## P-3 PENDULUM

Instructor's notes

VALUE OF THIS EXPERIMENT

The pedagogical purpose of this experiment is to illustrate how one investigates a natural phenomenon, isolates variables, and expresses the results with a mathematical equation (model). It is best scheduled early, before pendulums are discussed in lecture or text.

The graph on log-log paper should be quite straight. A linear regression of student data gives g to within 3% of the accepted value. This is not a shoddy experiment, and the results are good if the student takes reasonable care.

Results.

Period dependence on √L.
Value of constant n in T = KLn determined from graph. Should be near 0.5.
Value of constant K = 2π/√g, determined from graph.
Experimental determination of g.
Error limits for each of the above, with supporting experimental support for each value.

Graph.

Smooth curves (not "connect-the-dots")
Axis labels: quantity, symbol, units.
Descriptive title (not "T vs. L").
Genuine and appropriate graph paper.

Here many students try to avoid quantitative, mathematical and logical analysis, substituting vague qualitative prose. Graders must (unfortunately) read these very carefully to spot unsupported assertions, irrelevant assertions, meaningless statements, vague hand-waving, misconceptions about physics, failute to tell "how", "why" and "where from".

ERROR ANALYSIS

In this experiment students are not "given" the equation for the pendulum. The purpose is to try to discover the dependence of the period on L and g. So the error equation cannot be written in advance, but only when those relationships are discovered. The uncertainties will show up as repeated measurements are made, and when one constructs the graphs. The size of the uncertainties will be evident from the amount of scatter of points around the fitted curves). This will be considered below, after the graphs and data are analyzed.

(1) and (2) Look for a clear statement that the period depended on the mass, specifically how, and to what % uncertainty.

(3) Value of K = 2π/√g = 2.006 s2/m.

(4) Value of n, which should be very near 1/2.

(5) Dependence of the period on other variables? Dependence on θ should be noted.

(6) Clear statement that no mass dependence was observed.

(7) Clear statement that there was no dependence on the size of the bob.

(8) Two things must be addressed: The confirmation of the power 1/2, and the specific value of K being in agreement with K = 2π/√g.

(9) Determination of g. Should lie within 5% of accepted value.

(10) If no angle dependence was observed, the question about angle dependence obviously cannot be discussed. It's unlikely that the data whould show more than the second term of the series. However, if data was carefully taken, the period at an angle of 80° should be (1/4)sin(40°) which is 16% greater than the simple textbook formula (Eq. 6) predicts. This is easily observable. Students have observed it, consistently.

(11) How small must the amplitude be for the two equations to be in agreement to 1%? Look at the leading terms in parentheses in equation 7. The terms converge rapidly in size. The second term will be less than 1% of 1 when the angle is less than about 23°. This is probably smaller than the experimental uncertainties. The agreement is good to 5% for angles less than 53°. The agreement is good to 10% for angles up to 78°. So the angle dependence of the period should easily show up in this experiment for angles greater than 50°

7. QUESTIONS

(1) Use your graph to calculate the required length of suspension which would give the pendulum a period of 12 seconds.

About 36 meters! Used Eq. 6 solved for L = (T/2π)2g = (12/2π)2(9.8) = 35.8 m. The student, however, is asked to use the experimental graph, which would seem to require extrapolation way off the paper. This question is a test of the student's resourcefulness and ingenuity. A good way to do this is to create a new set of scales overlying the ones already there, by shifting both axes by the same number of decades.

(2) Why do you suppose we suggested that you use such a strange set of length values? (20, 40, 80, 150, 300 cm, etc.) What moral does this have for experimental analysis and strategy? If you had time to take twice as many measurements, what specific additional values would you choose?

These values are nearly equally spaced on the log paper, ensuring a uniform sample of values along the straight line. This choice is motivated by preliminary studies or analysis of the data tables, which suggest that T2 is proportional to L.

(3) Galileo Galilei (1564-1642) first noticed that the period of a pendulum was independent of amplitude, for small amplitudes. He observed a swinging chandelier in the Cathedral of Pisa [during a dull service, perhaps?] and timed the swings with his pulse. Later, in his Dialogue Concerning Two New Sciences (1638), he describes his understanding of the pendulum:

...Suppose, for example, that my friend counts 20 vibrations of the long cord during the same time in which I count 240 of my string which is one cubit in length; taking the squares of the two numbers, 20 and 240, namely 400 and 57600, then, I say, the long string contains 57600 units of such length that my pendulum will contain 400 of them; and since the length of my string is one cubit, I shall divide 57600 by 400 and thus obtain 144. Accordingly I shall call the length of the string 144 cubits.

Check Galileo's calculations against the results of your experiment. Has Galileo got his pendulum theory correct?

Solution:

Galileo has correctly done the proportion:

```
2                        2
T   L                  240
1   1                        57600   144 cubit
—— = ——       that is  ———— = ————— = —————————
2                        2    400     1 cubit
T    L                  20
2    2
```

There's no need to know what a cubit is to do this.

(4*) Continuing Galileo's discussion of the pendulum: In the same reference, Galileo writes "Nor will you miss it by as much as a hand's breadth, especially if you observe a large number of vibrations." Here Galileo is giving an estimate of the experimental error. Would you say his error estimate is reasonable? Explain, and state the conditions under which it might be reasonable, or unreasonable.

Solution:

My dictionary says that the cubit is an ancient unit of length equal to the length of the forearm, somewhere between 17 and 21 inches. 20 inches is an easy figure to work with. Work out the error equation:

```
2
L  = (T /T ) L
1     1  2   2

ΔL      ΔT      ΔT     ΔL
1       1       2      2
——  = 2 ——— + 2 ———  + ———
L       T       T      L
1       1       2      2
```

Galileo counted vibrations of both pendulums for the same length of time, 20 periods of the long pendulum in the time of 240 of the short one. This keeps the fractional error in T the same for both measurements. Now for L = 20 inches, T = 1.4 sec. approx, so 240 periods is 336 seconds or 5.6 minutes. [Galileo certainly wanted to get an accurate time measurement!] 144 cubits represents a length of 2880 inches, or 240 feet! One hand's breadth is about 3.5 inches, and 3.5in/2880in = 0.0012 or 0.12%. So Galileo must time swings long enough to get that precision. In the worst case the percent error is 4 times the timing error (assuming the error in measuring the one cubit to be negligible.) So it seems that the timing would have to be to 0.03%, or good to 1/10 second. Could Galileo have been a bit optimistic, considering he often timed things with his pulse?

But wait! Notice that Galileo didn't time the two pendulums independently, but simultaneously! He didn't use a watch, or his pulse, but merely counted. This is good experimental procedure. Count a number of swings of the slower (long) pendulum, while counting the number of swings of the shorter pendulum. So the error is, at worst, a fraction of a swing of the slower one, say one quarter of a swing at most. That's an error of 0.25/240 = 0.001. This gets doubled because of squaring, to 0.002. So the timing error contributes 0.2%. This is conservative, assuming that one can reliably judge when the slow pendulum is to the right or left of center, and whether it is going left or right, estimating the number of periods to 1/4 period.

This method also has the advantage of negligible timing error at the start of the swings, for the two pendula can be started simultaneously if the small pendulum is brought close to the large one. Galileo was no dummy when it came to experimentation!

(5*) You probably noticed that the amplitude decreased with time. The rate of decrease was greater when the initial amplitude was greater. Air drag on the bob is the likely cause of most of this. From your laboratory experience, observations, and results, make some tentative, reasonable hypotheses about the effects you would expect air drag to have on the pendulum's rate of amplitude decrease.

To start you off on an appropriate manner of analysis: consider which of the following one might expect air drag to depend on: bob mass, bob radius, bob speed, length of the swing arc.

Of those you choose, what mathematical form would the dependence take? For example, if the dependence were on bob radius, would it be KR, K/R, K(R1/2), KR2, or what? State your reasons for each hypothesis. See whether the hypothesis does predict what you observed, if the predicted size of the effect is larger than the experimental error.

This question is starred because it requires some serious digging into library resources, and into later chapters of your physics textbook.

• Bob mass. All else being equal, mass alone should have no effect on the period. It will affect the rate of decay of the amplitude. We expect that the average deceleration is proportional to 1/m (from F = ma) if the bob volume is held constant. Reason: The average force on the bob is approximately constant (for small angle swings), all else being equal.
• Bob radius. The bob radius affects the drag through Stokes' law. Find out about this through library research. Air drag depends on size of the bob, greater for larger bobs. Larger radius will cause faster amplitude decrease, but won't affect the period. However, if the radius is large compared to the suspension length, one must use the pendulum's radius of gyration rather than the length measured to the center of mass. Also, the large bob probably rotates as it swings, so its moment of inertia must be taken into account.
• The bob's volume is related to the radius. It is proportional to R3.
• Bob speed. Air drag is generally proportional to speed, or even to the square of speed, so larger swings will show more rapid decrease of amplitude.
• Length of the swing arc. This affects the period for large swing angles. The tangential component of force is not a linear function of angle, except at small angles. Length of the swing arc. We expect this to be the predominant effect, because the drag force does work F•dx during the swing. Note that sin2θ (in the first correction term) is the horizontal component of displacement squared. Equation 7 shows the effect of swing arc on period explicitly.

(6*) Very often the pendulum departs from motion in a single plane and begins to move in an oval path. Suppose this happened, in a pendulum swinging in an arc of 50 cm, and developed a sidewise component of motion of amplitude 10 cm. Would this affect your results? Woult this alter the measured period? If so, how much? If not, why?

It will not affect the period or alter the results for small angle vibrations. The reason is interesting. The period of a conical pendulum (with the bob moving in a horizontal circle) is easily calculated to be T = 2π√(L/g), the same as for the regular pendulum. This result is not dependent upon the amplitude, for small angles, just as in the plane pendulum. Now consider the superposition of two plane motions in mutually perpendicular planes. Applying superposition of the individual motions, one can treat the conical pendulum as the superposition of two identical plane-pendulum motions. The more general case of oval motion results when the two amplitudes are different, but in any case the periods are the same.

Reaction time

By the simple experiment of dropping a ruler between the fingers and catching it, one can find the reaction time to be between 0.1 and 0.2 second. Take the conservative 0.2 second. Then to get 1% error in a time, you'd need to time an interval of 0.2 × 100 = 20 seconds.

(NEW MATERIAL) Supplement to Wilson's lab manual. Wilson's is one of the few commercial manuals which takes the same approach to this experiment as we have here. But he doesn't ask for a plot on logarithmic graph paper, so these steps are added.

PLOTTING OF GRAPH—METHOD C. Steps 11, 12 and 13.

11. By the use of logarithmic graph paper we can arrive at the value of K and n by a more direct manner than by Method A or B. Read step 8 of Method B in order to understand the procedure and its mathematical justification.

12. Logarithmic graph paper saves us the work of calculating logarithms of our data. First you must determine the range of the data, and how many factors of 10 (cycles) it spans. If, for example, you used pendulum lengths, L, from 25 cm to 150 cm, you will need to plot this on a two cycle log scale. The graph paper has markings for 1, 10, 100 on the scale. Relabel them to read 10 cm, 100 cm and 1000 cm. You might wish to label the subdivisions within each cycle. The first major subdivision above 10 cm would be marked 20 cm, the next 30 cm, etc. The data on period ranges from about 1 second to over 2 seconds. This spans much less than one cycle. Its major divisions will be relabeled 1 and 10. So you will choose log paper which has two cycles on one axis (on which you will plot L) and one cycle on the other axis (on which you will plot T).

13. Plot your data on the logarithmic graph paper. Can you say that the data can be represented by a straight line? If so, draw, with a ruler, the best straight line which fits the data. Select two well separated points on the line (not necessarily data points), and designate them as P1 and P2 respectively. Then substitute their values into equation (4) and solve for n. (These are the only logs you'll have to take in this method.) Substitute this value of n into equation (1) along with any corresponding values for T and L and solve for k. Now re-write equation (1) in your data record with the proper values of k and n supplied so that T and L are the only variables. Show the computation as a part of your report.

13'. If it happens that you are using log paper in which each cycle on both axes has the same length (measured with a ruler), the slope calculation is even easier. Since lengths on the paper are proportional to the logarithms of the axis markings, the slope of a straight line is just the ratio of the lengths of the legs of a right triangle formed on the hypotenuse defined by two chosen points on that line.

## Actual data and Results

One way to deal with this data is by use of a graph, or a spreadsheet. The following results indicate what may be expected from a linear regression calculation on real data. The data was obtained following a commercial lab manual that allowed students to assume that the period is T = 2π√(L/g) and the objective was to determine the value of g, the acceleration due to gravity.

```Results from pendulum experiment May 31, 1990,
data from students at Lock Haven University.

Number Time   L       T    L^1/2                     Pendulum data
Swings       cm      sec                             Summer 1990

x      y       x^2     x*y     Variables

133  120   19    0.902   4.3589   0.81407  3.93284
118  120   25    1.017   5        1.03419  5.08475
80  100   31.7  1.25    5.6303   1.5625   7.03784
89  120   45    1.348   6.7082   1.81795  9.04477
62  100   60    1.613   7.746    2.60146  12.4935
77  120   60    1.558   7.746    2.42874  12.0716
63  120  110    1.905  10.488    3.62812  19.9773
43  100  126    2.326  11.225    5.40833  26.1046
49  120  150    2.449  12.247    5.9975   29.9938
36  100  199    2.778  14.107    7.71605  39.1854
31  100  247    3.226  15.716   10.4058   50.6975
18   60  264    3.333  16.248   11.1111   54.1603
19   60  268.5  3.158  16.386    9.9723   51.7452

Column sums: 13   26.86  133.61    64.4981  321.529
Labels:       N    Sumx   Sumy     Sum(x^2)  Sum(xy)

These values are the terms            4179.88  N*Sum(x*y)
in the linear regression              3589.08  Sumx*Sumy
formula for the slope                  590.801 Numerator
of a straight line (see below).        838.476 N*Sum(x^2)
721.621 (Sumx)^2
116.855 Denominator
5.05584 Slope, m
Results.  Not bad!                     1009.13 = g (experimental)
981 = g (Accepted value)
28.1296 Discreprancy
2.86744 % discrepancy
```

The least squares curve fit for the equation Y = mx + b are

ERRORS REVISITED

Another group of students, using similar data, used a graphical approach. They graphed √L vs. T. L is surely the most precise of the two, so treat the uncertainties as being entirely in the measurement of T. The deviations of T from the fitted line averaged about ± 0.1 second.

The slope, S = (ΔL)/T was 5.2 cm1/2/sec. S = (√g)/2π. So g = (2πS)2 = 1067 cm/s2, which, compared with 981 cm/s2, has a discrepancy of 8%.

From the scatter of points on the graph, they judged the slope to have an uncertainty of about 5%. That uncertainty is doubled when the slope is squared (giving 10%), so their value of g was reported as 1067 ± 106 cm/s2. The discrepancy is less than the estimated uncertainty. So this result is within expectations.

© 1998, 2004 by Donald E. Simanek.