# Calculating Expected Systematic Error in a Pendulum Experiment

I am a little confused by part c of problem 4.28 of Taylor's Introduction to Error Analysis book. A student measures the acceleration due to gravity by using a steel ball suspended by a light string. He records five different lengths (51.2, 59.7, 68.2, 79.7, 88.3)(all in cm) and five different periods (1.448, 1.566, 1.669, 1.804, 1.896)(all in seconds). He uses the formula

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

To calculate the mean and SDOM of the acceleration due to gravity, which I calculated to be $965.6\pm 1.5\,cm/s^2$. I have double checked this value, and am pretty sure of the result. The student is concerned that the accepted value $g=979.6\,cm/s^2$ is not contained within the calculated uncertainty, and looks for systematic error.

The question: how large would a systematic error in length l have to be so that the margins of the total error just include the accepted value of g? What I did was set

$$Avg(\frac{4\pi^2(l+\Delta l)}{T^2})+SDOM(g)=979.6$$ $$\rightarrow Avg(\frac{4\pi^2l}{T^2})+Avg(\frac{4\pi^2\Delta l}{T^2})+SDOM(g)=979.6$$

$$Avg(\frac{4\pi^2\Delta l}{T^2})=12.5$$ $$\rightarrow \Delta l=0.894$$

This is about 1.3% of the average length. However, in the book, it says that my answer should be approximately 1.5%. Am I doing something wrong? Is my procedure or calculation wrong, or I am over-analyzing the discrepancy between my value and the book's?

• Maybe calculation error: 979.6 - 965.64 = 13.96 ? – Vasiliy Sep 22 '13 at 21:31
• @Vasiliy I used the correct significant figures now, but I still only get 1.3% instead of the expected 1.5% – Bronzeclocksofbenin Sep 22 '13 at 23:45

Maybe this problem solve like this

Look at the Sec. 4.6 (Systematic Errors)

In part (c) Problem 4.28 We must find the systematic error of length of the pendulum ($\delta l_{sys}$). Given that the value of $\delta g_{tot}$ is $|g_{accepted} - \bar{g}|$.

Using the above argument and Eq. 4.26 $\left[ \delta g_{tot} = \sqrt{\delta g_{sys}^2 + \delta g_{ran}^2} \,\,\right]$, \begin{align} \delta g_{tot} &= \sqrt{\delta g_{sys}^2 + \delta g_{ran}^2} \\ |g_{accepted} - \bar{g}| &= \sqrt{\delta g_{sys}^2 + \sigma_{\bar{g}}^{\,\,\,2}} \\ \delta g_{sys} &= \sqrt{\mathrm{discrepancy}^2 - \sigma_{\bar{g}}^{\,\,\,2}} \end{align}

and propagation error techniques for the equation $g = 4\pi l/T^2$ gives the formula for $\delta g_{sys}$ \begin{align} \frac{\delta g_{sys}}{\bar{g}} = \sqrt{\left(\frac{\delta l_{sys}}{\bar{l}}\right)^2 + \left(2\frac{\delta{T_{sys}}}{\bar{T}}\right)^2} \end{align}

Since there was no problem with the measurement of the period $T$ ($\delta T_{sys} = 0$), we have the propagation error for $\delta g_{sys}$ \begin{align} \frac{\delta g_{sys}}{\bar{g}} = \frac{\delta l_{sys}}{\bar{l}} \end{align}

Therefore, the systematic of the length $l$ (in percentage unit) is $$\frac{\delta l_{sys}}{\bar{l}} = \frac{\sqrt{\mathrm{discrepancy}^2 - \sigma_{\bar{g}}}}{\bar{g}} \cdot 100\%$$

Substituting all the value has given and obtained from the previous question, we have the final result \begin{align} \frac{\delta l_{sys}}{\bar{l}} &= 1.43793098102868 \% \\ &\approx 1.4\% \end{align}