Transverse waves on a string - Harvard University

1 Chapter 4 Transverse waves on a stringDavid Morin, the previous three chapters, we built up the foundation for our study of waves . In theremainder of this book, we ll investigate various types of waves , such as waves on a string ,sound waves , electromagnetic waves , water waves , quantum mechanical waves , and so Chapters 4 through 6, we ll discuss the properties of the two basic categories of waves ,namelydispersivewaves, andnon-dispersivewaves. The rest of the book is then largely aseries of applications of these results. Chapters 4 through 6 therefore form the heart of non-dispersive system has the property that all waves travel with the same speed,independent of the wavelength and frequency.

2 These waves are the subject of this andthe following chapter (broken up into longitudinal and Transverse waves , respectively). Adispersive system has the property that the speed of a wavedoesdepend on the wavelengthand frequency. These waves are the subject of Chapter 6. They re a bit harder to wrap yourbrain around, the main reason being the appearance of the so-calledgroup velocity. As we llsee in Chapter 6, the difference between non-dispersive and dispersive waves boils down tothe fact that for non-dispersive waves , the frequency and wavelengthkare related by asimple proportionality constant, whereas this is not the case for dispersive outline of this chapter is as follows.

3 In section we derive the wave equation fortransverse waves on a string . This equation will take exactly the same form as the waveequation we derived for the spring/mass system in Section , with the only differencebeing the change of a few letters. In Section we discuss the reflection and transmissionof a wave from a boundary. We will see that various things can happen, depending onexactly what the boundary looks like. In Section we introduce the important conceptofimpedanceand show how our previous results can be written in terms of it. In we talk about the energy and power carried by a wave. In Section we calculate theform of standing waves on a string that has boundary conditions that fall into the extremes(a fixed end or a free end).

4 In Section we introduce damping, and we see how theamplitude of a wave decreases with distance in a scenario where one end of the string iswiggled with a constant The wave equationThe most common example of a non-dispersive system is a string with Transverse waves onit. We ll see below that we obtain essentially the same wave equation fortransversewaves12 CHAPTER 4. Transverse waves ON A STRINGon a string as we obtained for theN limit oflongitudinalwaves in the mass/springsystem in Section Either of these waves could therefore be used for our discussion ofthe properties of non-dispersive systems.

5 However, the reason why we ve chosen to studytransverse waves on a string in this chapter is that Transverse waves are generally easier tovisualize than longitudinal a string with tensionTand mass density (per unit length ). Assume thatit is infinitesimally thin and completely flexible. And assume for now that it extends in-finitely in both directions. We ll eventually relax this restriction. Considersmalltransversedisplacements of the string (we ll be quantitative about the word small below). Letxbethe coordinate along the string , and let be the Transverse displacement. (There s no deepreason why we re using for the displacement instead of the we used in Chapter 2.)

6 Ourgoal is to find the most general form of (x, t).Consider two nearby points separated by a displacementdxin the longitudinal directionalong the string , and by a displacementd in the Transverse direction. Ifd is small (moreprecisely if the sloped /dxis small; see below), then we can make the approximation thatall points in the string move only in the Transverse direction. That is, there is no longitudinalmotion. This is true because in Fig. 1 the length of the hypotenuse equalsdxd dx2+d 2 Figure 1 dx2+d 2=dx 1 +(d dx)2 dx(1 +12(d dx)2)=dx+d 12(d dx).(1)This length (which is the farthest that a given point can move to the side; it s generally lessthat this) differs from the length of the long leg in Fig.

7 1 by an amountd (d /dx)/2, whichis only (d /dx)/2 times as large as the Transverse displacementd . Since we are assumingthat the sloped /dxis small, we can neglect the longitudinal motion in comparison withthe Transverse motion. Hence, all points essentially move only in the Transverse can therefore consider each point to be labeled with a unique value ofx. That is,the ambiguity between the original and present longitudinal positions is irrelevant. Thestring will stretch slightly, but we can always assume that the amount of mass in any givenhorizontal span stays essentially see that by the phrase small Transverse displacements we used above, we meanthat theslopeof the string is small.

8 The slope is a dimensionless quantity, so it makessense to label it with the word small. It makes no sense to say that the actual transversedisplacement is small, because this quantity has strategy for finding the wave equation for the string will be to write down the trans-verseF=maequation for a little piece of string in the span fromxtox+dx. The situationis shown in (We ll ignore gravity here.) LetT1andT2be the tensions in the string atx + dxxT1T2 2 1 Figure 2the ends of the small interval. Since the sloped /dxis small, the slope is essentially equalto the angles in the figure. (So these angles are small, even though we ve drawn them withreasonable sizes for the sake of clarity.)

9 We can use the approximation cos 1 2/2 tosay that the longitudinal components of the tensions are equal to the tensions themselves,up to small corrections of order 2 (d /dx)2. So the longitudinal components are (essen-tially) equal toT1andT2. Additionally, from the above reasoning concerning (essentially)no longitudinal motion, we know that there is essentially no longitudinal acceleration of thelittle piece in Fig. 2. So the longitudinal forces must cancel. We therefore conclude thatT1=T2. Let s call this common , although the two tensions and their longitudinal components are all equal, thesame thingcannotbe said about the Transverse components.

10 The Transverse componentsdiffer by a quantity that isfirstorder ind /dx, and this difference can t be difference is what causes the Transverse acceleration of the little piece, and it can becalculated as THE WAVE EQUATION3In Fig. 2, the upward Transverse force on the little piece at its right end isTsin 1,which essentially equalsTtimes the slope , because the angle is small. So the upward forceat the right end isT (x+dx). Likewise, the downward force at the left end is T (x).The net Transverse force is thereforeFnet=T( (x+dx) (x))=T dx (x+dx) (x)dx T dxd2 (x)dx2,(2)where we have assumed thatdxis infinitesimal and used the definition of the derivative toobtain the third , the difference in the first derivatives yields the secondderivative.