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4. THE SIMPLE PENDULUM
By means of an empirical method you will deduce the relationship between the period of the simple pendulum, the length of the string, the mass of the object, and the angle the object is initially displaced.
Why Study the Simple Pendulum?
The simple pendulum is an ideal system on which to apply the empirical method, a method which embodies the very essence of science. The empirical method enables one to deduce a mathematical relationship which describes a phys- ical effect when a theory is unavailable. This is likely to happen most often for new discoveries. The empirical method begins with exploratory experimentation, follows with deduction based on the results, then more experimentation, and so on,
until the description sought for is found. Now a theory does exist for the simple pendulum, so the Theory Section that would ordinarily be placed at this point in the guidesheet is placed near the end; you are expected to pretend that the theory does not exist so you can experience the empirical approach for yourself. This was the approach used by Galileo more than 400 years ago on this very same problem. You are on your honor not to look at the theory until the proper time!
The Pendulum and Its Period
A simple pendulum consists of an object of point mass^1 m suspended on the end of a string of length l as sketched in Figures 1 and 2. When the object is displaced from its equilibrium position it swings back and forth with a well defined motion. The period o f the motion is, by definition, the time for one complete oscillation—the time for the object to
move from position A (Figure 1) to position B and back to position A again. It is natural to hypothes- ize that the period T of the pendulum depends in some way on the mass m , the length l and the angle θ the string is displaced from its equilibrium (vertical) position. It is the object of this experi- ment to uncover this relationship.
l
m T : Time for one oscillation
knot
toothpick wedge
string
O
A
B
Figure 1. The period of a simple pendulum depends in Figure 2. A way of supporting a simple some way on m, l, and θ. pendulum to constrain its motion to a plane.
The Experiment
Exercise 0. A Practice Run
Orientation Identify the apparatus: three metal cylindrical objects of different mass, a piece of string, a stand, a meter rule, a vernier caliper, a Prosonic model 1301 stopwatch (precision ±0.01s), a few toothpicks and a protractor for measuring angles.
First Run Most experiments in science benefit from practice runs to enable the experimenter to become familiarized with the equipment and to iron out bugs in procedures. This exercise is a practice run. Choose any one of the cylindrical objects and some arbitrary length of string. If you wish, set up the pendulum as shown in Figure 2. Measure the length carefully, keeping in mind that l is the dis- tance from the point of suspension to the center of mass of the object. (Estimating the position of the center of mass will involve some error; this is natural.) Set the object swinging with a small amp- litude, measure the time for some convenient num-
ber of oscillations (say 10) and from this total time T total calculate the period T , the time for one oscil- lation.^2 Ordinarily you would measure only one T total for each combination of variables. But first we must find the uncertainty in T.
Finding the Error in T How does one go about finding the error in T? One approach is to measure T a number of times, to form a sample, and then take one standard devia- tion of this sample as the error in a single measure- ment. The idea is illustrated in Figure 3. (Recall this subject was covered in Exercise 3 of the Orientation Workshop.) Of course, the result must be rounded, as usual, to one signaificant digit— here 0.02 s. Do this now for the mass and length you chose before. This standard deviation you can now use as the error in every subsequent measure- ment of T.
Period (s)
Standard Deviation>> 0.
Figure3. If the period is measured several times to form a sample, the standard deviation of the sample can be calculated (as here on an Excel spreadsheet). The standard deviation can then be inferred as the error in all subsequent measurements of the period.
Figure 5. The results of the fit whose graph is shown in Figure 4. The correlation coefficient is 0.64 indicating strongly that there is likely no dependence of T on m.
Length Dependence
Your data has probably indicated to you that T depends somehow on l. One good guess as to this dependence is the general function
T = kln^ …[1]
where, of course, T and l are variables and k and n are—as yet unknown—constants. Now eq[1] looks rather complicated. This func- tion can be simplified by the technique of linear- ization—here by taking the natural logarithm of both sides. The result is
ln( T ) = ln( k ) + n ln( l ), …[2]
which is now in the form of a straight line
Y = b + mX
with the following transformation of variables:
Y → ln (T) X → ln (l)
and where b (the intercept) = ln ( k ) and m (the slope) = n.
Therefore go ahead and enter your ( l i, T i) data into pro Fit, make the transformation and fit to a straight line. From the fit results find the values of k and n and their errors. An example of this activity is illustrated in Figures 6 and 7. You should now be able to state that the relationship
T = ( k ± ∆ k ) l (^ n ±∆^ n )^ …[3]
describes your data …or not. In Exercise 3 you can compare your results with the theory.
Angle Dependence Question:
? How would you go about studying the depen-
dence of T on θ?
- The Simple Pendulum
-2.0 -1.5 -1.0 -0.5 0.
Ln (l)
Ln (T)
Ln(T) vs Ln(l) for Simple Pendulum
Figure 6. A straight line fit to a typical set of pendulum data of T vs l.
Figure 7. An example of the results window of proFit for a typical dataset for a simple pendulum. The correlation coefficient is 0.999 indicating strongly a dependence of T on l.
d
dt
d θ
dt
d^2 θ
dt^2
and since the linear and angular accelerations are related by a = l α we can write
α = l
d^2 θ
dt^2
= – g sinθ ,
so that
d^2 θ
dt^2
g
l
sinθ = 0. …[4]
This is a differential equation of the second order. Its solution is very difficult to obtain. However, if θ is “small enough” we can make the following approximation
sinθ ≈ θ in radians.
Eq[4] then reduces to
d^2 θ
dt^2
g
l
θ = 0. …[5]
Eq[5] can be integrated to find a solution for θ or a solution can be guessed at. You should be able to show that a solution is 5
θ =θ 0 sin 2( π ft ), …[6]
where θo is the amplitude and f is the frequency in hertz (Hz). Substituting eq[6] into eq[5] satisfies the equation provided
f =
g
l
. …[7a]
Therefore, for a simple pendulum where θ is always “small”,
period = T =
f
l
g
. …[7b]
Note that the formula predicts that T is independent of m and independent of θ provided θ is “small enough”.
Videos and Physics Demonstrations on LaserDisc The Vision of Galileo , with David Suzuki, Tape # from Chapter 3 Linear Dynamics Demo 03-13 Galileo’s Pendulum Demo 03-14 Bowling Ball Pendulum
Activities Using Maple E04The Simple Pendulum In this worksheet three kinds of pendulums are examined: The Simple Pendulum, The Non-Linear Pendulum and The Damped Simple Pendulum. For a demo of aspects of the period of pendulums see the Maple worksheet Period of a Pendulum.
J Perz Stuart Quick 94
EndNotes for The Simple Pendulum (^1) By referring to the object as a point mass we are assuming the mass is concentrated in a point of negligible size.
This way we avoid having to deal with the object’s internal energy. (^2) This term has a precise meaning in physics. A period is not just some duration of time as it is in everyday usage;
in physics it is the time taken for one complete oscillation. (^3) A dependence probably exists if R is of the order 1, or conversely probably does not exist if R is about 0.7 or less
(this number is somewhat arbitrary). This procedure may fail if the sample is small. (^4) You will not be held responsible for the calculus here. The following analogies may be of help to you:
linear displacement x → angular displacement θ, linear velocity v → angular velocity ω linear acceleration a → angular acceleration α. Also,
v = d x d t
ω = dθ d t
a = d v d t
= d (^2) x d t^2
α = dω dt
= d (^2) θ d t^2
(^5) You may find these expressions useful: dθ d t
= 2 π f θo cos 2 π ft and d (^2) θ d t^2
= – 4π^2 f^2 θo sin 2 π ft.
When substituted into eq[5] these expressions will satisfy that equation provided the relationship eq[7a] holds.
