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Unformatted text preview: PHYS 302: Unit 5 Viewing Notes – Athabasca University PHYS 302: Vibrations and Waves – Unit 5 Viewing Notes
So far, we have looked at single systems with some outside source of forcing. Most realistic physical systems have
more than one part, so interaction between the parts can cause vibrations. This is known as coupling. For simplicity,
we will leave damping out for now. The first example, with pendulums connected by a spring, is a good
approximation of coupling without damping, since the damping is small. In the demo (2:30), oscillation seems to
“move” between the two pendulums. In normal modes each object is at the same frequency, and in the case of no
damping, the coupled objects can move in only two ways: either in phase or in antiphase (180º out of phase). For
two objects, we expect two normal modes; their frequencies are denoted ω- and ω+. In general, motion will occur in
a combination of normal modes, so for the two objects (denoted by subscripts 1 and 2) the positions are
x1 x0 cos(t ) x0 cos( t ) and x2 x0 cos(t ) x0 cos( t ) . The
constants x0- and x0+ show the amplitude of each normal mode and can be zero. Dr. Lewin excites first the – mode,
with identical motion of the pendulums, followed by the + mode, with opposed motion (10:20). Since the spring
does not do anything in the – mode, the frequency is simply that of a single pendulum 0 where
2 02 g
l . For the opposed mode, we need a force analysis. The spring is stretched by 2x, T=mg, and Fs=-2kx. k
02 g , and that of the spring and one mass would be s2 m .
2 NII (Newton’s second law) gives mx 2kx mg x or 2 s x 0 x 0 , which is so similar to previous
l The natural frequency of the pendulum gives DEs that we can solve immediately that 2s2 02 (15:20). As sample initial conditions (t=0), let x1=C and v1=0, and x2=0 and v2=0. This means that both masses are released from rest, but mass 1 is offset and mass 2 is
hanging vertically downward. Solving, including use of derivatives, gives both phases as 0. Then C=x0-+x0+. Also,
0= x0-+x0+. So x0-= ½ C and x0+= ½ C. Then x1 1 C cos t 1 C cos t and x2 1 C cos t 1 C cos t
(21:10). This is easily verified to meet the initial conditions. Using a trig identity, x1 C cos
and x2 C sin 2 2 t cos t
2 t sin t . This resembles beats with a fast term multiplying a slow term. Since the slow
2 terms are out of phase, oscillations will “move” from one mass to the other and back again. Dr. Lewin demonstrates
this and then repeats it with less coupling by moving the spring up on the supports of the pendulums.
A “recipe” is given for solving for normal modes:
1. Give each object a displacement from equilibrium. 2. Apply NII to each object (which for n objects gives n+1 unknowns including a single frequency). 3. Put in solutions x1=C1cosωt, x2=C2cosωt, etc., where phase is subsumed in the sign of the Cs. 4. Now solve for ratios of Cs and for ω. This will allow a general solution with some algebraic work (37:00). Break
The recipe is then applied to the situation of two pendulums connected by a spring (previously analyzed). Then, the
magnitude of the spring force is |Fs|=k(x2-x1). The NII DEs are m1 1 mg
x m2 2 mg
l k ( x2 x1 ) and k ( x2 x1 ) . Pay very careful attention to signs here. Now, we rearrange and use the notations introduced before that as 02 g and s2 l k
. This gives 1 (0 s ) x1 s x2 0 and
x 2 (02 s2 ) x2 s2 x1 0 . Again, you should check everything carefully. Note that the last terms are the
coupling terms. We can now propose the solutions x1=C1cosωt, x2=C2cosωt as in step 3. With only second
derivatives, the cos terms end up as common; hence, they can be cancelled, giving 1 PHYS 302: Unit 5 Viewing Notes – Athabasca University 2
2 2C1 (0 s2 )C1 s2C2 0 and 2C2 (0 s2 )C2 s2C1 0 . You should carefully check the signs again at this point (49:00). We now solve for the ratio C1/C2. For the first equation, the intermediate step is 2
2 2C1 (0 s2 )C1 s2C2 or ( 2 0 s2 )C1 s2C2 . From this equation and the similar second equation, C1
2 0 s
2 2 0 s s2 for ω, ω- and ω+. Multiplying through, (51:00). The part s2
2 0 s
2 2 0 s s2 must have two possible solutions s4 ( 2 02 s2 ) 2 . Taking the square root, 2 2 0 s2 s2 , where the ± is important. We can separate out ω for solution, and use an inverted ± to keep track of where this came from, finding
two solutions 2 02 s2 s2 . Here, all the algebra simplifies, and we get the 0 and 02 2s2 . Substituting back, we can now find C1
C2 1 and C1
C2 1 (56:20).
Now, Dr. Lewin discusses a seemingly simple (but asymmetric) case of the double pendulum. Here,
C2 1 2 2.4 for the lowest mode, and C1
C2 2.4 for the highest mode (where 1 is the top bob). A normal mode is a resonance frequency or natural frequency of the whole system, and demos follow (1:00:00).
First, Lewin shows the case with C1
C2 1 2 2.4 (lower mode), and then C1
C2 2.4 (higher mode). It is hard to determine the frequencies, but the higher mode clearly has higher frequency.
Now, Dr. Lewin introduces a more complex system, four springs and three cars, with the same spring constant k and
mass m. There must be an ω-, an ω+, and an ω++. The first of these, of frequency ω-, has all the objects in phase, and
the middle one must be at 2 larger displacement than the others (not shown). In the ω+ mode, the rightmost car
has the opposite displacement from the leftmost, and the middle one is at rest. In the ω++ mode, the outer two are in
phase and the central one in antiphase with displacement 2 of the others. If these (known) initial conditions are
set up on the air track, these modes will occur. The demonstration, impressively, matches what is expected. The
math is hard, however.
Next, Dr. Lewin examines some general characteristics of the triple pendulum. The lowest mode must have all bobs
in phase, while the highest has all bobs out of phase. The middle mode is harder to figure out since there are two
possibilities, so Dr. Lewin uses a demo. 2 ...
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This note was uploaded on 02/18/2011 for the course PHYS 320 taught by Professor Martinconner during the Spring '10 term at Open Uni..
- Spring '10