In the experiment of simple pendulum error in length of pendulum \((\ell)\) is \(5 \%\) and that of \(g\) is \(3 \%\) then find percentage error in measurement of periodic time for pendulum (a) \(4.2 \%\) (b) \(1.2 \%\) (c) \(2 \%\) (d) \(4 \%\)

Short Answer

Expert verified
The percentage error in the measurement of periodic time for the pendulum is \(4 \%\). The answer is (d) \(4 \%\).

Step by step solution

01

Use the formula for the period of a simple pendulum

The formula for the period \((T)\) of a simple pendulum is given by: \[ T = 2\pi \sqrt{\frac{\ell}{g}} \]
02

Compute the relative error in the period

To find the relative error in the periodic time, we need to compute the derivative of the period with respect to both \(\ell\) and \(g\). First, let's find the derivative of the period with respect to \(\ell\): \[ \frac{dT}{d\ell} = \frac{2\pi}{2 \sqrt{\ell g}} \] Now, we'll find the derivative of the period with respect to \(g\): \[ \frac{dT}{dg} = -\frac{\pi \sqrt{\ell}}{g^{\frac{3}{2}}} \]
03

Calculate the percentage error in the period

To find the percentage error in the period \((T)\), we will apply the formula for relative errors. If \(x\) and \(y\) are two variables, the relative error in \(z = f(x, y)\) is given by: \[ \frac{\Delta z}{z} = \left|\frac{\partial f}{\partial x}\frac{\Delta x}{x}\right| + \left|\frac{\partial f}{\partial y}\frac{\Delta y}{y}\right| \] Applying this formula to the period \((T)\), we get: \[ \frac{\Delta T}{T} = \left|\frac{dT}{d\ell}\frac{\Delta \ell}{\ell}\right| + \left|\frac{dT}{dg}\frac{\Delta g}{g}\right| \] Substituting the known values of percentage errors in \(\ell\) and \(g\) and the derivatives of the period with respect to \(\ell\) and \(g\): \[ \frac{\Delta T}{T} = \left|\frac{2\pi}{2 \sqrt{\ell g}}\frac{0.05 \ell}{\ell}\right| + \left|-\frac{\pi \sqrt{\ell}}{g^{\frac{3}{2}}}\frac{0.03 g}{g}\right| \] Simplifying the equation further: \[ \frac{\Delta T}{T} = \left|\frac{\pi}{\sqrt{\ell g}}(0.05)\right| + \left|\frac{\pi \sqrt{\ell}}{g^{\frac{3}{2}}}(0.03)\right| \] Notice that \(T = 2\pi \sqrt{\frac{\ell}{g}}\). Therefore, we can write the above equation as: \[ \frac{\Delta T}{T} = \frac{0.05}{2} + \frac{0.03}{2} \] Now, let's find \(\frac{\Delta T}{T}\): \[ \frac{\Delta T}{T} = 0.025 + 0.015 = 0.04 \] So, the percentage error in the measurement of periodic time for pendulum is \(4 \%\). The answer is (d) \(4 \%\).

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