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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 second-year 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 (Ai-Ar)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 non-inverted wave also gets through, somewhat like a free-end 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 ...
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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..

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