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Unformatted text preview: PHYS 302: Unit 7 Viewing Notes – Athabasca University PHYS 302: Vibrations and Waves – Unit 7 Viewing Notes
The air track, coupled with springs, supported longitudinal oscillations. Oscillation perpendicular to the alignment of
the oscillators is called transverse oscillation. We are already familiar with oscillation, and we already discussed
polarization in transverse oscillation. The algebra for both longitudinal and transverse oscillation is identical, but
since transverse motion is easier to visualize in some ways, the lecture examines transverse oscillations.
For modest amplitude, the tension along a beaded string is constant, and motion along the string can be neglected.
The motion is considered purely transverse: in the y direction. Although the tensions are the same in all segments of
the string, they change direction (slightly) at each bead. As a result, the components of tension acting in the y
direction on each bead differ slightly, giving a net force due to tension. With NII and the definition of sin, we have
at the pth bead: my p T sin p 1 T sin p T ( y p y p1 )
l T ( y p1 y p )
l . Dividing by m, we get the DE set 2
2
(one for each p from 1 to N) p 20 y p 0 ( y p 1 y p 1 ) 0 , with boundary conditions y0=0 and yN+1=0 at
y the fixed ends (9:20). We also put in T
02 ml , which not only makes sense dimensionally, but reflects reality— tighter strings will have higher frequency, and strings with more mass or longer length (between particles) will have
lower frequency. We now restrict to a twoparticle system where we can imagine the modes as being low—with
both particles on the same side—and high—with particles on opposite sides of the rest position. In this case, we get
two DEs: 2
2
2
2
1 20 y1 0 ( y2 y0 ) 0 and 2 20 y2 0 ( y3 y1 ) 0 . Since y0=0 and y3=0, these are,
y
y 2
2
in fact, 1 20 y1 0 y2
y use y p 2
2 0 and 2 20 y2 0 y1 0 (12:00). As a trial function for normal modes, we
y Ap cos t , and can immediately differentiate and substitute, giving the two equations 2
2 2 A1 20 A1 0 ( A2 A0 ) 0
2
2 2 A2 20 A2 0 ( A3 A1 ) 0 . In these, A0 and A3 (both 0) were put back in to help us see the general form Ap 20 Ap 0 ( Ap 1 Ap 1 ) 0 (15:15). There are N differential (actually linear) equations to
2 2 2 solve, which is difficult if N is large. Based on experience with normal modes, we feel as though they should be like (17:50). For p=0 C sin , zero at p=0 or p=N+1. The next mode, A C sin , sin curves with an increasing number of bumps. So we can try, for mode n, Ap , n Cn sin
or p=N+1, this is 0. For n=1, Ap ,1 1 p
N 1 pn
N 1 p ,2 2 2 p
N 1 has an extra 0 at p= ½ (N+1). And so it goes on. From the general form Ap 20 Ap 0 ( Ap 1 Ap 1 ) 0
2 , we can get Ap 1 Ap 1
Ap 2 2 2 0
2
0 2 2 (22:30). Trigonometry in the French textbook (p. 141) gives mode frequency n 20 sin 2( n1) (24:20). Adding in the amplitude already found, the displacement of particle p in mode n is
N y p ,n Ap ,n cos nt . In the case N=5 (N+1=6), 1 20 sin 2(6) 20 sin 12 0.510 . It is a useful exercise
to calculate the other four mode frequencies and compare them to what is done in the lecture. Note that the numbers
are not in a simple ratio. Superpositions of these modes do not repeat! The sketch (33:00) shows the particles lying
on sinusoids, BUT they are connected with straight lines, not arcs, and particles often do not reach the nominal
amplitude of their mode if they are not in the right place. In the computer demo that follows, individual modes are
shown, followed by superpositions (“cocktails”). The regular normal modes, when all added together, lead to a
complex or even chaotic motion (43:15). The longitudinal case is mathematically identical to the transverse case.
Next, Dr. Lewin moves on to continuous media—basically solid material with a huge number of individual atoms
(a string is a good example). The demo shows a disturbance moving down a string. Upon reflection at a solid end,
the disturbance comes back inverted. The analysis for a small length of string is similar to that for a bead on a string.
Again we assume that tension is constant, due to low amplitude, and we consider motion in the y direction. The
force acting on this short length is Fy T sin T sin( ) T T ( ) T (52:35). We 1 PHYS 302: Unit 7 Viewing Notes – Athabasca University (dm) T , and if we have a mass
y
y
per unit length of μ (pronounced mu or miu), then the mass is μΔx. Then x T . Returning to the
y
. You may not have recognized the
geometry, it is clear, by the definition of the tan function, that tan x
dy
. The “d” notation is for the
concept of a “partial derivative” and may instead have only seen tan x
derivative. The new symbol “ ” (pronounced die) is also for a derivative, but represents a derivative in a function
take the mass of the short length of string to be dm. NII can then be written of more than one variable. The derivative is to be taken while other variables held are fixed. In this case, the
displacement y of the string is a function both of position x along the string, and of time t. The forces acting at any
instant of time depend on the tension in the string and the string’s angle from equilibrium. So, in doing Newton’s II
Law, we wish to take the derivative of y only with respect to x, not taking into account that the displacement of the
string also changes with time. For this reason, we use the partial derivative with respect only to x, y
. The
x discussion near 54:20 involving tanθ is more complicated than it needs to be. Like sinθ, tanθ reduces to θ (in
radians) in the small angle approximation. Thus, we can simply write d 2 y
y and then take
. The NII
dx x 2
x d
y
T
. To stress that reflects time change only, we
x
dx
2 y
d 2 y 2 y 2 y write it as
, so with
,
. Dr. Lewin’s mention of “18.0 whatever people” refers to
t 2
dx x 2 T t 2 x 2
equation x T y can be rewritten T
y the MIT Math department. Mathematicians at MIT probably would not like this derivation of the wave equation any
better than the one Dr. Lewin does on the blackboard.
Our differential equations earlier on were solved (for the most part) by proposing solutions and finding that they
worked (possibly if some condition was forced on the variables involved). Now, for the first time, we are seeing a
partial differential equation relating derivatives of one function, but with respect to more than one of the variables it
depends on. Despite this, the solution of this equation is actually very easy. Any function of the form f(x±Ct) will
solve this equation! Using the chain rule, the second derivative with respect to time of this equation is just C2 times
the derivative with respect to the argument. With respect to x, it is just the same as the derivative with respect to the
argument. (Note, the argument itself is x±Ct). This will satisfy the wave equation if
2 C T
m . Since T is a force 2 with units kg·m /s , and it is clear that the units of C are m/s, C must be a speed (58:45). As such, we can use v as a
symbol instead of C. Then we can write the wave equation as 1 2 y 2 y . The function f(xvt) will have the
v 2 t 2 x 2 same value whenever xvt has the same value. Since t (the time) always increases, this means we get the same value
of f whenever x has also increased to offset the greater value of vt. For the opposite case, f(x+vt), x+vt must have the
same value to give the same f. Since t always increases, this means that we get the same value of f whenever x has
decreased to offset the lesser value of vt. So, the – sign case corresponds to a function moving towards +x, and the +
sign case corresponds to a function moving towards –x.
Carefully follow the discussion around 1:05:00. How can one generate an exact negative pulse to compensate for
one coming in? In simply holding the end in a fixed position, the hand had to generate the forces to make this
negative pulse! This explains why a “mountain comes back as a valley” (1:07:30). If a negative force was not
applied to keep the far end fixed, the situation would be different. That is, if the end could move up and down freely,
no inversion would occur. The presence or absence of a fixed end represents a very different boundary condition. If
the end was free to move up and down, then the slope of the string (at its end) must be zero. If not, the massless end
would undergo infinite acceleration, which is impossible. This concept is best demonstrated with a set of coupled
torsional oscillators (1:15:00). Note that the open end case gives twice the amplitude at the time of reflection
compared to the amplitude of the incoming wave. 2 ...
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 Spring '10
 martinconner

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