{-\frac{1}{2}}\frac{1}{g}\] Simplifying the equation, we get: \[\frac{dT}{d\ell} = \frac{\pi}{\sqrt{g\ell}}\]

The formula for the relative error is: \[\frac{\Delta T}{T} = \left|\frac{dT}{d\ell}\right|\frac{\Delta\ell}{\ell}\] Substitute the obtained result of \(\frac{dT}{d\ell}\) and the given formula for the period of oscillation \(T\) into the relative error formula: \[\frac{\Delta T}{T} = \left|\frac{\pi}{\sqrt{g\ell}}\right|\frac{\Delta\ell}{\ell}\cdot\frac{1}{2\pi\sqrt{\frac{\ell}{g}}}\] Simplify the equation: \[\frac{\Delta T}{T} = \frac{1}{2}\frac{\Delta\ell}{\ell}\]

The obtained equation for the relative error is: \[\frac{\Delta T}{T} = \frac{1}{2}\frac{\Delta\ell}{\ell}\] This matches with option (a). Thus, the correct answer is: (a) \(\frac{1}{2}(\frac{\Delta\ell}{\ell})\)