Question

The following observations we're made during an experiment to find the value of g using simple pendulum l=90.0 cm

Time t for 20 vibrations =36 sec

Find percentage error in measurement of g given that T=2π√l/g length is being measured to an accuracy of 0.1 cm and time 0.2 sec

Answer 1

We know that

$T = 2\pi\sqrt{\dfrac{l}{g}} \Longrightarrow g = \dfrac{4\pi^2l}{T^2}.$

T is equal to 36:20 = 1.8 seconds, so

$g = \dfrac{4\pi^2 \cdot0.9}{1.8^2} \approx 11.0\,\mathrm{m/s^2}.$ The uncertainty of period will be $\dfrac{0.2}{20} = 0.01\,\mathrm{s}.$

Next, we should obtain the formula for calculating the percentage error (see part 5 in http://lectureonline.cl.msu.edu/~mmp/labs/error/e2.htm)

$\delta g = \sqrt{\left(\dfrac{\partial g}{\partial T}\delta T \right)^2 + \left(\dfrac{\partial g}{\partial l}\delta l \right)^2} = \sqrt{\left(-\dfrac{8\pi^2l}{T^3}\cdot {\delta T} \right)^2 + \left(\dfrac{4\pi^2}{T^2}\cdot{\delta l} \right)^2} = \sqrt{\left(-\dfrac{8\pi^2\cdot0.9}{1.8^3}\cdot{0.01} \right)^2 + \left(\dfrac{4\pi^2}{1.8^2}\cdot{0.001} \right)^2} \approx 0.12\,\mathrm{m/s^2}.$

The percentage error will be $\dfrac{0.12}{11.0} \approx 1.1\%.$

The simplier way to calculate the percentage error is to use the formula

$\dfrac{\delta l}{l} + 2\dfrac{\delta T}{T} = \dfrac{0.001}{0.9} + 2\dfrac{0.01}{1.8} = 1.2\%.$