PHYS 302: Unit 1 Viewing Notes – Athabasca University
PHYS 302: Vibrations and Waves – Unit 1 Viewing Notes
This lecture gives some simple examples of periodic motion, and then the lecturer demonstrates the rather complex
motion of the Euler’s disk. Wilczek, mentioned in the lecture, is an MIT Nobel Prize winning physicist. Note: The
German language name Euler is pronounced oiler.
In the lecture, the unit of frequency is the hertz (Hz), and 1 Hz is one vibration per second. You should be able to
hear most of the frequencies demonstrated, but it is unclear when Dr. Lewin really turned off the machine!
Dr. Lewin introduces the term simple harmonic oscillation (SHO). When he refers to 8.01, this indicates an
introductory mechanics course at MIT. The SHO equation is x  x0 cos(t   ) (at ~15:00 in the lecture). The
argument of a trigonometric function is an angle, which we measure in radians. Thus, the product ωt is an angle,
with ω (omega) as the angular frequency, measured in radians/second (different from Hz; there are 360º=2π radians
in one turn, so ω is 2π times the frequency in Hz). If T=2π/ω, replacing ωt by ω(t+T) gives the same value, since
the argument of the cos function increases by 2π, so T is the period of oscillation; f=1/T is the frequency in Hz. The
concept of projection of circular motion onto an axis to give SHO along that axis is important. Unfortunately, the
angular velocity is also called ω, which may create some confusion. φ is the angle at time t=0. The overall argument
of the cos function is called the phase, so φ is referred to as the initial phase.
mx  kx , which is a differential equation (DE). If
x  x  0 , the SHO solution x = x0 cos(ωt+φ) does solve this DE, provided that  2  mk . This
rewritten as 
(19:00) Springs suspending a mass obey Hooke’s law F=-kx or
k
m
means

k
m
, or T  2
m
k
. The demo verifies this, in a way (26:20). Note that k is not known, and that the
behaviour of the equation with a change in mass is explored. It is actually fairly simple to determine k for a spring
using Hooke’s Law. The whole equation does work, but it is not demonstrated in the video. Let’s look at this
equation intuitively. In the period form, it states that a long period arises from a large mass or from a weak spring.
Think about why this makes sense. In the actual demonstration (26:20), pay attention to the use of errors and the use
of multiple oscillations to increase the timing accuracy. The fact that the results obtained do NOT agree with initial
expectations demonstrates the scientific method, which leads to improvement of an initially inadequate theory.
French (pp. 60-61) shows how to account for the mass of the spring, and finds that
T  2
m  M3
k
and that the
inclusion of mass M of the spring fixes up the initial perceived problem.
Complex notation is very useful. If j  1 , then a set of Cartesian axes can be laid out, labelled as Re and Im
(Real and Imaginary, respectively), and the equivalents of x and y coordinates keep track of multiplication by j (for
the y part). It easily follows that a complex number on a unit circle is z=cosθ+jsinθ. The lecture gives the Euler
formula cosθ+jsinθ=ejθ without proof (42:30). This famous formula means that complex exponents correspond to
moving through an angle on the complex unit circle. This result dates from 1748! For a general point in the complex
plane, z=A ejθ. Clearly, j=ej(π/2±2πn) from the geometry and periodicity, if n is an integer.
Break (50:33)
For a pendulum, T
 2
l
g
, so a long pendulum should have a longer period. This is both intuitive and readily
demonstrated (52:30). Dr. Lewin discusses the small angle approximation, meaning that in radians, sinθ = θ.
Another approximation, that the string is of negligible mass, is in fact very accurate (55:20). While a correction is
necessary for the non-negligible mass of a spring, it is easy to make a light, strong string, so its mass rarely needs to
be considered. Since g is known, and l is easy to measure, the periods of pendulums can easily be compared to
predictions. The small angle approximation sinθ→θ is shown (57:20) with careful attention to experimental errors.
Although the video includes a clip from the MIT Physics 8.01 lectures (58:20), don’t try this at home! Since mass is
not in the formula, we do not expect any dependence on mass. This is well demonstrated by adding a mass to a
pendulum. Next, Dr. Lewin considers the physical pendulum, in which the mass is not necessarily concentrated at
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PHYS 302: Unit 1 Viewing Notes – Athabasca University
one point, using angular motion equations from the prerequisite mechanics course. The angular motion analogue of
F=ma is τ=Iα, with τ the torque, I the moment of inertia, and α the angular acceleration. While the video uses an
arrow above the symbol to denote a vector, we use boldface text when mentioning vectors in text format. Sometimes,
when the vector appears in a larger equation, we do use an arrow above a symbol, even in text. Both notations are
fairly common. Please be careful to always pay attention to the difference between vectors and scalars. It is also
fairly common for the length of a vector to be indicated by the symbol for the vector written without an arrow or
bolding. If the term cross product is unfamiliar to you, simply note that the angular acceleration and torque are
vectors (much like F and a in Newton’s law) and are perpendicular to the plane in which the force is applied. The
torque is τ=r×F (1:04:00), where the × denotes the cross product (rather than ordinary multiplication). The choice of
origin is important in angular motion problems, and in this case a clever choice is made to eliminate the torque from
the force present at the point of suspension. It is hoped that you have seen this method applied in an introductory
mechanics course. By defining the suspension point of the pendulum as the origin, r to that point is zero, so there is
no torque! For the force applied a distance b from this origin (the mg of the centre of mass) the magnitude of the
cross product is bmgsinθ. From the angular equivalent of Newton’s law, this is Iα. Taking into account the direction
of the torque and that α is the second time derivative of the angle θ (actually also a vector, a detail we will ignore),
we get bmg sin   I P (1:06:20), and in the small angle approximation sinθ→θ, we get the DE
 
bmg
 0 (1:07). This is SHM with solution θ = θ0 cos(ωt+φ).
IP
Note the discussion of angular frequency versus angular velocity, which (unfortunately) have the same symbol ω.
We get T  2
IP
bmg
, which is tested for a hoop of mass m and radius R. No one in the MIT class seems to recall
how to calculate the moment of inertia (1:12:00). If it is about the centre, it would be MR2. (You can do this
calculation from the prerequisite course). By the parallel axis theorem, IP= MR2 + MR2 = 2 MR2, and then
T  2
2 mR 2
Rmg
 2
2R
g
. You may not have seen this, but it is in some ways intuitive (the MIT students seem to
have forgotten it from their earlier course). Note that m should not appear, and it does cancel out, helping to
convince us that the answer is plausible (1:14:00). The value found for the period is the same as that of a pendulum,
in which all the mass is at the bottom. In testing, the formula works within the bounds of experimental error (just).
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