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\textbf{Practical Set-up Notes and Lab-Specific Advice}
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With the introduction to the lab taking a maximum of 15 minutes, it It is
recommended that demonstrators recommend to divide up the 3 hours so that the first two parts take
~40 45 minutes, and the third takes
~90 90 minutes. Mark each section as you go to avoid a scramble at the end of the lab, and also to correct students who are not calculating uncertainties correctly,
etc.\\
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Each semester on the Intro lab we have a miscommunication issue between the demonstrators. Please, come at least 15 minutes before the lab and discuss (with your colleagues) what the answers are for the practical questions and how they are solved, especially if it is your first time or you have forgotten the approach from the last time when you did demonstration.\\
\phantom{1}\\ Do \textbf{not} argue with your colleague in front of the students! Even if you think that your colleague is wrong. Please have a word with him/her when he/she walks away from the student. Many times I have seen how demonstrators correct each other in front of the students, and students always get confused as they don’t know who is right and who is wrong. Sometimes they ask the same question to different demonstrators and when they get different answers they don’t know what to think. If they say that another demonstrator told you something else and you think it is wrong, ask them to wait, have a word with your colleague and then explain the agreed-upon approach to the student
together!\\
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Finally, make the distinction between random errors and measurement uncertainty. Just because a student places \textit{the same weight on the same scale} 10 times does not mean they can increase their precision using $\sigma_{\bar{X}} = \frac{\sigma_{X}}{\sqrt{N}}$ - if the base uncertainty of the scale is much larger than any gaussian fluctuations, it is difficult (or impossible) to justify applying normal-distribution tricks such as adding variances rather than absolute uncertainties. together!
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\textbf{Practice Problem - Approach and Solutions}
\begin{enumerate}
...
\begin{itemize}
\item{Find mean radius by dividing mean diameter by 2:
\begin{align*}
\mu_D
= \frac{\sum{D_i}}{N} &=
\frac{\sum{D_i}}{N} = \frac{24.75+24.82+24.87+24.79+24.95+24.71+24.81+24.65+24.69}{9}\\
&= \frac{223.04}{9} = 24.78\text{mm}\\
\text{Therefore, } \mu_r &= \frac{\mu_D}{2} = \frac{24.78}{2}\\
&= 12.39\text{mm}\\
...
Based on data from Reserve Bank Coinage Specifications[1], the densities for 10 cent, 20 cent, 50 cent, 1 dollar and 2 dollar coins should come out as 6328, 6901, 6113, 7027, 6715 kg/m3 respectively. Having said this, values should roughly fall within 6000-8000 $kgm^{-3}$. Diameters and masses are pretty consistent between testing one coin and five coins, but thicknesses are very dependent on exactly where you test the coins. Students will probably need to measure the thickness of the coin at the *rim* to get a density in the Specification range. By measuring thicknesses at the center of each coin, the calculated densities are closer to 7500$kgm^{-3}$.\\
\begin{table} \begin{table}[!h]
\begin{tabular}{c|c|c|c|c|c}
& 10c & 20c & 50c & \$1 & \$2 \\ \hline
Thickness($\pm0.01$mm) & 1.37 & 1.48 & 1.39 & 2.59 & 2.37 \\ \hline
...
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It is assumed that all of the measurements lie within the scale uncertainty of the calipers - if this is not the case, then adding the relative errors in quadrature, rather than linearly, is an acceptable approach.
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\begin{table} \begin{table}[!h]
\begin{tabular}{ c|c|c|c|c|c }
$\rho=\frac{M}{V}$ & 10c & 20c & 50c & \$1 & \$2 \\ \hline
V (m$^3$) & $4.51\times10^{-7}$ & $5.47\times10^{-7}$ & $6.69\times10^{-7}$ & $1.07\times10^{-6}$ & $1.312\times10^{-6}$ \\ \hline
$\frac{\delta V}{V}$ \% & 0.828 & 0.768 & 0.8002 & 0.473 & 0.7986 \\ \hline
$\rho$ ($\frac{kg}{m^3}$) & 7538.8 & 7495.4 & 7623.3 & 7476.6 & 7469.5 \\ \hline
$\frac{\delta\rho}{\rho}$ \% & 3.77 & 3.207 & 2.761 & 1.723 & 1.819 \\
\end{tabular}
\caption{Volumes and densities for single coins}
\end{table}
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In short, you should expect the uncertainties for the volume to be around $0.5-1\%$ and the density to be around
$2-5\%$.\\ $2-5\%$.
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For the last question of part 1,
if we assume that it is the
random (gaussian) variations between coins key thing for students to realise is that
causes if 10 coins were used, each relative uncertainty would decrease by a factor of 10, which would ultimately propagate to 10 times the
difference in our measurements, rather than precision. It is not necessary for students to derive this, but they should be able to explain their reasoning.\\
Basically, if $r_i$ is the
scale relative uncertainty of
the calipers/micrometers, one measurement, then
with $n$ coins, this will become $\frac{r_i}{n}$. From there, a quick derivation:
\begin{align*} & 10=\dfrac {r_{old}} {r_{new}}\\ & =\dfrac {\sqrt {\sum r_{i}^{2}}} {\sqrt {\sum \left( \dfrac {r_{i}} {n}\right) ^{2}}}\\ & =n\end{align*} thus n=10 coins.\\
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Here is a good example for this question:\\
\textbf{Example}\\
(a) If 10 is added to each number in a set, what is the
standard error of change in the
calculated mean (as compared with mean? What is the change in the
true mean), calculated from $N$ independent measurements, standard deviation?\\
(b) If each number in a set is
given multiplied by
\begin{equation*}
\sigma_{\bar{X}} = \frac{\sigma_X}{\sqrt{N}}
\end{equation*}
where $X$ 10, what is the
set of measurements, $\bar{X}$ is their mean value, $\sigma_{X}$ change in the mean? What is the
change in the standard
deviation of deviation?\\
\textbf{Solution}\\
(a) Since each value increases by 10, the
measurements from mean increases by 10. Adding 10 to each number does not change the
calculated mean, distances between the numbers and
$\sigma_{\bar{X}}$ the mean. Since the standard deviation is
dependent on those distances, the standard
error deviation does not change. If you are not sure about the answer, you can plug in some numbers and try it. The mean of
{1, 2, 3} is 2. The mean of {11, 12, 13} is 12. The mean increased by 10.The distances between {1, 2, 3} and their mean are 1. The distances between {11, 12, 13} and their mean are also 1. The standard deviation did not change.\\
(b) Since each value is multiplied by 10, the
calculated mean
relative to is multiplied by 10. Multiplying each number by 10 changes the
true mean. In order to reduce distances between the numbers and the mean by a factor of 10. Since the standard deviation is dependent on those distances, the standard
error deviation changes by a factor of
10, we need increase 10. If you are not sure about the
number $N$ answer, you can Plug In some numbers and try it. The mean of
measurements taken to $N= 10^2 = 100$. Similarly, {1, 2, 3} is 2. The mean of {10, 20, 30} is 20. The mean increased by
taking 10 measurements, we reduce a factor of 10. The distances between {1, 2, 3} and their mean are 1. The distances between {10, 20, 30} and their mean are 10. Since the distances changed by a factor of 10, the standard
error deviation also changed by a factor of
$\sqrt{10} \approx 3.16$. \\ 10.\\
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In part 3 (measuring reaction time), the free-fall distance associated with a reaction time of 0.18s is 16.2cm. This should be what people aim for, but it is difficult to test this in a reliable manner in the lab, so expect results to vary wildly - results will most likely be within the range of 10cm - 40cm.\\
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\textbf{Key Discussion Points}
Again, it It is worth emphasising that students should `add errors in quadrature' when the errors are generated by random fluctuations. If the fundamental uncertainty of the measuring equipment is the limiting factor, the quadrature approach is no longer appropriate. This is covered in appendix A4.1 under the discussion of dependent and independent
quantities.\\
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Section 1.3 - Archimedean method of determining volume by placing the coin in water and measuring the volume of displaced water would be good example. A reasonable way to do this is to use a graduated (marked) cylinder. Pour water into the graduated cylinder until it reaches a known level (seen by markings on the cylinder’s surface). Add the object to the water and record the new water level. The difference between the new water level and the original level will be the object's volume. This measurement is taken in milliliters, which are interchangeable with cubic centimeters. Once volume is determined, then weigh the object on a scale and do the same calculations as before. \\
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Section 5.1 - Students should use $\frac{\mu_1 - \mu_2}{\sigma}$ formula\\
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Section 5.2 - Ideally students should give a rough probability of difference occurring due to random error using the table from A.5.3\\
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Section 5.3 - Firstly, students should distinguish between measuring their own reaction, and the average reaction time. Secondly, students should realise that their experiment cannot prove or disprove the reference value, only give evidence in support of or against it. Thirdly, students should realise that taking everyone’s value in the class would introduce sampling bias.\\
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\textbf{References}
