F=ma/Toolkit: Before You Start

Data Analysis: Graphs & Uncertainty

Pulling a relationship out of a data table, straightening a curved graph, and what error bars are telling you.

Sometimes on exams25 min read30 problems

You have a table of numbers. Somewhere in it is a law of physics, and your job is to get the law out. A pendulum was timed at five lengths. A cart rolled down a ramp past five photogates. A detector counted clicks every two minutes for ten minutes. None of those tables announces what it means. Stare at raw columns long enough and you can convince yourself of almost anything.

The last lesson was about attacking a problem on paper: name what you know, name what you want, find the bridge. Data is the same job with one extra difficulty. The numbers are not exact. Every one of them is a little wrong, and you need an answer anyway, plus an honest statement of how wrong the answer might be. This lesson is the toolkit: turn the table into a straight line, read the physics off the slope and the intercept, and attach an uncertainty that means something.

Slope and area are the two things a graph gives you

Start with the graph you already half-know. Plot velocity against time. Between two instants the velocity climbs by $\Delta v$ while the clock advances by $\Delta t$ , and acceleration is defined as exactly that ratio, so the slope of a $v$ – $t$ graph is the acceleration. Steeper line, bigger acceleration. A flat line means no acceleration at all.

Now cut the same graph into thin vertical strips. Over a strip so narrow that $v$ barely changes, the distance covered is $v\,\Delta t$ , which is the height of the strip times its width, its area. Add up every strip and you get the total displacement. So the area under a $v$ – $t$ graph is the displacement. Slope and area: those two readings cover most of what any graph in a lab is for.

Figure 1. The slope triangle gives the acceleration on the rising segment; the shaded trapezium gives the total displacement. Notice that you never needed a formula: a ruler and the area of a trapezium were enough.

Read that area as a trapezium: parallel sides of $8.0$ s and $4.0$ s, height $4.0$ m/s, so area $=\tfrac12(8.0+4.0)(4.0) = 24$ m. Chopping it into a triangle, a rectangle and another triangle gives the same 24 m. Use whichever shape is easier.

The same two readings work on any axes. On a force-extension graph the slope is the spring constant, since $F = kx$ means $k = \Delta F/\Delta x$ . On a voltage-current graph the slope is the resistance. On a position-time graph the slope is the speed. Before you read anything off a graph, say out loud what its slope means physically.

Quick check

A resistor's voltage is plotted on the vertical axis against current on the horizontal axis, and the points lie on a straight line through the origin. What does the area between the line and the current axis represent?

Pick one option first.

Make the graph straight before you read it

A straight line is the only shape your eye judges reliably. You can spot one point that has strayed off a line. You cannot spot one point that has strayed off a curve, because you do not know what the curve should have been. So the first move with any dataset is to plot it in whatever variables make the expected relation a straight line.

Take the pendulum. Theory says $T = 2\pi\sqrt{L/g}$ , where $T$ is the period in seconds, $L$ the length in metres and $g$ the acceleration of free fall. Plot $T$ against $L$ and you get a gentle curve that a dozen different laws would fit about equally well. Square both sides instead:

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

That is $y = mx$ with $y = T^2$ , $x = L$ and $m = 4\pi^2/g$ . Plot $T^2$ against $L$ and the prediction is a straight line through the origin. Its slope measures $g$ , and the fact that it comes out straight is itself evidence that the square-root law is right.

The general move: rearrange the expected relation into $y = mx + b$ where $x$ and $y$ are things you can compute from your columns. Four patterns cover almost everything you will meet.

Squared variable. $v^2 = v_0^2 + 2ax$ is already linear in $v^2$ against $x$ : slope $2a$ , intercept $v_0^2$ .
Reciprocals. The thin lens equation $1/u + 1/v = 1/f$ becomes $1/v = -\,1/u + 1/f$ . Plot $1/v$ against $1/u$ : slope $-1$ and intercept $1/f$ . The physics you want sits in the intercept here, not the slope.
Power law with unknown exponent. If $y = Ax^{n}$ , take natural logs of both sides: $\ln y = \ln A + n\ln x$ . On a log-log plot the slope is the exponent $n$ and the intercept is $\ln A$ .
Exponential. If $y = A e^{-t/\tau}$ , then $\ln y = \ln A - t/\tau$ . On a semi-log plot ( $\ln y$ against $t$ , with $t$ not logged) the slope is $-1/\tau$ .

Those last two are different plots and telling them apart is a result in itself. A power law straightens on log-log. An exponential straightens on semi-log. If you are unsure which law you have, try both and see which one comes out straight.

Figure 2. The same five pendulum timings, plotted as T to a power of your choosing against L. Drag the power from 1 to 2 and watch the worst departure from a straight line collapse from a few percent to essentially zero. Then change gravity: the points stay straight, only the slope moves.

Worth remembering

Decide what the slope and the intercept mean physically before you fit anything. Half the linearising problems you will see hide the answer in the intercept, and if you only ever look at slopes you will miss them.

Worked example 1: a curve straightened, twice

A cart is released from rest on a ramp and photogates measure its speed at four positions: $(0.25\ \text{m},\ 1.24\ \text{m/s})$ , $(0.50\ \text{m},\ 1.75\ \text{m/s})$ , $(0.75\ \text{m},\ 2.15\ \text{m/s})$ , $(1.00\ \text{m},\ 2.48\ \text{m/s})$ . Find the acceleration.

Speed against position curves, so do not plot that. From rest, $v^2 = 2ax$ , so square the speeds and plot $v^2$ against $x$ :

$1.24^2 = 1.538$ , $1.75^2 = 3.063$ , $2.15^2 = 4.622$ , $2.48^2 = 6.150$ , all in m²/s².

Those four numbers against $0.25, 0.50, 0.75, 1.00$ do lie on a line through the origin. Take the slope from the two end points, which are the furthest apart and so the least affected by the wobble in any single reading:

$m = \frac{6.150 - 1.538}{1.00 - 0.25} = \frac{4.612}{0.75} = 6.149\ \text{m/s}^2$

The slope is $2a$ , so $a = 6.149/2 = 3.07$ m/s². Two things to notice. The squaring was not a trick, it came from the kinematics you already know. And the line passing through the origin confirms the cart really did start from rest; a positive intercept would have meant it was given a push.

Worked example 2: the answer lives in the intercept

A converging lens forms real images. Measured object and image distances are $(0.200\ \text{m},\ 0.300\ \text{m})$ and $(0.300\ \text{m},\ 0.200\ \text{m})$ . Find the focal length.

Write the lens equation in straight-line form: $1/v = -1/u + 1/f$ . Compute the reciprocals.

For the first pair: $1/u = 1/0.200 = 5.000\ \text{m}^{-1}$ and $1/v = 1/0.300 = 3.333\ \text{m}^{-1}$ . For the second: $1/u = 3.333\ \text{m}^{-1}$ , $1/v = 5.000\ \text{m}^{-1}$ .

Slope first, as a check on the model:

$m = \frac{5.000 - 3.333}{3.333 - 5.000} = \frac{1.667}{-1.667} = -1.00$

Theory says the slope must be exactly $-1$ , and it is, so the data is behaving. Now the intercept. Using the first point,

$b = \frac{1}{v} - m\cdot\frac{1}{u} = 3.333 - (-1.00)(5.000) = 8.333\ \text{m}^{-1}$

and $b = 1/f$ , so $f = 1/8.333 = 0.120$ m, which is 12.0 cm. If you had only hunted for a slope you would have found $-1$ , a pure number with no physics in it, and missed the lens entirely.

Quick check

Terminal speeds of coffee filters are plotted as the log of speed against the log of mass, and the points lie on a line of slope 0.5. What does that say about how terminal speed depends on mass?

Pick one option first.

Worked example 3: a background that ruins the straight line

A detector watches a radioactive source and records total count rates: $(0\ \text{min},\ 425\ \text{s}^{-1})$ , $(3\ \text{min},\ 225\ \text{s}^{-1})$ , $(6\ \text{min},\ 125\ \text{s}^{-1})$ . With the source taken away it still reads $25\ \text{s}^{-1}$ from cosmic rays and the walls. Find the half-life.

Try the semi-log plot on the raw numbers first, to see the failure. $\ln 425 = 6.052$ , $\ln 225 = 5.416$ , $\ln 125 = 4.828$ . The drops are $0.636$ then $0.588$ , not equal, so the points are not on a line, and a plain exponential does not fit. That is the signature of an added constant.

Subtract the background, because only the source part decays: $425-25 = 400$ , $225-25 = 200$ , $125-25 = 100$ . Now $\ln 400 = 5.991$ , $\ln 200 = 5.298$ , $\ln 100 = 4.605$ . The drops are $0.693$ and $0.693$ , identical. Straight line.

The slope is

$-\lambda = \frac{4.605 - 5.991}{6.0 - 0} = \frac{-1.386}{6.0} = -0.2310\ \text{min}^{-1}$

and the half-life is the time to fall by a factor of two, so $\lambda t_{1/2} = \ln 2$ and

$t_{1/2} = \frac{0.693}{0.2310} = 3.00\ \text{min}$

You could have seen it without logs at all: the corrected rates halve every three minutes. The logarithm is there so the eye can check that they keep halving at the same rate all the way along.

Where this goes wrong

An exponential sitting on a constant offset still looks like a decay, and its semi-log plot still slopes downwards. It is only slightly bent, and if you fit a line through it anyway you get a half-life that is too long and no warning that anything happened. Whenever a decay is measured by a detector, ask what the reading is with the source gone, and subtract it first.

Uncertainty: the number after the plus-or-minus

A measurement with no uncertainty attached is not a measurement, it is a rumour. Two different things can go wrong.

Random error scatters points on both sides of the truth. Timing by hand, you are sometimes early and sometimes late. Averaging beats it down. Systematic error shifts every point the same way: a ruler with a worn end, a stopwatch that starts late, a length measured to the top of a pendulum bob instead of its centre. Averaging does nothing to it. A thousand repeats with a bad ruler gives a beautifully precise wrong answer.

For a single measurement quoted as $L = 12.4 \pm 0.3$ cm, the $0.3$ cm is the absolute uncertainty and the ratio $\delta L / L = 0.3/12.4 = 0.024$ , or $2.4\%$ , is the relative uncertainty. The same relation read backwards turns a percentage specification into a number: a balance good to $1.6\%$ reading $4.75$ kg has $\delta m = 0.016 \times 4.75 = 0.076$ kg.

Repeat a measurement $N$ times and you get a spread. The standard deviation $s$ measures how far a typical single reading falls from the mean:

$s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{N-1}}$

But you do not report a single reading, you report the mean, and the mean is better than any one reading. Its uncertainty is the standard error

$\sigma_{\bar{x}} = \frac{s}{\sqrt{N}}$

Try five timings: $2.31$ , $2.28$ , $2.35$ , $2.30$ , $2.26$ s. The mean is $11.50/5 = 2.300$ s. The deviations are $+0.01, -0.02, +0.05, 0.00, -0.04$ , and their squares sum to $0.0001+0.0004+0.0025+0+0.0016 = 0.0046$ . Divide by $N-1 = 4$ to get $0.00115$ , take the root: $s = 0.034$ s. Then $\sigma_{\bar x} = 0.034/\sqrt{5} = 0.034/2.236 = 0.015$ s. Report $2.300 \pm 0.015$ s.

That $\sqrt{N}$ is also the answer to "how many trials do I need". Rearranged, $N = (s/\sigma_{\bar x})^2$ . To pull a spread of $0.24$ s down to a standard error of $0.05$ s you need $N = (0.24/0.05)^2 = 4.8^2 = 23.04$ . A count has to be a whole number, and $23$ would leave the standard error just above target, so round up: 24 trials. Notice how brutal the square root is: cutting the uncertainty in half costs four times the work.

Figure 3. Each dot is one timing. Drag N up and the dots stay just as scattered, but the shaded band (one standard error either side of the mean) closes in on the true value. Try going from N = 8 to N = 32 and check that the band narrows by about a factor of two, not four.

Carrying uncertainty through a calculation

Uncertainties combine, and they do not simply add. Two independent errors are unlikely to both be at their worst in the same direction, so you add them in quadrature, which means squaring, adding, and taking the root.

For a sum or difference $z = x \pm y$ , the absolute uncertainties combine:

$\delta z = \sqrt{(\delta x)^2 + (\delta y)^2}$

For a product or quotient $z = xy$ or $z = x/y$ , the relative uncertainties combine:

$\frac{\delta z}{z} = \sqrt{\left(\frac{\delta x}{x}\right)^2 + \left(\frac{\delta y}{y}\right)^2}$

And a power multiplies the relative uncertainty by the exponent: if $z = x^n$ then $\delta z/z = |n|\,\delta x/x$ . That last rule tells you where to spend your effort. In $g = 4\pi^2 L/T^2$ an error in $T$ counts double, so time twenty swings and divide by twenty before you worry about the ruler.

Worked example 4. Five pendulum timings are fitted with $T^2$ against $L$ and the least-squares fit reports a slope $m = 4.037 \pm 0.025$ s²/m. What is $g$ , and how well do you know it?

From $T^2 = (4\pi^2/g)L$ , the slope is $4\pi^2/g$ , so

$g = \frac{4\pi^2}{m} = \frac{39.478}{4.037} = 9.779\ \text{m/s}^2$

Since $g$ is a constant divided by $m$ , it is a quotient, and the relative uncertainty passes straight through:

$\frac{\delta g}{g} = \frac{\delta m}{m} = \frac{0.025}{4.037} = 0.00619$

$\delta g = 0.00619 \times 9.779 = 0.061\ \text{m/s}^2$

Report $g = 9.78 \pm 0.06$ m/s². Round the uncertainty to one significant figure, then round the value to the same decimal place. Writing $9.7792 \pm 0.06$ claims four digits you have not earned.

One more number you will be asked for: the percent difference from an accepted value, which is just $|{\text{measured}-\text{accepted}}|/\text{accepted}$ as a percentage. Here $|9.779-9.80|/9.80 = 0.0021$ , or $0.2\%$ . Compare that with your own $\pm 0.06$ : the disagreement is smaller than your uncertainty, so the experiment agrees with the accepted value.

Where this goes wrong

Forcing a fit through the origin because theory says the intercept is zero throws away the best warning you have. Fit both the slope and the intercept, then ask whether the intercept is consistent with zero. If a $T^2$ against $L$ line hits the vertical axis well above zero, every length was measured short by the same amount, and the size of that missing length is $b/m$ , which you can now report as a finding instead of a mistake.

Recognising that this is the technique

You are being asked a data-analysis question, and should reach for this toolkit, whenever one of these shows up.

A table of pairs and the word "plot", "graph", "slope" or "intercept". Even if the word is absent, four or five $(x, y)$ pairs in the problem statement means fit, not solve. You are not meant to substitute one pair into a formula.

A relation that is not linear, plus an unknown constant. Then the question is which combination of the variables to put on the axes. Rearrange to $y = mx + b$ first and everything else follows.

An unknown exponent. "Assuming $v = Am^{p}$ , find $p$ " always means logs of both columns, then the slope. If the independent variable appears in an exponent instead (a decay, a discharge, a cooling curve) log only the dependent variable.

A plus-or-minus anywhere in the given data. Then the answer wants an uncertainty too, and you should track relative uncertainties through the algebra rather than the absolute ones, except across sums and differences.

Words like "background", "offset", "zero error", or a description of how the measurement was made that sounds slightly wrong. Those are systematic-error problems in disguise, and the fix is either to subtract a constant before fitting or to read the constant off the intercept.

The equipment in these questions changes every time (lenses, springs, filters, counters) but the moves do not. Get the relation into a straight line, decide in advance what the slope and the intercept mean, take the slope from well-separated points, and never quote a number without saying how well you know it.

Next you will put the graph-reading half of this to work on real motion, where the slopes and areas stop being an exercise and start being positions, velocities and accelerations: Motion in One Dimension.

Problems

Work these on paper, then check your answer here.

Acceleration from a velocity-time graph

EasyWritten for this guideslopekinematics

A glider on an air track is tracked by a motion sensor. Its velocity-time graph is a straight line rising from $2.0$ m/s at $t = 0$ to $14.0$ m/s at $t = 4.0$ s. What is the acceleration of the glider, in m/s $^2$ ?

m/s^2

Distance from the area under a v-t graph

EasyWritten for this guideareakinematics

A cart's velocity-time graph has three straight segments: it rises from $0$ to $6.0$ m/s between $t = 0$ and $t = 3.0$ s, stays at $6.0$ m/s until $t = 7.0$ s, then falls linearly back to $0$ at $t = 9.0$ s. How far does the cart travel in the $9.0$ s, in metres?

Spring constant from the slope of F versus x

EasyWritten for this guideslopespring

A student hangs masses from a spring and records the extension $x$ and the applied force $F$ : $(0.050\ \text{m},\ 2.0\ \text{N})$ , $(0.100\ \text{m},\ 4.0\ \text{N})$ , $(0.150\ \text{m},\ 6.0\ \text{N})$ , $(0.200\ \text{m},\ 8.0\ \text{N})$ . What is the spring constant, in N/m?

N/m

Speed from a position-time graph

EasyWritten for this guideslopekinematics

A trolley moves at constant speed. Its position is recorded as $(1.0\ \text{s},\ 0.45\ \text{m})$ , $(2.0\ \text{s},\ 0.90\ \text{m})$ , $(3.0\ \text{s},\ 1.35\ \text{m})$ , $(4.0\ \text{s},\ 1.80\ \text{m})$ . What is its speed, in m/s?

m/s

Percent difference from an accepted value

EasyWritten for this guideuncertaintypercent

A pendulum experiment gives $g = 9.62$ m/s $^2$ . The accepted value is $9.80$ m/s $^2$ . What is the percent difference between the measured and accepted values? Give the answer in percent.

Relative uncertainty of a length

EasyWritten for this guideuncertaintyrelative

A rod is measured as $L = 12.4 \pm 0.3$ cm. What is the relative (fractional) uncertainty in $L$ , expressed in percent?

Absolute uncertainty from a percent specification

EasyWritten for this guideuncertaintyrelative

A balance is specified to be accurate to $1.6\%$ of the reading. It shows a mass of $4.75$ kg. What is the absolute uncertainty in the mass, in kg?

Standard error of the mean

EasyWritten for this guideuncertaintystatistics

Five timings of the same pendulum give periods $4.82$ , $4.90$ , $4.75$ , $4.88$ , $4.85$ s. What is the standard error of the mean period, in seconds?

Reading an intercept by extrapolation

EasyWritten for this guideintercept

Hot water cools and its temperature falls linearly over the first few minutes. Readings are $(2.0\ \text{min},\ 68.0\ ^\circ\text{C})$ , $(4.0\ \text{min},\ 64.0\ ^\circ\text{C})$ , $(6.0\ \text{min},\ 60.0\ ^\circ\text{C})$ , $(8.0\ \text{min},\ 56.0\ ^\circ\text{C})$ . Extrapolating the straight line back, what was the temperature at $t = 0$ , in $^\circ$ C?

degC

10.

Resistance from a V-I graph

EasyWritten for this guideslopecircuits

A resistor obeys Ohm's law. Measurements give $(0.15\ \text{A},\ 1.8\ \text{V})$ , $(0.30\ \text{A},\ 3.6\ \text{V})$ , $(0.45\ \text{A},\ 5.4\ \text{V})$ , $(0.60\ \text{A},\ 7.2\ \text{V})$ . What is the resistance, in ohms?

ohm

11.

Linearizing the pendulum: T squared versus L★

NormalWritten for this guidelinearizationpendulum

A simple pendulum is timed at four lengths: $(0.400\ \text{m},\ 1.269\ \text{s})$ , $(0.600\ \text{m},\ 1.555\ \text{s})$ , $(0.800\ \text{m},\ 1.795\ \text{s})$ , $(1.000\ \text{m},\ 2.007\ \text{s})$ . Plotting $T^2$ against $L$ gives a straight line through the origin. What value of $g$ does the slope give, in m/s $^2$ ?

m/s^2

12.

Power law from a log-log plot

NormalWritten for this guidelog-logpower-law

The orbital periods of four moons are measured against their orbital radii: $(1.00,\ 1.00)$ , $(2.00,\ 2.83)$ , $(4.00,\ 8.00)$ , $(8.00,\ 22.63)$ , where the radius is in arbitrary units and the period in days. Assume $T = A r^{n}$ . What is the exponent $n$ ?

13.

Capacitance from a semi-log discharge plot★

NormalWritten for this guidesemi-logcircuitsexponential

A capacitor discharges through a $200$ k $\Omega$ resistor. The voltage across it is $(0\ \text{s},\ 12.00\ \text{V})$ , $(2\ \text{s},\ 7.28\ \text{V})$ , $(4\ \text{s},\ 4.41\ \text{V})$ , $(6\ \text{s},\ 2.68\ \text{V})$ , $(8\ \text{s},\ 1.62\ \text{V})$ . What is the capacitance, in microfarads?

14.

Half-life from a semi-log decay plot

NormalWritten for this guidesemi-logdecay

A radioactive source gives activities (in counts per second) $(0\ \text{h},\ 800)$ , $(1.5\ \text{h},\ 566)$ , $(3.0\ \text{h},\ 400)$ , $(4.5\ \text{h},\ 283)$ , $(6.0\ \text{h},\ 200)$ . What is the half-life, in hours?

15.

Ohm's law with a contact-resistance intercept

NormalWritten for this guideinterceptcircuits

A student measures the voltage across a length of resistance wire while varying the current, but the voltmeter leads also pick up a constant contact voltage. The data are $(0.10\ \text{A},\ 0.70\ \text{V})$ , $(0.20\ \text{A},\ 1.10\ \text{V})$ , $(0.30\ \text{A},\ 1.50\ \text{V})$ , $(0.40\ \text{A},\ 1.90\ \text{V})$ , $(0.50\ \text{A},\ 2.30\ \text{V})$ . What is the resistance of the wire, in ohms?

ohm

16.

Planck's constant from stopping voltage versus frequency

NormalWritten for this guideslopephotoelectric

In a photoelectric experiment the stopping voltage is measured against light frequency: $(5.0\times10^{14}\ \text{Hz},\ 0.069\ \text{V})$ , $(6.0\times10^{14}\ \text{Hz},\ 0.483\ \text{V})$ , $(7.0\times10^{14}\ \text{Hz},\ 0.897\ \text{V})$ , $(8.0\times10^{14}\ \text{Hz},\ 1.311\ \text{V})$ . Take $e = 1.602\times10^{-19}$ C. What value of Planck's constant does the slope give, in J $\cdot$ s?

J*s

17.

Least-squares slope of resistance versus temperature

NormalWritten for this guideleast-squaresslope

A metal resistor's resistance is measured at five temperatures: $(20\ ^\circ\text{C},\ 10.0\ \Omega)$ , $(40\ ^\circ\text{C},\ 10.8\ \Omega)$ , $(60\ ^\circ\text{C},\ 11.7\ \Omega)$ , $(80\ ^\circ\text{C},\ 12.4\ \Omega)$ , $(100\ ^\circ\text{C},\ 13.1\ \Omega)$ . What is the least-squares slope, in $\Omega/^\circ$ C?

ohm/degC

18.

Uncertainty in a density from a quotient

NormalWritten for this guideuncertaintypropagation

A metal block has mass $m = 250 \pm 2$ g and volume $V = 32.0 \pm 0.5$ cm $^3$ . What is the uncertainty in the density $\rho = m/V$ , in g/cm $^3$ ?

g/cm^3

19.

Propagating uncertainty through a power

NormalWritten for this guideuncertaintypropagation

A single-swing pendulum measurement gives $L = 1.000 \pm 0.005$ m and $T = 2.006 \pm 0.004$ s, and $g = 4\pi^2 L / T^2$ . What is the relative uncertainty in $g$ , in percent?

20.

Combining independent uncertainties in quadrature

NormalWritten for this guideuncertaintyquadrature

A distance is found by adding two independently measured lengths, $d = L_1 + L_2$ , with $\delta L_1 = 0.3$ cm and $\delta L_2 = 0.4$ cm. What is the uncertainty in $d$ , in cm?

21.

Work from the area under a force-displacement graph

NormalWritten for this guideareawork

A force acting along the direction of motion rises linearly from $0$ to $20$ N as an object moves from $x = 0$ to $x = 0.40$ m, then stays at $20$ N until $x = 1.00$ m. How much work is done over the full $1.00$ m, in joules?

22.

Is the point consistent with the fit?

NormalWritten for this guideerror-barsconsistency

A best-fit line through calibration data is $y = 2.00x + 1.00$ (with $x$ in metres and $y$ in newtons). A new measurement at $x = 3.00$ m gives $y = 7.60 \pm 0.30$ N. By how many standard deviations does this point differ from the fitted line?

23.

Uncertainty in g from the uncertainty in a slope★

HardWritten for this guideleast-squaresuncertaintypendulum

A pendulum is timed at five lengths: $(0.400\ \text{m},\ 1.272\ \text{s})$ , $(0.600\ \text{m},\ 1.552\ \text{s})$ , $(0.800\ \text{m},\ 1.798\ \text{s})$ , $(1.000\ \text{m},\ 2.003\ \text{s})$ , $(1.200\ \text{m},\ 2.203\ \text{s})$ . A least-squares fit of $T^2$ against $L$ gives slope $m = 4.037$ s $^2$ /m with standard uncertainty $\delta m = 0.026$ s $^2$ /m. What is the resulting uncertainty in $g$ , in m/s $^2$ ?

m/s^2

24.

Linearizing with v squared versus x

HardWritten for this guidelinearizationkinematics

A cart is released on a ramp and its speed is measured by photogates at five positions: $(0.20\ \text{m},\ 1.12\ \text{m/s})$ , $(0.40\ \text{m},\ 1.50\ \text{m/s})$ , $(0.60\ \text{m},\ 1.80\ \text{m/s})$ , $(0.80\ \text{m},\ 2.06\ \text{m/s})$ , $(1.00\ \text{m},\ 2.29\ \text{m/s})$ . What is the acceleration, in m/s $^2$ ?

m/s^2

25.

Terminal-velocity exponent from a log-log fit

HardWritten for this guidelog-logpower-law

Coffee filters of different masses are dropped and their terminal speeds recorded: $(1.0\ \text{g},\ 0.80\ \text{m/s})$ , $(2.0\ \text{g},\ 1.13\ \text{m/s})$ , $(4.0\ \text{g},\ 1.60\ \text{m/s})$ , $(8.0\ \text{g},\ 2.26\ \text{m/s})$ . Assuming $v_t = A m^{p}$ , what is the exponent $p$ ?

26.

Focal length from the intercept of a 1/v versus 1/u plot

HardWritten for this guidelinearizationinterceptoptics

A converging lens forms real images. The object distance $u$ and image distance $v$ are measured: $(0.150\ \text{m},\ 0.301\ \text{m})$ , $(0.200\ \text{m},\ 0.199\ \text{m})$ , $(0.300\ \text{m},\ 0.152\ \text{m})$ , $(0.400\ \text{m},\ 0.132\ \text{m})$ . Plotting $1/v$ against $1/u$ gives a straight line. What focal length does the intercept give, in centimetres?

27.

Half-life with a constant background★

HardWritten for this guidesemi-logdecaybackground

A detector records total count rates from a decaying source: $(0\ \text{min},\ 525\ \text{s}^{-1})$ , $(2\ \text{min},\ 379\ \text{s}^{-1})$ , $(4\ \text{min},\ 275\ \text{s}^{-1})$ , $(6\ \text{min},\ 202\ \text{s}^{-1})$ , $(8\ \text{min},\ 150\ \text{s}^{-1})$ . With the source removed, the detector reads a steady background of $25\ \text{s}^{-1}$ . What is the half-life of the source, in minutes?

min

28.

How many trials to reach a target uncertainty

HardWritten for this guideuncertaintystatistics

Repeated timings of a falling ball have a standard deviation of $0.24$ s about the mean, no matter how many trials are taken. How many trials are needed for the standard error of the mean to be $0.05$ s or smaller?

29.

Uncertainty in Planck's constant from a fitted slope

Very HardWritten for this guideleast-squaresuncertaintyphotoelectric

A photoelectric experiment gives stopping voltages $(5.0\times10^{14}\ \text{Hz},\ 0.070\ \text{V})$ , $(6.0\times10^{14}\ \text{Hz},\ 0.480\ \text{V})$ , $(7.0\times10^{14}\ \text{Hz},\ 0.900\ \text{V})$ , $(8.0\times10^{14}\ \text{Hz},\ 1.310\ \text{V})$ , $(9.0\times10^{14}\ \text{Hz},\ 1.725\ \text{V})$ . Take $e = 1.602\times10^{-19}$ C. Using the standard-error formula for a least-squares slope, $\sigma_m = \sqrt{\dfrac{1}{N-2}\dfrac{\sum r_i^2}{S_{xx}}}$ where $r_i$ are the residuals, what is the uncertainty in $h$ , in J $\cdot$ s?

J*s

30.

A systematic offset hiding in an intercept

Very HardWritten for this guideinterceptsystematicpendulum

A student measures pendulum length from the support to the top of the bob, so every length is short by the same unknown amount $d$ . The data are $(0.300\ \text{m},\ 1.135\ \text{s})$ , $(0.500\ \text{m},\ 1.447\ \text{s})$ , $(0.700\ \text{m},\ 1.703\ \text{s})$ , $(0.900\ \text{m},\ 1.925\ \text{s})$ , where the first number is the recorded length $L$ and the second the period. What is $d$ , in centimetres?

Solutions

1. Acceleration from a velocity-time graph Answer: 3 m/s^2

On a $v$ - $t$ graph the slope is the acceleration. $a = \frac{\Delta v}{\Delta t} = \frac{14.0 - 2.0}{4.0 - 0} = 3.0\ \text{m/s}^2.$ The nonzero intercept ( $2.0$ m/s) is the initial velocity and does not affect the slope.

2. Distance from the area under a v-t graph Answer: 39 m

Displacement is the area under the $v$ - $t$ curve. Triangle: $\tfrac12(3.0)(6.0) = 9.0$ m. Rectangle: $(4.0)(6.0) = 24$ m. Second triangle: $\tfrac12(2.0)(6.0) = 6.0$ m. Total $= 39$ m.

3. Spring constant from the slope of F versus x Answer: 40 N/m

Hooke's law $F = kx$ means a graph of $F$ against $x$ is a line through the origin with slope $k$ . Using the end points, $k = \frac{8.0 - 2.0}{0.200 - 0.050} = \frac{6.0}{0.150} = 40\ \text{N/m}.$ The points are exactly collinear, so any pair gives the same slope.

4. Speed from a position-time graph Answer: 0.45 m/s

On an $x$ - $t$ graph the slope is the velocity. $v = \frac{1.80 - 0.45}{4.0 - 1.0} = \frac{1.35}{3.0} = 0.45\ \text{m/s}.$

5. Percent difference from an accepted value Answer: 1.8 %

Percent difference is the discrepancy divided by the accepted value. $\frac{|9.62 - 9.80|}{9.80} \times 100\% = \frac{0.18}{9.80}\times 100\% = 1.8\%.$

6. Relative uncertainty of a length Answer: 2.4 %

Relative uncertainty is $\delta L / L$ , independent of units. $\frac{0.3}{12.4} = 0.0242 \Rightarrow 2.4\%.$ Absolute uncertainty carries units; relative uncertainty does not.

7. Absolute uncertainty from a percent specification Answer: 0.076 kg

Absolute uncertainty $=$ relative uncertainty $\times$ measured value. $\delta m = 0.016 \times 4.75\ \text{kg} = 0.076\ \text{kg}.$

8. Standard error of the mean Answer: 0.026 s

The uncertainty in a mean is $\sigma/\sqrt{N}$ , not $\sigma$ . The mean is $4.84$ s and the sample standard deviation (dividing by $N-1 = 4$ ) is $\sigma = 0.0587$ s. Then $\sigma_{\bar T} = \frac{0.0587}{\sqrt{5}} = 0.026\ \text{s}.$ Repeating measurements shrinks the uncertainty in the mean, but not the spread of individual readings.

9. Reading an intercept by extrapolation Answer: 72 degC

The intercept is the fitted line evaluated at $x = 0$ , found by extrapolation. The slope is $(56.0-68.0)/(8.0-2.0) = -2.0\ ^\circ$ C/min, so $T(0) = 68.0 - (-2.0)(2.0) = 72.0\ ^\circ\text{C}.$

10. Resistance from a V-I graph Answer: 12 ohm

$V = IR$ , so a plot of $V$ against $I$ is a line through the origin with slope $R$ . $R = \frac{7.2 - 1.8}{0.60 - 0.15} = \frac{5.4}{0.45} = 12\ \Omega.$

11. Linearizing the pendulum: T squared versus L Answer: 9.8 m/s^2

$T = 2\pi\sqrt{L/g}$ is not linear in $L$ , but squaring gives $T^2 = (4\pi^2/g)L$ , so plot $T^2$ versus $L$ . The values of $T^2$ are $1.610$ , $2.418$ , $3.222$ , $4.028$ s $^2$ ; the slope is $(4.028-1.610)/(1.000-0.400) = 4.03$ s $^2$ /m. Then $g = 4\pi^2/4.03 = 9.80$ m/s $^2$ .

12. Power law from a log-log plot Answer: 1.5

$T = Ar^n$ gives $\ln T = n \ln r + \ln A$ , so the slope of $\ln T$ versus $\ln r$ is $n$ . Using the first and last points, $n = \frac{\ln(22.63/1.00)}{\ln(8.00/1.00)} = \frac{3.119}{2.079} = 1.50.$ This is Kepler's third law, $T \propto r^{3/2}$ .

13. Capacitance from a semi-log discharge plot Answer: 20 uF

$V = V_0 e^{-t/RC}$ , so $\ln V = \ln V_0 - t/(RC)$ : plot $\ln V$ against $t$ and the slope is $-1/(RC)$ . Here $\ln(1.62/12.00)/8 = -0.2502$ s $^{-1}$ , so $RC = 4.00$ s and $C = \frac{4.00}{2.00\times 10^{5}} = 2.0\times10^{-5}\ \text{F} = 20\ \mu\text{F}.$

14. Half-life from a semi-log decay plot Answer: 3 h

$A = A_0 e^{-\lambda t}$ , so $\ln A$ versus $t$ is a straight line of slope $-\lambda$ . From $t = 0$ to $t = 6.0$ h, $\ln(200/800) = -1.386$ , so $\lambda = 0.2310$ h $^{-1}$ and $t_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{0.2310} = 3.0\ \text{h}.$ You can also read it straight off: the activity halves from $800$ to $400$ in $3.0$ h.

15. Ohm's law with a contact-resistance intercept Answer: 4 ohm

The systematic offset shifts the intercept but not the slope, so take the slope and ignore the intercept. $R = \frac{2.30 - 0.70}{0.50 - 0.10} = \frac{1.60}{0.40} = 4.0\ \Omega.$ The $0.30$ V intercept is the spurious contact voltage; dividing $V/I$ at a single point would wrongly give $7.0\ \Omega$ at $0.10$ A.

16. Planck's constant from stopping voltage versus frequency Answer: 6.63 × 10^-34 J*s

$eV_0 = hf - W$ , so $V_0 = (h/e)f - W/e$ and the slope of $V_0$ versus $f$ is $h/e$ . $\text{slope} = \frac{1.311 - 0.069}{3.0\times10^{14}} = 4.14\times10^{-15}\ \text{V}\cdot\text{s},$ so $h = (4.14\times10^{-15})(1.602\times10^{-19}) = 6.63\times10^{-34}$ J $\cdot$ s. The intercept gives the work function, about $2.00$ eV.

17. Least-squares slope of resistance versus temperature Answer: 0.039 ohm/degC

The least-squares slope is $m = S_{xy}/S_{xx}$ with $S_{xy} = \sum (x_i-\bar x)(y_i-\bar y)$ and $S_{xx} = \sum(x_i-\bar x)^2$ . Here $\bar T = 60\ ^\circ$ C, $\bar R = 11.6\ \Omega$ , $S_{xx} = 4000$ , $S_{xy} = 156$ , so $m = 0.0390\ \Omega/^\circ$ C. (The endpoint slope, $3.1/80 = 0.0388$ , is close but not identical.)

18. Uncertainty in a density from a quotient Answer: 0.14 g/cm^3

For a product or quotient, relative uncertainties add in quadrature. $\frac{\delta\rho}{\rho} = \sqrt{\left(\frac{2}{250}\right)^2 + \left(\frac{0.5}{32.0}\right)^2} = \sqrt{0.0080^2 + 0.0156^2} = 0.0176.$ With $\rho = 250/32.0 = 7.81$ g/cm $^3$ , $\delta\rho = 0.0176 \times 7.81 = 0.14$ g/cm $^3$ . The volume dominates the error budget.

19. Propagating uncertainty through a power Answer: 0.64 %

A quantity raised to the power $n$ contributes $n$ times its relative uncertainty. $\frac{\delta g}{g} = \sqrt{\left(\frac{\delta L}{L}\right)^2 + \left(2\frac{\delta T}{T}\right)^2} = \sqrt{(0.0050)^2 + (2\times 0.00199)^2} = 0.0064,$ i.e. $0.64\%$ . Even though $\delta T/T$ is smaller than $\delta L/L$ , the factor of $2$ makes the period the larger contributor.

20. Combining independent uncertainties in quadrature Answer: 0.5 cm

For a sum or difference of independent quantities, absolute uncertainties add in quadrature. $\delta d = \sqrt{0.3^2 + 0.4^2} = 0.5\ \text{cm}.$ Note this is less than the naive sum $0.7$ cm: independent errors partly cancel.

21. Work from the area under a force-displacement graph Answer: 16 J

Work is the area under the $F$ - $x$ graph. Triangle: $\tfrac12(0.40)(20) = 4.0$ J. Rectangle: $(0.60)(20) = 12.0$ J. Total $W = 16$ J.

22. Is the point consistent with the fit? Answer: 2

Compare the discrepancy to the error bar, not to the value itself. The fit predicts $y = 2.00(3.00) + 1.00 = 7.00$ N, so the discrepancy is $0.60$ N and $\frac{0.60}{0.30} = 2.0\ \text{standard deviations}.$ A $2\sigma$ disagreement is marginal: about a $5\%$ chance of occurring by luck alone, so the point is suspicious but not decisive.

23. Uncertainty in g from the uncertainty in a slope Answer: 0.062 m/s^2

$g = 4\pi^2/m$ depends on $m$ to the power $-1$ , so the relative uncertainties are equal: $\delta g/g = \delta m/m$ . Here $g = 4\pi^2/4.037 = 9.78$ m/s $^2$ and $\delta m/m = 0.026/4.037 = 0.0064$ , so $\delta g = 0.0064 \times 9.78 = 0.062\ \text{m/s}^2.$ The measured $g = 9.78 \pm 0.06$ m/s $^2$ is consistent with $9.80$ .

24. Linearizing with v squared versus x Answer: 2.49 m/s^2

$v$ against $x$ is curved, but $v^2 = v_0^2 + 2ax$ is linear in $x$ with slope $2a$ . The values of $v^2$ are $1.254$ , $2.250$ , $3.240$ , $4.244$ , $5.244$ m $^2$ /s $^2$ ; the least-squares slope is $4.99$ m/s $^2$ , so $a = \frac{4.99}{2} = 2.49\ \text{m/s}^2.$ The intercept $v_0^2 = 0.25$ gives the launch speed $v_0 \approx 0.50$ m/s.

25. Terminal-velocity exponent from a log-log fit Answer: 0.5

Take logs so the power law becomes a line: $\ln v_t = p \ln m + \ln A$ , and $p$ is the slope. $p = \frac{\ln(2.26/0.80)}{\ln(8.0/1.0)} = \frac{1.039}{2.079} = 0.50.$ Since $v_t \propto m^{1/2}$ , the drag force must scale as $v^2$ (quadratic drag), not as $v$ : linear drag would give $p = 1$ .

26. Focal length from the intercept of a 1/v versus 1/u plot Answer: 9.97 cm

$1/u + 1/v = 1/f$ rearranges to $1/v = -(1/u) + 1/f$ , so a plot of $1/v$ against $1/u$ has slope $-1$ and intercept $1/f$ . The data give $1/u = 6.667, 5.000, 3.333, 2.500$ m $^{-1}$ and $1/v = 3.322, 5.025, 6.579, 7.576$ m $^{-1}$ ; the least-squares fit gives slope $-1.006$ (confirming the model) and intercept $10.03$ m $^{-1}$ , so $f = \frac{1}{10.03} = 0.0997\ \text{m} = 9.97\ \text{cm}.$

27. Half-life with a constant background Answer: 4 min

Only the source decays exponentially, so subtract the background before taking logs; otherwise $\ln(\text{total})$ curves and gives a wrong slope. The net rates are $500, 354, 250, 177, 125$ s $^{-1}$ , which halve every $4$ min. Formally $\ln(125/500) = -1.386$ over $8$ min gives $\lambda = 0.1733$ min $^{-1}$ and $t_{1/2} = \ln 2/\lambda = 4.0$ min. Fitting the raw totals would give about $4.6$ min.

28. How many trials to reach a target uncertainty Answer: 24

$\sigma_{\bar t} = \sigma/\sqrt N$ , so reducing the uncertainty by a factor $k$ costs $k^2$ trials. Setting $0.24/\sqrt N \le 0.05$ gives $N \ge \left(\frac{0.24}{0.05}\right)^2 = 23.04,$ so $N = 24$ trials. Note this only shrinks the random error; a systematic timing offset would survive any number of repeats.

29. Uncertainty in Planck's constant from a fitted slope Answer: 1.3 × 10^-36 J*s

Fit $V_0 = (h/e)f - W/e$ , get $\sigma_m$ from the scatter of the residuals, then scale by $e$ . The fit gives $m = 4.1400\times10^{-15}$ V $\cdot$ s with $S_{xx} = 1.0\times10^{29}$ Hz $^2$ and residuals of order $10^{-3}$ V, giving $\sigma_m = 8.16\times10^{-18}$ V $\cdot$ s. Then $h = me = 6.63\times10^{-34}$ J $\cdot$ s and $\sigma_h = \sigma_m e = 1.3\times10^{-36}\ \text{J}\cdot\text{s},$ i.e. $h = (6.632 \pm 0.013)\times10^{-34}$ J $\cdot$ s, about $0.2\%$ .

30. A systematic offset hiding in an intercept Answer: 1.97 cm

The true length is $L + d$ , so $T^2 = \frac{4\pi^2}{g}(L + d) = mL + b$ with $m = 4\pi^2/g$ and $b = md$ . The offset lives entirely in the intercept, so $d = b/m$ . Fitting $T^2 = 1.288, 2.094, 2.900, 3.706$ s $^2$ against $L$ gives $m = 4.029$ s $^2$ /m and $b = 0.0794$ s $^2$ , so $d = \frac{0.0794}{4.029} = 0.0197\ \text{m} = 1.97\ \text{cm}.$ The slope still gives the correct $g = 4\pi^2/4.029 = 9.80$ m/s $^2$ : a constant offset in $x$ biases the intercept, not the slope.