Question

A student measures the time T for one complete oscillation of a pendulum of length l.
Her results are shown below.
l/m-> 0.420+_ 0.001
T/s-> 1.3+_ 0.1
She uses the formula
T=2π√l÷g
to calculate the acceleration of free fall g.
What is the best estimate of tje percentage uncertainty in the value of g?
A)0.02%
B)4%
C)8%
D)16%

Answer 1

Expressing $g$ from the formula, obtain:

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

Let's differentiate $g$:

$$dg = \dfrac{4\pi^2}{T^2}dl - \dfrac{8\pi^2l}{T^3}dT$$

where $dl = 0.001m$ is the uncertainty in length and $dT = 0.1s$ is the uncertainty in time. The percentage uncertainty in $g$ is then

$$\delta=\left|\dfrac{dg}{g}\right| = \left|\dfrac{dl}{l} - 2\dfrac{dT}{T}\right|$$

The maximum uncertainty occurs when $dT<0$. Then:

$$\delta_{max}=\left|\dfrac{dl}{l}\right| + 2\left|\dfrac{dT}{T}\right|\\ \delta_{max}=\left|\dfrac{0.001}{0.420}\right| + 2\left|\dfrac{0.1}{1.3}\right| \approx 0.156 \approx 16\%$$

Answer. 16%.