Corrected exercises: measurement, uncertainty and graph analysis in the lab (203-SN1-RE)

The smallest division is $1$ mm, so $\Delta L=0.5$ mm $=0.05$ cm. Each end of the object is read to within half a division.

$\dfrac{\Delta L}{L}=\dfrac{0.05}{23.47}=2.1\times 10^{-3}$, that is $0.21$ %.

The uncertainty is rounded to ONE significant figure (sometimes two when that figure is a $1$): $0.0467\to 0.05$. The value is then rounded to the same decimal place as the uncertainty: $9.81234\to 9.81$. We write $g=(9.81\pm 0.05)$ m/s². Writing $9.81234\pm 0.0467$ claims to know thousandths when the hundredths are already uncertain.

$T=\dfrac{15.3}{10}=1.53$ s and $\Delta T=\dfrac{0.2}{10}=0.02$ s. $T=(1.53\pm 0.02)$ s.

The starting and stopping uncertainty ($0.2$ s) does not depend on the duration measured: over ten oscillations it is spread over ten periods, and the uncertainty on one period is divided by ten. With a single oscillation we would have $\Delta T/T=13$ %; with ten, $1.3$ %.

$\Delta x=47.8-12.3=35.5$ cm.

$0.1+0.1=0.2$ cm: $\Delta x=(35.5\pm 0.2)$ cm. Each reading can be off in either direction; in the worst case, the two errors add up.

$\Delta T=23.1-22.4=0.7$ °C, with an uncertainty of $0.2$ °C: $(0.7\pm 0.2)$ °C.

$\dfrac{0.2}{0.7}=0.29$, that is $29$ %, whereas each reading was uncertain by only $0.4$ %. A DIFFERENCE of two nearby quantities inflates the relative uncertainty: this is the first reflex to have when designing an experiment.

Because nothing guarantees that the two errors cancel. If $x_{1}$ is read too low and $x_{2}$ too high, the difference is overestimated by both errors at once. The rule $\Delta z=\Delta x+\Delta y$ covers this worst case.

$V=\pi\left(\dfrac{d}{2}\right)^{2}h=\pi\times 1.25^{2}\times 6.40=31.4$ cm³.

$\rho=\dfrac{m}{V}=\dfrac{85.2}{31.42}=2.712$ g/cm³.

$\rho=\dfrac{4m}{\pi d^{2}h}$: the diameter is SQUARED, so its relative uncertainty counts twice. $\dfrac{\Delta\rho}{\rho}=\dfrac{0.1}{85.2}+2\times\dfrac{0.01}{2.50}+\dfrac{0.05}{6.40}=0.12\,\%+0.80\,\%+0.78\,\%=1.70$ %. The exact factors, $\pi$ and $4$, contribute no uncertainty.

$\Delta\rho=0.0170\times 2.712=0.046$, rounded to $0.05$: $\rho=(2.71\pm 0.05)$ g/cm³.

The interval $[2.66, 2.76]$ contains $2.70$: the measurement is consistent with aluminum, and with aluminum alone. Notice that the balance, although the most “visible” instrument, weighs the least in the uncertainty: it is the caliper that would need improving.

$g=\dfrac{4\pi^{2}\times 0.800}{1.795^{2}}=9.802$ m/s².

$T$ is in the DENOMINATOR: $g$ is largest when $T$ is the SMALLEST. So you take $L_{max}=0.802$ m with $T_{min}=1.790$ s. Pairing $L_{max}$ with $T_{max}$ is the classic error: it makes the two effects cancel.

$g_{max}=\dfrac{4\pi^{2}\times 0.802}{1.790^{2}}=9.882$ m/s² and $g_{min}=\dfrac{4\pi^{2}\times 0.798}{1.800^{2}}=9.723$ m/s².

$\Delta g=\dfrac{9.882-9.723}{2}=0.079$ m/s², so $g=(9.80\pm 0.08)$ m/s².

$\dfrac{\Delta g}{g}=\dfrac{0.002}{0.800}+2\times\dfrac{0.005}{1.795}=0.25\,\%+0.56\,\%=0.81$ %, that is $\Delta g=0.079$ m/s². The two methods agree for small uncertainties; the extreme-values method remains usable when the formula is too complicated for the rules to apply.

$R=\dfrac{12.0^{2}\times\sin 80°}{9.81}=14.46$ m.

$\dfrac{\partial R}{\partial v_{0}}=\dfrac{2v_{0}\sin 2\theta}{g}=\dfrac{2R}{v_{0}}$, so $\dfrac{2\times 14.46}{12.0}\times 0.1=0.24$ m. This is the power rule again: $v_{0}^{2}$ doubles the relative uncertainty of $v_{0}$.

$\dfrac{\partial R}{\partial\theta}=\dfrac{2v_{0}^{2}\cos 2\theta}{g}$, and $\Delta\theta=1°=0.01745$ rad: $\dfrac{2\times 144\times\cos 80°}{9.81}\times 0.01745=0.089$ m. Leaving $\Delta\theta=1$ in the formula, in degrees, would give $5.1$ m, an absurd uncertainty.

$\Delta R=0.24+0.09=0.33$ m, rounded to $0.3$: $R=(14.5\pm 0.3)$ m.

$\cos 90°=0$: the contribution of the angle vanishes. At $45°$ the range is at its MAXIMUM, so it is insensitive to first order to a small error in the angle: that is the meaning of a zero derivative.

$\bar t=\dfrac{3.628}{8}=0.4535$ s.

Deviations from the mean, in ms: $-1.5$, $-5.5$, $7.5$, $1.5$, $-4.5$, $4.5$, $-2.5$, $0.5$. Sum of squares: $138$ ms². $s=\sqrt{\dfrac{138}{7}}=4.4$ ms. We divide by $N-1$: a mean computed from the data themselves “uses up” one degree of freedom.

$\dfrac{4.44}{\sqrt{8}}=1.6$ ms, so $\bar t=(0.4535\pm 0.0016)$ s, which we write $(0.454\pm 0.002)$ s.

$\dfrac{4.4}{1}\approx 4$. The release of the ball, its passage through the beam and small bumps vary from one trial to the next: the resolution of the display is NOT the uncertainty of the measurement. The thousandths digit is already uncertain.

The uncertainty of the mean decreases as $\dfrac{1}{\sqrt{N}}$: to divide it by two you need four times as many measurements, that is $32$. Repeating helps, but less and less, and never corrects a systematic error.

Squaring gives $T^{2}=\dfrac{4\pi^{2}}{g}L$. The graph of $T^{2}$ against $L$ is a straight line through the origin, with slope $\dfrac{4\pi^{2}}{g}$. Plotting $T$ against $L$ would give a square-root curve, from which nothing precise can be read by eye.

Values of $T^{2}$: $0.810$, $1.613$, $2.403$, $3.240$, $4.040$ s². The least-squares line has slope $a=4.04$ s²/m (spreadsheet or calculator).

$g=\dfrac{4\pi^{2}}{a}=\dfrac{39.48}{4.044}=9.76$ m/s².

$\dfrac{9.76-9.81}{9.81}=-0.5$ %. Using the WHOLE data set through a slope does better than any single measurement, because random errors partly cancel along the line.

It should be zero: a pendulum of zero length has no period. A clearly nonzero intercept would point to a SYSTEMATIC error on $L$, for example a length measured to the hook rather than to the centre of the mass: $T^{2}=\dfrac{4\pi^{2}}{g}(L+\ell)$ shifts the line without changing its slope. This is another advantage of the slope: it is insensitive to that shift.

Precision is set by the UNCERTAINTY, not by the display. $9.81234$ m/s² from a calculator is no more precise than $9.81$ if the uncertainty is $0.05$: the digits beyond that are noise.

Repeating reduces the effect of RANDOM errors, as $\dfrac{1}{\sqrt{N}}$. A systematic error (a poorly calibrated ruler, parallax always on the same side, a constant reaction time) is identical in every measurement and in the mean.

For a product or a quotient, we add the RELATIVE uncertainties. Adding absolute uncertainties of different quantities (grams and centimetres) would not even make dimensional sense.

The criterion is not closeness but the interval: $[9.67, 9.77]$ does not contain $9.81$. The gap, $0.09$ m/s², is almost twice the uncertainty: the measurement is NOT consistent, and you should look for a systematic error or an underestimated uncertainty.

It all depends on what you want to conclude. $5$ % is enough to tell aluminum ($2.70$ g/cm³) from iron ($7.87$ g/cm³), but not to tell $9.78$ from $9.81$ m/s². An uncertainty is judged against the question being asked.

$h=\dfrac{g}{2}t^{2}$: you plot $t^{2}$ on the horizontal axis, and the slope is $\dfrac{g}{2}$. Values of $t^{2}$: $0.1024$, $0.2043$, $0.3058$, $0.4083$ s².

Regression: $a=4.91$ m/s², intercept $-0.002$ m, negligible.

$g=2a=9.81$ m/s².

Extreme slopes: $\Delta a=\dfrac{4.97-4.84}{2}=0.065$ m/s², so $\Delta g=2\Delta a=0.13$ m/s² and $g=(9.8\pm 0.1)$ m/s². This is the honest way to carry the uncertainty of each point through to the slope.

$t^{2}$ is a power: $\dfrac{\Delta(t^{2})}{t^{2}}=2\dfrac{\Delta t}{t}=2\times\dfrac{0.003}{0.639}=0.94$ %, so $\Delta(t^{2})=0.0094\times 0.408=0.0038$ s². The horizontal bars, not the vertical ones, dominate here: it is the timing that limits the experiment.

$F=0.300\times 9.81=2.94$ N and $\dfrac{F}{x}=\dfrac{2.943}{0.148}=19.9$ N/m, in agreement with the slope of the regression.

$T=\dfrac{15.5}{20}=0.775$ s, with $\Delta T=\dfrac{0.2}{20}=0.010$ s.

$k_{d}=\dfrac{4\pi^{2}m}{T^{2}}=\dfrac{4\pi^{2}\times 0.300}{0.775^{2}}=19.7$ N/m.

$\dfrac{\Delta k}{k}=\dfrac{0.001}{0.300}+2\times\dfrac{0.010}{0.775}=0.33\,\%+2.58\,\%=2.9$ %, so $\Delta k_{d}=0.6$ N/m: $k_{d}=(19.7\pm 0.6)$ N/m.

Yes: the intervals $[19.8, 20.2]$ and $[19.1, 20.3]$ overlap. Two measurements of the same quantity are consistent when their uncertainty intervals have a part in common. The static method is the more precise here; the dynamic one is limited by the timing, whose uncertainty counts twice.

$\Delta v=0.0832\to 0.08$, to the hundredth; $v=3.46$ m/s. Result: $(3.46\pm 0.08)$ m/s.

$\Delta F=0.46\to 0.5$, to the tenth; $F=12.3$ N. Result: $(12.3\pm 0.5)$ N.

$\Delta E=27\to 30$ J, to the tens; $E=1234.5\to 1230$ J. It is better to write $(1.23\pm 0.03)\times 10^{3}$ J so that the zeros are not mistaken for significant ones.

$\Delta a=0.0013$ starts with a $1$: we keep two figures, $0.0013$, to the ten-thousandth; $a=0.0453$ m/s². Result: $(0.0453\pm 0.0013)$ m/s², or $(4.53\pm 0.13)\times 10^{-2}$ m/s².

The value is given to the tenth while the uncertainty announces the thousandth: the value was rounded too early, or the uncertainty is made up. You must go back to the data and write the value to the thousandth, for example $T=(2.003\pm 0.004)$ s; you never “complete” a value with zeros that you did not measure.

$v=\dfrac{d}{\Delta t}=\dfrac{0.0200}{0.0125}=1.60$ m/s. Both quantities must be in SI units BEFORE the division.

$\dfrac{\Delta v}{v}=\dfrac{0.05}{2.00}+\dfrac{0.1}{12.5}=2.5\,\%+0.8\,\%=3.3$ %, so $v=(1.60\pm 0.05)$ m/s.

$K=\dfrac{1}{2}mv^{2}=\dfrac{1}{2}\times 0.250\times 1.60^{2}=0.320$ J.

$\dfrac{\Delta K}{K}=\dfrac{0.001}{0.250}+2\times 3.3\,\%=0.4\,\%+6.6\,\%=7.0$ %, and $\Delta K=0.022$ J: $K=(0.32\pm 0.02)$ J. The squared speed doubles its relative uncertainty.

The width of the flag: its $2.5$ % weighs three times as much as the electronic timing, and it counts twice in the energy. A flag measured with a caliper ($\pm 0.005$ cm) would bring the uncertainty on $K$ down to $2.5$ %.

$\dfrac{338-343}{343}=-1.5$ %.

$\dfrac{5}{4}=1.25$: the gap exceeds the uncertainty by a quarter.

No: $[334, 342]$ does not contain $343$. A relative difference of $1.5$ % looks small, but it is the stated uncertainty that judges, not the impression. Two leads: a systematic error (poorly estimated temperature, forgotten end correction of the tube), or an underestimated uncertainty.

Yes: $[330, 370]$ contains $343$.

The first one, despite its verdict. Its uncertainty is five times smaller: it is able to reveal a systematic error of $1.5$ %, which the second, too wide, is unable to see. Being “consistent” with a large uncertainty is not a virtue.

$e=\dfrac{5.10}{500}=0.0102$ cm $=0.102$ mm.

Dividing by an EXACT number divides the absolute uncertainty by that same number: $\dfrac{0.05}{500}=1\times 10^{-4}$ cm $=0.001$ mm. $e=(0.102\pm 0.001)$ mm.