This preview shows page 1. Sign up to view the full content.
Unformatted text preview: PHYS 302: Unit 8 Viewing Notes – Athabasca University PHYS 302: Vibrations and Waves – Unit 8 Viewing Notes
The textbook treated normal modes immediately upon derivation of the wave equation, whereas the video lectures
focused on progressive (i.e., travelling) waves. Now the two converge as we consider normal modes in continuous
media. One approach could be to take the results from N beads on a string and let N go to infinity. This is easy to
envision qualitatively (1:20), but quantitatively, we will use an approach based on travelling waves and boundary
conditions. A few definitions are needed. The wavelength λ (pronounced lambda) is the distance a disturbance
travels in one oscillation time, as it propagates with speed v T as described by y A sin[ 2 ( x vt )] . Note that this is a solution of the wave equation, since any function f(x±Ct) is a solution. This particular function
describes a periodic wave as opposed to just a pulse. If, for a fixed time t, one advances by a distance x in the
direction of propagation of a wave, the disturbance will change as a sinusoid. The periodic nature of the wave is
clear, since if one advances by a distance λ, the phase will change by 2π, so the same disturbance will repeat. If one
sits at a fixed point and lets the wave move past as time t naturally increases, the pattern repeats after one period P
(Note: the notation T was previously used for some other periodic motions, but here we want to retain T as the
tension). Thus, if t increases by an amount P, the phase also increases by 2π. Since the overall argument for phase is
2
2
2
2 ( x vt ) , for the change in phase 2 ( x v(t P )) ( x vt ) vP . This immediately gives vP . We can describe this by saying that the wavelength is the distance traversed by the wave in one period
(4:20). Since 2π radians of phase go by in one period lasting time P, the rate at which phase changes (the angular frequency) is 2P . From this, we extend vP to v 2 and this gives 2 v (4:20) (and v ). We f
can introduce the wave number k 2 ; then kv . Note the discrepant definition of k mentioned relative to the French textbook Also, for your information, “Bekefi and Barrett” is a textbook used in the MIT 8.03 course but not
used in this course. Use of k brings a more symmetric form of the wave equation: y ( x, t ) A sin( kx t ) .
We now consider two identical waves propagating in opposite directions. The two such waves are
y1 ( x, t ) A sin(kx t ) and y2 ( x, t ) A sin(kx t ) , which superpose by simple addition. You may recall
that the sum of two sins is the sin of half the sum of the arguments times the cos of half the difference. This is
analogous to the identity proved for cos on page 26 of French and mentioned in the Reading Notes for Unit 2.
Taking the cos case and adding π/2 to all the arguments will give you the sin. Note that Dr. Lewin arbitrarily chose
to work with sin here; cos could have been equally well used, so the similar identity must work. Anyway, the
important result arises that y y1 y2 2 A sin( kx) cos(t ) . This result is important, because the spatial and
temporal variations are now completely separated (7:45) into a dependence on sin( kx ) and on cos( t ) ,
respectively. The overall pattern does not move! It is called a standing wave. We now consider fixing the ends of the string at x=0 and x=L. Since the spatial part does not move, we satisfy that
the ends are fixed (y=0) at both of these points only if the argument of the sin makes the spatially dependent
sin( kx ) term zero. At x=0, this is automatic for sin functions (probably why they were chosen). At x=L, we must
have kx=nπ and x=L, so we really need kL=nπ, or
Further, since v in Hz is then fn 2 2 and
nv
2L kn kv , we have n v n
L 2
kn v . We have labelled the wavenumber for this mode as n. 2
kn 2L
n , or directly, n k n v n v
L . The frequency . These modes end up being integral numbers of half sinusoids, and they oscillate at the frequency unique to that mode. One half wavelength between fixed ends is called the fundamental or first harmonic.
A full wavelength is the second harmonic (etc.). The frequencies go up linearly, proportional to the mode number,
that is, ωn=nω1. In music, the harmonics are called overtones. Dr. Lewin alludes to quantum mechanics, where
similar standing waves play an important role, and the modes are called eigenstates (16:00). The supplemental
readings and videos on Quantum Mechanics accompanying this unit explore these ideas further. Understanding
quantum mechanical waves is part of our course objectives. The general solution for the nth mode is
yn An sin( n x ) cos(nt ) (16:30). By driving at appropriate frequencies, the resonances corresponding to the
L
modes can be excited (we have not yet discussed excitation, but when the modes are found, little driving is required, 1 PHYS 302: Unit 8 Viewing Notes – Athabasca University so the hand is hardly moving). The demo shows a number of modes on ropes, excited both by hand and with a
motor.
The same considerations apply to sound, a longitudinal oscillation in which the pressure increases or decreases
(slightly) with regard to the ambient pressure. To emphasize that we are addressing a different physical variable
while using the same equation structure, the variable is changed: n An sin( n x ) cos(n t ) . For sound, the
L
velocity is very nearly fixed at 340 m/s near ground level, and this value is used in n n v . If instead of
L displacement, we talk about pressure, then the boundary conditions dictate that the pressure change is greatest at the
ends and pn ( x, t ) Pn cos( n x ) cos(n t ) . In the demonstration, 803 Hz is based on the MIT class number 8.03,
L
but 2409 Hz was actually used. Then, v
f 340
2409 0.14m . In the demo, the nodes (no sound) are clear. We can measure the speed of sound accurately, since f is precisely known, and the λ/2 distance between nodes can be fairly
accurately measured too (37:00).
Waves move energy despite the fact that no mass moves in the direction of the wave (even in a longitudinal wave
the particles do not move far—they oscillate around their equilibrium position). Initially, we consider the kinetic
energy (KE). In a short length of string, for a transverse wave, there is a y velocity and this gives KE of
2
dEkin 1 (dm)v y 1 dx( y ) 2 (40:10). Taking the derivative,
2
2
t
y
t t A sin[ k ( x vt )] Akv cos[ k ( x vt )] . So, the element of KE is dEkin 1 A2 k 2 v 2 cos 2 [ k ( x vt )]dx (40:50). At this point we must consider how we obtain the KE over one
2
whole wavelength where a similar element of KE applies at each point. We do this with an integral, which you may
not have covered in detail in your prerequisite calculus class (41:30). We will now expand on this topic, since Dr.
Lewin assumes that the MIT students have this knowledge. In fact, the element’s KE never appeared explicitly in
the video lecture, since Dr. Lewin went directly to the integral Ekin applying to the whole length of rope in one
wavelength. Let us write the element’s KE and consider that this is simply the element i of many (N) elements. Then
we have dEkin ,i 1
2 A2 k 2 v 2 cos 2 [k ( xi vt )]dx . Bear in mind that v is the propagation speed, not the transverse speed. The transverse speed, which contributes to the KE, is simply reflected in the cos term. The only
dependence is on position (at a given time) which is why x now has a subscript. Clearly, if there are N small
elements in a wavelength, the total KE is the sum Ekin N dE
i 1 kin ,i . Since each i advances us along the rope, and we go one wavelength, each dE contributes to raising the total amount of KE Ekin in that length of rope (they are all
positive due to the square). The KE is a curve for which the slope (always positive) makes the curve rise. What this
means is that with respect to the total KE Ekin, dE is a slope. So, if KE as a function of amount of x advanced along
the curve, dE would be the slope or derivative of that curve. In that respect, the Ekin function is the antiderivative of
dE. Ekin for a whole wavelength is, however, not a function, it is a number. Doing an evaluation over the interval
gives us a total quantity. It turns out that the evaluation involves simply taking the difference Ekin(λ)Ekin(0) (we will
not show this), which then gives a number Ekin. Unlike a derivative, an antiderivative cannot always be found easily.
Much of a secondyear calculus course (not a prerequisite for this course) focuses on how to find antiderivatives.
Here, we simply state that if you can find the function whose derivative is the function for an element, you have
found the antiderivative. The operation we want to perform is called integration, which results in an integral. When
we integrate a function over an interval, the result is called a definite integral (as is the case here). An indefinite
integral is a function, which we referred to above as the antiderivative. The notation for a definite integral is similar to that for a sum: Ekin 1 A2 k 2 v 2 cos 2 [k ( xi vt )] dx (41:50). Like in a sum, any constant terms can be
2
0 moved outside the integral sign to give exactly what is written on the board in the lecture: Ekin A k v
1
2 2 22 cos [k ( x vt )] dx . Recalling the strong connection of trig functions to exponentials, we
2 i 0 2 PHYS 302: Unit 8 Viewing Notes – Athabasca University will note that the exponential function has both the easiest derivative and the easiest antiderivative. If
, then the antiderivative e ax d
dx e ax ae ax dx 1 e ax . Using the complex exponential expansion of cos, squaring it, using this
a antiderivative, and looking carefully (with a sketch) at how the oscillating terms cancel out, it is actually quite easy
to show that the integral needed is just λ/2, as Dr. Lewin states (42:00). Factoring in that v 2 find that in one wavelength, Ekin 1
4 A2 k 2 T 1 A2 2 4 2T A2 2T T and k 2 , we (43:20). Note that the energy is proportional to the amplitude squared.
Potential energy also exists due to the need to stretch the string to get the shape. French does this exercise in the
textbook on page 238. It turns out that the potential energy per wavelength is the same as the kinetic energy. Thus, Etrw 2 A2 2T , where trw stands for travelling wave. A standing wave is equivalent to two travelling waves of half the amplitude, coming in opposite directions (50:20). In travelling waves, energy flows in the direction of
propagation, and power needs to be put in. The power is Etrw Power 2 2 2A T 1
P 2 A2 2T . However, since the period is v/λ, Power injected in each oscillation, that is, 22 2A T v (54:20). This assumes that the power is added in and goes away (maybe on an infinitely long rope). Recall that in the standing wave demonstration, no
energy needed to be added once the oscillation was going. Unlike in a travelling wave, there is no net energy flow. Break to 55:10
In the case of two ropes joined, the tension must be the same in each rope. If one rope has mass density μ1 and the
other μ2, then the propagation velocities are v1 T 1 and v2 T 2 . If the junction between ropes is at x=0, with the incoming (incident) wave from the left, there will be a reflected wave which must move towards the left and a
transmitted wave that must move towards the right (i.e., like the incident wave). The boundary conditions mean that
both y and y’ must be continuous at the junction. While the incident and reflected waves move in opposite
directions, they have the same wavelength since they are in the same medium: yi Ai sin(t k1 x) and yr Ar sin(t k1 x) . The transmitted wave, however, is in the second medium, so its k is different:
yt At sin(t k2 x) . There is no particular reason to subscript the frequency, which is the same in both media
(strings): k 1 v1 k2 v2 (1:00:00). The boundary conditions now allow for determining the ratios of amplitudes. At the junction, the sin term is the same for all waves, and the continuous string allows equating the y’s to give
Ai+Ar=At. Taking the x derivative for each gives another pair of functions, which also must be continuous, with the
resulting cos terms being the same at the junction, so that Aik1+Ark1=Atk2 (1:02:10). From k 1 v1 k2 v2 we
can replace k with v to get (AiAr)v2=Atv1. Then the ratios R Ar
Ai v2 v1
v1 v2 and Tr At
Ai 2 v2
v1 v2 (1:03:45; note, there is a later—1:07:00—correction to what is written on the board to give the form shown in these notes). To consider
limited cases, first put μ2=∞, which must be a fixed end. Then R=1 and Tr=0, as expected. If v1<v2, that is, μ1> μ2,
R>0, and Tr>0, the wave that comes back is not inverted, and some noninverted wave also gets through, somewhat
like a freeend case. With two identical strings, we expect nothing to happen: from the formulas, R=0 and Tr=1.
Leading to the demo, try v1=2v2. Then R=1/3 and Tr=+2/3, and the wavelength will change in the transmitted wave.
This is demonstrated. A free (open) end can be put in if μ2 goes to 0 (i.e., v2 goes to infinity). Then R=1, Tr=0. The
consequences of this are left for you to ponder… 3 ...
View Full
Document
 Spring '10
 martinconner

Click to edit the document details