Longitudinal waves - Harvard University

1 Chapter 5 Longitudinal wavesDavid Morin, Chapter 4 we discussed transverse waves , in particular transverse waves on a string. We llnow move on to Longitudinal waves . Each point in the medium (whatever it consists of)still oscillates back and forth around its equilibrium position, but now in the longitudinalinstead of the transverse direction. Longitudinal waves are a bit harder to visualize thantransverse waves , partly because everything is taking place along only one dimension, andpartly because of the way the forces arise, as we ll see. Most of this chapter will be spenton sound waves , which are the prime example of Longitudinal outline of this chapter is as follows. As a warm up, in Section we take anotherlook at the Longitudinal spring/mass system we originally studied in Section , where weconsidered at the continuum limit (theN limit). In Section we study actualsound waves . We derive the wave equation (which takes the same form as all the other waveequations we ve seen so far), and then look at the properties of the waves .

2 In Section apply our knowledge of sound waves to musical Springs and masses revisitedRecall that the wave equation for the continuous spring/mass system was given in Eq. ( )as 2 (x, t) t2=E 2 (x, t) x2,(1)where is the Longitudinal position relative to equilibrium, is the mass density, andEisthe elastic modulus. This wave equation is very similar to the one for transverse waves ona string, which was given in Eq. ( ) as 2 (x, t) t2=T 2 (x, t) x2,(2)where is the transverse position relative to equilibrium, is the mass density, andTisthe equations take exactly the same form, so all of the same results hold. However, thefact that is a Longitudinal position in the former case, whereas it is a transverse positionin the latter, makes the former case a little harder to visualize. For example, if we plot fora sinusoidal traveling wave (either transverse or Longitudinal ), we have the picture shown inFig. 1. The interpretation of this picture depends on what kind of wave we re talking ABCDEF igure 112 CHAPTER 5.

3 Longitudinal WAVESFor atransversewave, is the transverse displacement, so Fig. 1 is what the stringactuallylookslike from the side. The wave is therefore very easy to visualize you justneed to look at the figure. It s also fairly easy to see what the various points in Fig. 1 aredoing as the wave travels to the right. (Imagine that these dots are painted on the string.)PointsBandDare instantaneously at rest, pointsAandEare moving downward, andpointCis moving upward. To verify these facts, just draw the wave at a slightly later result is shown in Fig. 2, with the new positions of the dots being represented by grayx wave at alater timeABCDEF igure 2dots. Remember that the points keep their same Longitudinal position and simply move upor down (or not at all). They don t travel longitudinally along with the , for alongitudinalwave, is the Longitudinal displacement, so although Fig. 1is a perfectly valid plot of , it doesnotindicate what the wave actually looks like.

4 Thereis no transverse motion, so the system simply lies along a straight line. What changes isthedensityalong the line. You could therefore draw the wave by shading it as in Fig. 3,xA B C D EFigure 3but this is a bit harder to draw than Fig. 1. For a Longitudinal wave, the statements in thepreceding paragraph about the motion of the various points in Fig. 1 are still true, providedthat downward is replaced with leftward, and upward is replaced with what do things actuallylooklike along the 1-D line? In particular, how does Fig. 3follow from Fig. 1?At pointsBandDin Fig. 3, the density of the masses equals the equilibrium density,because nearby points all have essentially the same displacement (see Fig. 1). But at pointsAandE, the density is a minimum, because points to the left of them have a negativedisplacement, while points to the right have a positive displacement (again see Fig. 1). Theopposite is true for pointC, so the density is maximum there.

5 Various properties of thewave are indicated in Fig. 4. You should stare at this figure for a while and verify all of thestated properties. We ll talk more about the relation among the various quantities when wediscuss Fig. 8 later on when we get to sound + , v = 0, max -a, avg max - , v = 0, max +a, avg = 0, max +v, a = 0, max + = 0, max -v, a = 0, max - x positive positionpositive velocitypositive acceleration rightward travelingFigure 4In the Fig. 4, the relation between ,v, andais the same as always, namely,ais 90 ahead ofv, andvis 90 ahead of . But you should think about how these relate to thedensity . For example, from the preceding paragraph, the (excess) is proportional to thenegative of the slope (see Problem [to be added] for a rigorous derivation of this fact). Butwe already know thatvis proportional to the negative of the slope; see Eq. ( ). Therefore,the (excess) is proportional tov. For aleftwardtraveling wave, the same statement SOUND WAVES3 is still true, but nowvis proportional to the slope (with no negative sign).

6 So the (excess) is proportional to can double check that this result makes sense with the following reasoning. Sincethe (excess) is proportional tov, we can take the derivative of this statement to say that / x v/ x(the word excess is now not needed). But since a traveling wave takesthe form of (x, t) =f(x ct), the velocityv= / talso takes the functional form ofg(x ct). Therefore, we have v/ x= (1/c) v/ t. The righthand side of this is just theaccelerationa, so the / x v/ xstatement becomes x a= a x.(3)Does this make sense? It says, for example, that if the density is an increasing function ofxat a given point, then the acceleration is negative there. This is indeed correct, becausea larger density means that the springs are more compressed (or less stretched), which inturn means that they exert a larger repulsive force (or a smaller attractive force). So if thedensity is an increasing function ofx(that is, if / x >0), then the springs to the right of agiven region are pushing leftward more than the springs to the left of the region are pushingrightward.

7 There is therefore a net negative force, which means that the accelerationaisnegative, in agreement with Eq. (3). Sound NotationSound is a Longitudinal wave, in both position and pressure/density, as we ll see. Sound canexist in solids, liquids, and gasses, but in this chapter we ll generally work with sound wavesin air. In air, molecules push and (effectively, relative to equilibrium) pull on each other, sowe have a sort of spring/mass system like the one we discussed main goal in this section is to derive the wave equation for sound waves in air. We llfind that we obtain exactly the same type of wave equation that we had in Eq. (1) for thespring/mass system. The elastic modulusEappears there, so part of our task below willbe to find the analogous quantity for sound waves . We ll consider only one-dimensionalwaves here. That is, the waves depend only onx. waves like this that are uniform in thetransverseyandzdirections are called plane waves . To emphasize the 1-D nature of the wave, let s consider a tube of air inside a cylindricalcontainer, with cross-sectional areaA.

8 Fig. 5 shows a given section of air at equilibrium,and then also at a later time. Let the ends of this section be located atxandx+ xatequilibrium, and then atx+ (x) andx+ x+ (x+ x) at a later time, as shown. Sothe function measures the displacement from we define by (x+ x) (x) + , then is how much more the rightboundary of the region moves compared with the left boundary. The molecules of air are inthermal motion, of course, so it s not as if the molecules that form the boundary at positionxin the first picture correspond to the molecules that form the boundary at positionx+ (x)in the second picture. But we ll ignore this fact and just pretend that it s the same molecules,for ease of discussion. It doesn t actually matter. Note that ( / x) xfor small x, by definition of the derivative. In actual sound waves in air, is much less than other words, / xis very 5. Longitudinal waves (equilibrium)(later)xx+ xx+ x+ (x+ x)x+ x+ (x)+ x+ x+ (x)Figure 5A note on terminology: We re takingxto be the position of a given molecule at equilib-rium.

9 So even after the molecule has moved to the positionx+ (x), it is still associatedwith the same value ofx. Soxis analogous to the indexnwe used in Section and thebeginning of Section The movement of the particle didn t affect its labelnthere, andit doesn t affect its obtaining the wave equation, we ll need to get a handle on the pressure at the two endsof the given section of air, and then we ll figure out how these pressures cause the sectionto move. Let the pressure in the tube at equilibrium bep0. At sea level, the atmosphericpressure happens to be about lbs. per square inch. The first picture in Fig. 6 showsthe pressures at the two ends at equilibrium; it is simplyp0at both ends (and everywhereelse).(equilibrium)(later)xp0p 0p(x)=p0+ p(x)p(x+ x) = p0+ p(x+ x)p0+ p(x)+ px+ xFigure 6 What about at a later time? Let p(x) be theexcesspressure (abovep0) as a functionofx. (Remember thatxlabels the equilibrium position of the molecules, not the presentposition.)

10 The total pressure at the left boundary of the section is thenp0+ p(x). However,this total pressure won t be too important; the change, p(x), is what we ll be concernedwith. At the right boundary, the total pressure is, by definition,p0+ p(x+ x). If wedefine pby p(x+ x) p(x) + p, then pis how much the pressure at theright boundary exceeds the pressure at the left boundary. Note that p ( p/ x) xfor small x. In practice, pis much smaller thanp0. And pis infinitesimally small,assuming that we have picked xto be infinitesimally small. The pressures at a later timeare summarized in the second picture in Fig. SOUND The wave equationHaving introduced the necessary notation, we can now derive the wave equation for soundwaves. The derivation consists of four main steps, so let s go through them strategy will be to find the net force on a given volume of air, and then write down theF=maequation for that the volume changes:First, we need to determine how the volume of a gaschanges when the pressure is changed.