ch 15 1d waves (bad?) - physics2000 - relativity first · chapter 15 one dimensional ... had an...

Chapter 15OneDimensionalWave Motion

CHAPTER 15 ONE DIMENSIONAL WAVEMOTION

In the last few chapters we have followed the straight-forward procedure of identifying the forces on anobject, setting the vector sum of the forces equal to themass times acceleration, and solving the resultingequation. We began with problems like those involvingcircular motion where we knew the acceleration andcould solve the equation immediately, the conicalpendulum being an example. With oscillatory motionwe ended up with a differential equation whose solu-tion had to be guessed. The observation that oscilla-tory motion looks like circular motion viewed sidewayshelped greatly in this guess.

For damped and forced harmonic motion, it was nothard to write the differential equations, but the solu-tions involved mathematical functions and techniquesthat may not have been not familiar to the reader. Inthis chapter we are dealing with the subject of wavemotion, where it turns out that the differential equationdescribing the motion has derivatives in both time andspace. Setting up and solving such an equation re-quires mathematical discussions that are best left to amore advanced level course. Fortunately we can studythe physics of wave motion without working withdifferential equations.

If we went through the effort to derive the differentialequation for wave motion, we would end up with whatis called a wave equation. Once you have a waveequation, you can guess a solution and plug in your

guess just as we did for the simpler equations foroscillatory motion. For oscillatory motion, when weplugged in our guess sin (ωt) , we ended up with asimple equation ω= k/m for the frequency of theoscillation. For wave equations, if you plug in a guessrepresenting a wave traveling through the medium,you end up with a simple equation for the speed of thewave.

There are some famous wave equations in physics. In1860 James Clerk Maxwell combined the equations forelectricity with those for magnetism and, to his sur-prise, ended up with a wave equation. He initially hadno idea what the wave was, but he could calculate thespeed of the wave. Whatever wave he was dealing withtravelled at a speed of 3×108 meters per second or 1foot per nanosecond. As he knew of only one thing thattravelled at that speed—light—he concluded that hehad an equation for light waves and that the theoryleading to this equation was the theory of light. He haddiscovered that light was an electric and magneticphenomena.

In 1925, Louis De Broglie explained some bafflingphenomena in atomic physics by proposing that elec-trons have a wave nature. Erwin Schrödinger thenwent further and derived a wave equation for theelectron, an equation known as Schrödinger’s equa-tion that serves as the theoretical foundation foralmost all of chemistry.

15-2 One Dimensional Wave Motion

In 1929 Paul Dirac constructed a relativistic generali-zation of Schrödinger’s equation. The problem withDirac’s equation was that it had solutions for twodifferent kinds of wave, one representing the electronand the other an unknown particle of the oppositecharge. A particle similar to the electron but oppositein charge, the positron, was observed in a cloudchamber experiment carried out by Carl Anderson in1933.

It turns out that the relativistic wave equation for allelementary particles has two solutions, one solutionlike the electron representing matter, the other, like thepositron, representing antimatter. And the wave equa-tions predict that if a matter particle encounters itscorresponding antimatter particle, the two particlescan annihilate each other. There is an entire world ofantimatter, the existence of which was predicted byDirac’s wave equation.

With wave equations playing such an important role inphysics, one might think it is unfortunate that we are notprepared to derive and solve wave equations. Actuallythat is a blessing. There are certain general, simpleprinciples that apply to all forms of wave motion,principles that allow you to understand and predictmany features of the behavior of waves. These prin-ciples apply not only to waves like water and soundwaves whose behavior can be deduced from Newtonian

mechanics, but to light and electron waves whereNewtonian mechanics does not apply. Thus by learn-ing these general principles of wave motion, you aredeveloping a foundation in physics that goes beyondNewtonian mechanics.

The two basic principles of wave motion we will discussin this text are the principle of superposition and theHuygens principle. The principle of superposition isa fancy way of saying that waves add. If two waves aremoving through each other, they produce a total wavethat is the sum of the two waves. Since waves can havenegative amplitudes (troughs) this addition of wavescan produce cancellation. Two waves running intoeach other can, under the right circumstances, canceleach other out. This cancellation is clearly a wavephenomena, particles are not expected to do that.

The other general principle of wave motion is Huygensprinciple, which tells us how waves spread out in space.In this and the next chapter we will focus our attentionon one dimensional waves which do not spread out.Thus we do not need Huygens principle at this point.Later in Chapter 33, we discuss two and three dimen-sional waves and phenomena such as interference anddiffraction. In that chapter we do everything using theprinciple of superposition and the Huygens principle.In that chapter no calculus is used and we obtain resultsthat apply to a broad spectrum of phenomena, even tosub atomic particles where concepts like velocity andacceleration no longer have meaning.

15-3

WAVE PULSESWe begin our discussion of wave motion with the wavepulses we described in Chapter 1 on Special Relativity.To create a wave pulse on a stretched rope, you flick theend of the rope and a pulse travels down the rope asshown in Figure (1-4) reproduced here. This is calleda transverse wave because the particles in the ropemove perpendicular or transverse to the direction ofmotion of the wave pulse.

With a stretched Slinky we were able to observe twodifferent kinds of wave motion, the transverse waveseen in Figure (1-5) and a compressional wave seen inFigure (1-6). The compressional wave is also called alongitudinal wave because the particles in the spring

are moving longitudinally or parallel to the direction ofmotion of the wave pulse.

Sound waves are usually compressional waves travel-ing through matter. A sound wave pulse in air can beviewed as a region of compressed gas where themolecules are closer together as shown in Figure (1). Itis the region of compression that moves through the gasin much the same way as the region of compressed coilsmoves along the Slinky as seen in Figure (1-6).

To create the compressional wave on the Slinky wepulled back on the end of the Slinky and let go. Thisgives a small impulse directed down the Slinky. Inmuch the same way we can use a loudspeaker cone tocreate the pressure pulse in the air column of Figure (1).Here the impulse can be provided by applying a voltagepulse to the speaker causing the speaker cone to sud-denly jump forward. (If the speaker cone suddenlyjumps back, you get a pulse consisting of a region oflow pressure traveling down the tube.)

A transverse or sideways force in the medium tends torestore the medium to its original shape. For a trans-verse wave on a stretched rope, the tension on the ropeprovides the restoring force. For waves on the surfaceof a liquid, gravity or surface tension supplies therestoring force. But for waves passing through the bulkof a liquid or a gas, there are no transverse restoringforces and the only kind of waves we get are thecompressional sound waves.

a)

b)

c)

d)

Figure 1-4Wave traveling down a rope.

Figure 1-5Transverse wave on a Slinky

Figure 1-6Compressional wave on a Slinky.

Figure 1A sound wave pulse traveling down through a tubeof air. The pulse consists of a region of compressedair where the air molecules are closer together.This region of compression moves through the gasmuch as the region of compressed coils moves alongthe Slinky in Figure 1-6.

Vwave

speaker


The main difference between a liquid and a solid is thatin a liquid the molecules can slide past each other, whilein a solid the molecules are held in place by molecularforces. These forces which prevent molecules fromsliding past each other can also supply a transverserestoring force allowing a solid to transmit both trans-verse and compressional waves. An earthquake, forexample, is a sudden disruption of the earth that pro-duces both transverse waves called S waves and longi-tudinal or compressional waves called P waves. Thesewaves can easily be detected using a device called aseismograph which monitors the vibration of the earth.It turns out that the S and P waves from an earthquaketravel at different speeds, and will thus arrive at aseismometer at different times. By measuring thedifference in arrival time and knowing the speed of thewaves, you can determine how far away the earthquakewas.

Exercise 1The typical speed of a transverse S wave through theearth is about 4.5 kilometers per second, while thecompressional P wave travels nearly twice as fast,about 8.0 kilometers per second. On your seismo-graph, you detect two sharp pulses indicating theoccurrence of an earthquake. The first pulse is from theP wave, the second from the S wave. The pulses arrivethree minutes apart. How far away did the earthquakeoccur?

(Building a seismograph is a favorite high school sci-ence fair project. Basically you suspend a large massfrom springs and have a pen which is attached to themass draw a line on moving stripchart paper as shownin Figure (2). When the earth shakes, the stripchartshakes with the earth, but the mass remains more or lessstationary. The result is a squiggly line on the stripchartwhose amplitude is the amplitude of vibration of theearth.

SPEED OF A WAVE PULSEOne way to predict the speed of a wave is to set up thedifferential equation for the wave, plug in a travelingwave solution and let the equation tell you the speed.Without the wave equation we can in some casesdeduce the speed of the wave using clever tricks. Oneexample is the transverse wave on a rope, whose speedwe will calculate now. Another is the speed of Maxwell’swave of electric and magnetic forces which we willdiscuss in Chapter 32.

To calculate the speed of a transverse wave on a rope,consider a wave pulse moving down a rope at velocityv as shown in Figure (3a). To analyze the pulse,imagine that you are running along with the pulse at thesame velocity v. From your point of view, shown inFigure (3b), the pulse is at rest and the rope is movingback through the pulse at a speed v.

Now look at the top of the wave pulse. For anyreasonably shaped pulse, the top of the pulse will becircular, fitting around a circle of radius r as shown inFigure (3c). This radius r is also called the radius ofcurvature of the rope at the top of the pulse.

Finally consider a short piece of rope of length at thetop of the pulse as shown in Figure (3d). If this pieceof rope subtends an angle 2θ on the circle, as shown,then = 2 rθ and the mass m of this section of rope is

m = µ = µ2rθ massof short

sectionof rope(1)

where µ is the mass per unit length of the rope.

The net force on this piece of rope is caused by thetension T in the rope. As seen in Figure (3d), the ends

Figure 2Sketch of a simple seismograph for detectingearthquake waves. When the earth shakes, the masstends to remain at rest, thus the pen records the relativemotion of the stationary mass and shaking earth.

heavy masswith penattached

earth jiggling

rotating drum withstripchart paper

15-5

of the piece of rope point down at an angle θ. Thus thetension at each end has a downward component

T sin (θ) for a total downward force Fy of magnitude

Fy = 2T sin (θ) ≈ 2Tθ

downwardcomponentoftension force

(2)

If we keep the angle θ small, just look at a very smallsection of the rope, then we can approximate sin (θ) byθ as we did in Equation 2.

The final step is to note that this section of rope ismoving at a speed v around a circle of radius r. Thus weknow its acceleration; it is accelerating downward,toward the center of the circle, with a magnitude v2/r.

ay =

v2

r

downwardaccelerationofsectionof rope

(3)

Applying Newton’s second law to the downward com-ponent of the motion of the section of rope, we get usingEquations 1, 2 and 3

Fy = may

2Tθ = µ2rθ v2

r

(4)

Both the variables r and θ cancel, and we are left with T = µv2

v = T

µspeed of a wave pulseon a rope with tension T,mass per unit length µ

(5)

A result we stated back in Chapter 1.

Figure 3dThe ends of the rope point down at an angle θθ ,giving a net restoring force Fy = 2 T sin θθ .

r θ θ

θθTsin θTsin θ

Τ

= 2 r θ

Τ

Figure 3aWave pulse, and an observer, movingto the right at a speed v.

v (pulse)

v

v (rope)

rFigure 3cAssume that the top of the pulse fitsover a circle of radius r.

Figure 3bFrom the moving observer’s point of view, thepulse is stationary and the rope is movingthrough the pulse at a speed v.

v (rope)


DIMENSIONAL ANALYSISIn the above derivation of the speed of a transversepulse on a rope, we avoided solving a differentialequation by observing that the rope at the top of thepulse, from the moving observer’s point of view, wasmoving with circular motion whose acceleration weknow. For other kinds of wave pulses, particularly thecompressional pulse seen in Figure (1-6), we do nothave a simple circular motion, and a non calculusderivation of the wave speed becomes even moreconvoluted than the derivation we just went through.We could do it, but it is not worth the effort, especiallysince there are more straightforward ways of predictingwave speeds when one has the differential equation forthe wave motion.

What we will do instead is use a technique calleddimensional analysis to predict the speed of the wave.With dimensional analysis, you do not work out equa-tions. Instead you determine what the relevant vari-ables are, and then combine those variables in such away that the dimensions are correct. If you haveselected the correct variables, you get an answer that iscorrect to within a constant factor, and sometimes thecorrect answer.

To see how dimensional analysis works, let us firstapply it to the example we just worked out—to find thespeed of a transverse wave pulse on a rope. (For clarity,we will italicize the variable names. We will also useMKS units.) The first step is to do some experiments tofind out what variables the speed depends upon. Youchoose a rope, stretch it, and soon discover that thespeed of the pulse depends upon the tension T. Thus Tis one of the variables. Then you try two ropes of thesame length but different mass m, and discover that youget different wave speeds for the same tension. Thusthe mass m is one of the relevant variables. Anotherexperiment with 2 ropes of the same mass but differentlengths, gives different wave speeds. Thus the ropelength L is also important. Further experiments indi-cate that the speed of the pulse does not depend uponsuch variables as the color of the rope, the material fromwhich it is constructed, or the time of day. Thus youconclude that the relevant variables and their dimen-sions are

T

kg m(meter)

sec2, m kg, L m (6)

From these variables we have to construct the velocity.

v msec (7)

The only variable with the dimensions of seconds in itis the tension T, thus T must be included in our formulafor v. To get rid of kilograms, we must divide T by mto give

Tkg msec2 *

1m kg

= Tm

msec2

We are getting there, but we must have the same powerof meters and sec in order to get a velocity. If wemultiply T/m by L meters, we get

Tm

msec2 * L m = T L

mm2

sec2

Finally we get the correct dimensions by taking thesquare root, giving

v = T Lm

msec = T

µmsec (8)

where we noted that µ = m /L is the mass per unitlength.

Equation 8 tells us that no matter what the theory is, ifthe only relevant variables are T, m and L, the speed ofthe wave must be proportional to T µT µ for the dimen-sions to work out. We may have missed a factor of1/2 or 2π, but the functional dependence must be right.

Let us now use dimensional analysis to predict thespeed of the compressional Slinky pulse shown inFigure (1-6), or any compressional pulse on a stretchedspring. Since a stretched spring has a tension T and amass per unit length µ, one might guess that T µT µcould also be the formula for the compressional wave.However compressional and transverse waves do nothave the same speed. Even more important, you can getdifferent wave speeds for the same value of T µT µ, byusing different springs. It turns out that the tension T isnot a relevant variable.

Compressional waves depend upon the stiffness of amaterial, not the tension. For example a compressionalsound pulse will travel down a steel rod whether or notthe rod is under tension. Pulling on the ends of the steelrod does not noticeably change the speed of the soundpulse. Increasing the tension in a spring stretches thespring and therefore changes the mass per unit length.

15-7

It is the change in mass per unit length, not the changein tension, which affects the speed of the compres-sional pulse.

What variable is related to the inherent stiffness of aspring? The one that comes to mind is the springconstant k that appears in Hooke’s law

F = – kx ;

k = Fx

newtonsmeter Hooke′s law

(9)

The stiffer the spring, the greater the spring constant k.

Suppose we decide, after enough experimentation, thatthe relevant variables for the compressional pulse on aspring are the spring constant k, spring mass m andspring length L. We obtain the dimensions of k fromHooke’s law,

k newtons

meter = kkg ⋅ m sec2kg ⋅ m sec2

m = kkg

sec2

thus we have to construct a quantity with dimensionsm/sec from the variables

kkg

sec2 , M kg, L m

The only way we can do it is to divide k by m to get ridof kilograms and multiply by L2 to get

kL2

mm2

sec2

Taking the square root gives a quantity with the dimen-sions of a velocity

v = kL2

m = kLµ

speed of acompressionalwave on a spring

(10)

where again µ = m /L is the mass per unit length. Thisis our prediction for the speed of a compressional waveon a spring. The actual speed could differ by a constantfactor like 2, but it must have this functional depen-dence if we are correct in our assumption that the onlyrelevant variables are k, m and L.

In the formula v = kL/µ , the appearance of theproduct kL, rather than k alone may at first seemsurprising. But it turns out that the inherent stiffness of

a spring is proportional to kL and not just k, with theresult that the speed of the pulse is related to thestiffness, as we suspected.

To see why the inherent stiffness is related to kL,imagine that we wind a long spring and cut it in half tocreate two identical springs of length L1 . As shown inFigure (4a), if we apply a force F to one of the springs,and measure the distance ∆x that the spring stretches,we can use Hooke’s law to calculate that the springconstant k1 is given by

k1 =

F∆x

(11)

Now attach the two springs back together and stretchthe combination with the same force F as shown inFigure (4b). Since each spring feels the same force F,each stretches a distance ∆x , and the pair stretch adistance 2 ∆x . Thus from Hooke’s law the k2 of thecombination is given by

k2 =

F2∆x

=12

F∆x

=k1

2(12)

where we used k1 = F/∆x from Equation 11.

Figure 4bMeasuring the spring constantof two connected springs.

2 L1

F

2F = k (2∆x) ; k = F/2∆x k = k /22 12

(unstretched)

(stretched)∆x 2∆x

Figure 4aMeasuring the spring constant of a spring.

L1

1

F

F = k ∆x ; k = F/ x1

∆x

(unstretched)

(stretched)


When we attach two identical springs together, we endup with a longer spring but we are not changing theinherent stiffness of the spring. More importantly wedo not change the speed of the wave pulse. Connectingthe two springs and keeping the tension F the samemerely gives the pulse a longer distance to travel.

Note, however, that when we attach the two springs, thespring constant is cut in half, but the length is doubled,with the result that the product kL is unchanged.Explicitly we have

k2L2 =

k1

22L1 = k1L1 (13)

It should now appear more reasonable that the speed ofthe wave pulse should be given by kL/µ . Both thequantity kL, and the mass per unit length µ are inherentproperties of the spring that do not depend on the lengthof the spring. It is thus reasonable that the speed of thewave pulse should also involve only these variables.

Project SuggestionWe have spent some time discussing the speed ofpulses on a stretched spring for two reasons. One is thatwe used these pulses as our main example of wavemotion in our introduction to special relativity in Chapter1. The second is that measuring the speed of pulses ona spring makes a nice project, not much equipment isneeded, and you can fairly easily measure the variablesneeded to test Equation 10.

An additional advantage is that Equation 10 might ormight not be right. Since it was derived by dimensionalanalysis, it could be off by a constant factor like 1/2 or

2π . Therefore you have the challenge of determiningwhether or not there are some missing constant factors.

We expect that the wave speed should be proportionalto kL /µ , and if this does not turn out to be correct, wehave made some mistake in our analysis of what vari-ables are important. For example, in our analysis, wesaid nothing about the unstretched length L0 of thespring. Should L0 also appear in the formula for vwave?The way to find out is to do some experiments.

The experiments are made a bit easier by noting that

v =

kLµ =

kL2

m= L

km

where we used µ = m/L .

SPEED OF SOUND WAVESThe quantity kL that appeared in our formula for thespeed of a wave pulse is essentially the stiffness of a unitlength spring. By stiffness we mean the ratio of theforce applied to stretch a unit length of spring, to theamount of stretch ∆x that we get. The same ideas alsoapply to stretching a steel rod or any one dimensionalobject that has an elasticity and obeys Hooke’s law.

In engineering texts, the force applied to a unit length,area, or volume is given the generic name stress, andthe resulting displacement that the stress causes iscalled a strain. The ratio of the stress to the strain, iscalled the modulus. For our spring, the stress is thetension force F, and the strain is the change in length perunit length, or ∆x/L . The ratio of the stress to the strain,

F/(∆x /L) = FL /∆ x is called Young’s modulus. FromHooke’s law, F/∆ x = k , thus Young’s modulus is

FL /∆ x = kL, the quantity we have been discussing.

When we have a compressional wave in a gas, we canthink of the compression as being caused by a pressurepulse that travels through the gas. In the region wherethe gas is compressed, there is a slight excess pressure.The speed of the wave pulse depends upon the responseof the gas to this excess pressure.

Using the engineering terminology, the excess pres-sure ∆P represents the stress and the fractional changein volume, ∆V/V the corresponding strain. In this casethe ratio of the stress ∆P to the strain ∆V/V is called thebulk modulus B.The formula for B is thus

B = ∆P/ ∆V /V bulk modulus (15)

In Chapter 17 we will discuss the concept of a pressurein a gas, and see how changes in pressure are related tochanges in volume. Until we get to that chapter, anydetailed discussion of the concept of bulk modulus ispremature. What we will do now is assume that it is thebulk modulus B essentially represents the “stiffness” ofthe gas and should appear in the formula for the speedof a sound wave. We will then use dimensionalanalysis to figure out what the formula should be.

15-9

From this table we see that the speed of sound isconsiderably higher in the light gasses helium andhydrogen than in the more dense gas air. (The differ-ence in density is why helium and hydrogen filledballoons float in air.) The compressibility or bulkmodulus is usually the same for all gases at a givenpressure, thus the higher speed in hydrogen or heliumis due to the lower density, as we would expect from theformula v = B ρB ρ .

Exercise 2Steel is much stiffer than aluminum. (You make muchbetter springs from steel than aluminum.) Yet the speedof sound is greater in aluminum than steel. Why?

Exercise 3You tap the end of a 10 meter long steel rod with ahammer. How long before the tap can be detected atthe other end of the rod?

Exercise 4A dimensional analysis problemthat you should attempt now.

When working with the theory of electric and magneticphenomena, using the MKS system of units, one en-counters two rather mysterious constants labeled ε0(epsilon naught) and µ0 (mu naught). The constant ε0

appears in the formula for electric forces, and µ0 in theformula for magnetic forces. These constants have thefollowing dimensions

ε0

coulomb2

seconds2

kilogram meter 3 (18)

µ0kilogram meter

coulomb2

(19)

where a coulomb is a unit of electrical charge. Thenumerical values of ε0 and µ0 are to be found on theinside cover of this text along with other importantphysical constants.

(a) what combination of the constants ε0 and µ0 havethe dimensions of a velocity?

(b) from the numerical value of this velocity, what do youthink it is the velocity of?

In the ratio B = ∆P/ ∆V /V , the denominator ∆V /Vis dimensionless, thus B has the dimensions of pressurewhich is a force per unit area.

B newtonsmeter2 = B

kg msec2 m2 = B

kgsec2 m

(15)

To construct a quantity involving B that has the dimen-sions of a velocity, we have to get rid of the kilogramsby dividing by some quantity related to the mass of thegas. The only reasonable choice is the gas density

ρ kg/m3, thus we now have

B kg/(sec2 m)ρ kg/m3 = B

ρm2

sec2 (16)

which is the square of a velocity. Taking the squareroot, we get

v = B

ρspeed of asound wave (17)

a result we stated back in Chapter 1.

Equation 17 holds not only for a gas, but also forcompressional sound waves in a liquid and a solid.Liquids and solids, being far more incompressible thana gas, have a much greater bulk modulus B andtherefore higher speeds of sound. The speed of soundin various substances is given in Table 15-1.

Substance Speed of Soundin meters/sec

Gases (at atmospheric pressure)Air at 0° C 331Air at 20° C 343Helium at 20° C 965Hydrogen at 20° C 1284

Liquids

Water at 0° C 1402Water at 20° C 1482Sea water at 20° C 1522

Solids

Aluminum 6420Steel 5941Granite 6000Nuclear matter near c

Table 15-1


Conceptually we can separate all wave motion into twoclasses. There are the relatively smooth waves that canpass through each other like the circular ripples ofFigure (1-2), and the relatively wild waves that crest,crash and change their shapes. The relatively smoothwaves all obey what is called a linear wave equation.The properties of these waves are well understood andtheir behavior easy to predict. The wild waves obeynonlinear wave equations. We know very little aboutthe behavior of nonlinear waves, and in most cases findit very difficult to make predictions about their behav-ior. (Ocean waves, for example, are linear until theystart to crest. When you see whitecaps, the waves havebecome nonlinear.)

In this text we will restrict our discussion to the smooth,linear waves that behave like the circular ripples.Fortunately, most kinds of wave motion we encounterin nature, including almost all examples of light wavesand the probability waves of quantum mechanics, arelinear and therefore relatively easy to analyze. Butthere are growing applications for nonlinear waves,particularly in the field of laser optics.

LINEAR AND NONLINEARWAVE MOTIONFew sights are more awesome than the crashing ofocean rollers on a rocky beach during a storm. Thewaves seen in Figure (3) of Chapter 1, produced byhurricane Bertha hundreds of miles out to sea, werecrashing against the rocky shores of Mt. Desert Island,Maine, in July 1990. Hundreds of tourists and localtelevision station reporters were at the beach to observethe event. At one spot, called ‘Thunder Hole’, thecrashing waves created a loud boom and a geyser ofwater that went 40 or 50 feet in the air.

A very different sight are the circular ripples emergingfrom where raindrops have hit a puddle of water, seenin Figure (1-2), reproduced here. The special feature ofthese ripples is that they maintain their identity as theymove through each other. They are still circular waveseven after moving through other waves.

a)

b)

c)

d)

Figure 5Two wave crests running into each otheradd up to produce a bigger crest.

Figure 1-2Rain drops creating circular waves on a puddle.

Figure 1-3This ocean wave from Hurricane Bertha (July 31, 1990).

15-11

THE PRINCIPLE OFSUPERPOSITIONFigure (1-2) illustrates one aspect of linear wave mo-tion. The waves can move through each other andemerge undisturbed. The waves are still circular andunbroken after they have crossed.

There is another simple feature of this wave motion thatis a bit harder to see from that picture. While the wavesare crossing, they produce a wave whose height is thesum of the heights of the individual waves. If two crestsare moving through each other, the crests add toproduce a higher crest. Two troughs produce a deepertrough, and a crest and a trough will tend to cancel asthey move through each other.

This adding of the heights of crossing waves is moreeasily illustrated for the case of one dimensional wavepulses traveling down a rope. In Figure (5), two similarcrests add together to produce a doubly high crest for aninstant. In Figure (6) we see that a similar shaped crestand trough will cancel at the instant they are together.Figure (7) illustrates the idea that as any two waveshapes move through each other, they produce a waveshape whose height at any point along the rope is thesum of the heights of the individual waves movingthrough each other.

The concept that waves can maintain their identity asthey move through each other, and that they produce aresultant wave whose height or amplitude is the sum ofthe heights or amplitudes of the individual waves, thisconcept is known as the principle of superposition. Inmore colloquial language, the principle of superposi-tion says that waves add. The principle of superposi-tion is one of the key concepts of linear wave motion.It distinguishes linear from nonlinear wave motion.When nonlinear waves interact, you get somethingdifferent than the simple sum of the two waves.

Before leaving our discussion of the principle of super-position, we wish to take one further look at part (c) ofFigure (6). That is the point where an equal shapedcrest and trough are right on top of each other, preciselycancelling each other out. This kind of cancellation ofwaves is a common feature of wave motion. In fact, itis what distinguishes wave motion from what we havebeen calling particle motion. If two particles run intoeach other, they do not cancel like the waves of Figure(6). They bounce or crash but not cancel.

a)

b)

c)

d)

Figure 6When a crest meets a trough, there is ashort time when the waves cancel.

Figure 7In general, for linear wave motion. We obtainthe shape of the resulting wave by adding theamplitudes of the individual waves.

v

v


SINUSOIDAL WAVESIf you ask someone to describe wave motion, they arelikely to picture water waves and sketch a curve thatlooks like a sine wave. A sine wave represents just oneof many possible shapes for a wave. But it is animportant shape because it is often seen in nature andit is easy to handle mathematically. We will see shortlythat any arbitrary wave shape can be constructed fromsine waves, thus the sine wave can be thought of as abasic building block of wave motion.

To relate the mathematical sine function to wavemotion, recall our definition of sine function shownback in Figure (14-4). The point of that figure is that thesine function is the sideways projection of circularmotion. As the arrow rotates at an angular velocity ω,the angle θ that the arrow has rotated increases as

θ = ωt . On the right we have graphed the height of therotating arrow as a function of the angle θ = ωt toobtain a sine curve.

To actually create the sine wave shape seen in Figure(14-4), you can start shaking one end of a long rope asshown in Figure (8). If you move your hand up and

down with a sinusoidal oscillation, a sinusoidal shapedwave will start traveling down the rope, at a speed

vwave = T/µ . This creates an example of what iscalled a traveling sine wave.

The problem with creating traveling sine waves on arope, is that the wave reaches the end of the rope,reflects, and moves back through the incoming wave,complicating the situation. A better example of travel-ing sine waves can be seen on the surface of a lake orthe ocean where there is plenty of room for the wavesto move before they strike an object or a shore.

There are two distinct ways to view a traveling sinewave. One is to move along with the wave. Then allyou see is a stationary sinusoidal shape. The other is tostand still and let the wave pass by you. Then you willsee the wave oscillate up and down as successive crestsand troughs pass by you. This is illustrated in Figure (9)where we have sketched a traveling sinusoidal waterwave passing a fixed post in the water. If you movealong with the wave, then the shape of the wave doesnot change. But if you look at the post, the level of thewater is moving up and down with a sinusoidal oscil-lation.

Figure 8Sine wave created on astretched rope.

θ0

π/2

π

3π/2

π2

3π2

2ππ

0 θ = ωt

1

−1ω

ω vwave

x

λ

Figure 14-4Definition of thefunction

sinθθ = sin (ωω t).

15-13

Wavelength, Period, and FrequencyWe usually describe a wave in terms of its wavelengthλ , frequency f or period T. The easy way to rememberhow to go back and forth between these quantities, is touse dimensions.

When we view the shape of a traveling sine wave, thepredominant feature is the wavelength, the distance λbetween crests shown in Figure (8). Considering thatone full cycle of the wave fits between the crests, wecan assign the dimensions of meters per cycle to λ .

λ meterscycle

wavelength

When we let the wave pass by us and view the up anddown motion of the surface, we see an oscillationwhose period is T seconds per cycle and frequency isf cycles per second.

Since the period T is the length of time it takes onewavelength λ of the wave to pass by at a speed vwave ,we have (distance = speed times time)

λ meterscycle

= vwavemeterssecond

T secondcycle (20)

By assigning the dimensions meter/cycle to λ andsec/cycle to T, we can get the relationship λ = vwaveTfrom dimensions without having to memorize formu-las, or even having to think very much.

Figure 9Traveling sine wave on the surfaceof water. The sine wave shapemoves as a unit along the surfaceat a speed vwave . But if we look at afixed post in the water, the waterlevel at the post oscillates up anddown with a sinusoidal oscillation.

As an example of using dimensions to derive a formula,let us see if we can get a formula for the frequency f ofa wave of wavelength λ . The idea of using dimensionsis to try something, then see if the dimensions match. Ifthey don't match, change the formula until they do. Asa guess, let us try the formula

f = vwaveλ guess

Putting in dimensions, we have

f

cyclessec = vwave

meterssec × λ meters

cycle

= vwaveλ meters2

sec cycle

Clearly the dimensions do not match. We have tochange the formula so that meters cancel and we getcycles upstairs. This can be done if we move λdownstairs, giving

f

cyclessec =

vwave meters/sec

λ meters/cycle

=vwave

λcycles

sec (21)

which works. Thus the correct formula is f = vwave/λ ,a result that can be a bit tricky to figure out other ways.

ω t = 0

vwave

λ0

ω t =

ω t =

ω t =

ω t =

π

π2

3π2

2π

post


Angular Frequency ωωIn Figure (14-4), we reminded ourselves that a sinusoi-dal oscillation is equivalent to the sideways view ofcircular motion. If the vector on the left side of Figure(14-4) is rotating with an angular velocity ω radiansper cycle, we get one rotation, one period T of theoscillation, when the angle θ = ωt goes from 0 at t = 0,to 2π at t = T. Thus at the end of one period,

θ = ωT = 2π.

Again we can avoid memorizing new formulas byusing dimensions. We note that 2πis the number ofradians in a full circle or cycle. Thus we will assign toit the dimensions

2πradians

cycle(23)

Then to find the formula, for example, for the wave'sangular frequency ω in terms of the wave's period T,we have

ω radians

sec =2π radians / cycle

T sec / cycle

=2πT

radianssec

(23)

Exercise 5 (Do this one now.)For a traveling sine wave moving at a speed vwave , usedimensions to find

(a) λ in terms of vwave and ω

(b) vwave in terms of λ and T

(c) T in terms of λ and ω

(d) f in terms of ω

Spacial Frequency kWhen we let a traveling wave pass by us, we observea sinusoidal oscillation in time. This oscillation can bedescribed in terms of the number of seconds in eachcycle (T seconds/cycle), in terms of the frequency

(f cycles/second) or the angular frequency (ω radians/second) .

If instead we look at the whole wave at one instant oftime, or move along with the wave, we see a sinusoidaloscillation in space. We have described this spacialoscillation in terms of the wavelength, the number ofmeters in each cycle (λ meters/cycle). What we aremissing is a spacial analogy to frequency, the numberof cycles or radians per meter.

By dimensions we immediately see that

1λ meters

cycle

= 1λ

cyclesmeter

is the special analogy to the time frequency f, and that

2π radians/cycle

λ meters/cycle=

2πλ

radiansmeter

is the spacial analogy to the angular frequency ω .

In physics texts, it is not common to use a specialsymbol to designate the spacial frequency 1/λ(cycles/meter), but it is standard practice to designatethe angular spacial frequency 2π/λ (radians/meter) bythe letter k

kradiansmeter ≡ 2π

λradiansmeter

spacialfrequency (24)

15-15

The standard name for k is the lackluster expressionwave number, which says very little about the quantity.Instead we will refer to k as the spacial frequency of thewave. The higher the spacial frequency k, the moreradians we get in a meter, just as the higher the timefrequency ω , the more radians we get in one second.

When you first study wave motion, it may be anirritating complication to have two kinds of frequency,f cycles/sec and ω radians/sec (or two spacial frequen-cies 1/λ cycles/meter and k radians/meter). Why notstick with cycles which are much easier to visualizethan radians? The answer is that in the formulas for sinewaves, the sine function basically has to be expressedin terms of an angle, as in sin θ , and radians are anangle. To convert time t to an angle, we multiply by

ω as in

θ radians = ωradianssec × t sec

= ωt radians (25a)

while to convert the distance x to an angle we multiplyby k as in

θ radians = k radiansmeter × x meters

= kx radians (25b)

Using Equations 25a or 25b, we can express the singlefunction sin θ either as sin ωt , a sine wave in timeshown in Figure (10a), or as sin kx, a sine wave in spaceshown in Figure (10b). From these graphs we can seethat when ωt gets up to 2π, we have one period T, andwhen kx gets up to 2π we have one wavelength λ .

Exercise 6 (Try this now.)

You have a traveling sine wave moving at a speed vwave .Using dimensions find the formula for vwave in terms ofthe wave's time frequency ω and spacial frequency k.

Figure 10Sine waves in time and space.

ωt

ωT = 2π

y

π

a) Time: sin(θ) = sin(ωt)

2π

(one period)

kx

kλ = 2π

y

π

b) Space: sin(θ) = sin(kx)

2π

(one wavelength)

t = T

x = λ


Traveling Wave FormulaThus far we have formulas for a time varying sine wave

sin ωt , and a space varying wave sin kx. Now we wanta formula for a traveling sine wave whose amplitudevaries in both space and time. The answer turns out tobe

y = sin θ = sin kx – ωt travelingsine wave (26)

What we will do is show that this formula represents asine wave moving down the x axis.

Figure (11) shows a sinusoidal shape that is movingdown the x axis at a speed vwave . If we describe thewave by the function sin θ , then it is the origin

sin θ = 0 that moves down the axis at a speed vwave .Thus what we need is a formula for θ so that when weset θ = 0 , that point does move down the x axis at thedesired speed.

The answer we gave in Equation 26 suggests that thecorrect formula for θ is

θ = kx – ωt (27)

Setting θ = 0 we get

θ = 0 = kx – ωt ; kx = ωt

x = ωk

t (28)

But if the θ = 0 point travels at a speed vwave , then aftera time t, it has traveled a distance x given by

x = vwavet (29)

Comparing Equations (28) and (29), we see that the

point θ = 0 moves along the x axis at a speed

vwave = ωk

(30)

If you did Exercise 6, you recognize that the quantity

ω/k has the dimensions of a velocity

ωk

radian/secradian/meter

= ωk

metersec = vwave (31)

Thus the origin does move down the x axis at a speed vwave , and the formula kx – ωt , is our desired trav-

eling wave formula.

Exercise 7

Explain what kind of a wave is represented by theformula y = sin(kx + ωt) .

θ = 0

v t

θ = 0

θ = 0

wave

v wave

v wave

v wave

y = sin(θ)

Figure 11The cycle begins at θ = 0.

15-17

Phase and AmplitudeThe Equation 26 for a traveling wave can be general-ized by noting that the wave can have an arbitraryamplitude A, and an arbitrary constant phase angle φto give

y = A sin kx – ωt + φ (32)

The amplitude A just makes the sine wave bigger orsmaller, and the phase angle φ shifts the sine wave tothe left or right.

To see precisely how the phase angle φ shifts the sinewave, we have in Figure (12) compared sin θ and

sin θ + φ . The function sin θ + φ crosses zerowhen the angle θ + φ = 0 or at θ = – φ. Thus addinga phase angle φ shifts the sine wave back a distance

– φ radians. If, for example, we set φ= π/2 , the waveis shifted back 1/4 of a wavelength, and we haveconverted a sine wave into a cosine wave.

0θ

y = sin (θ)

(θ) = 0

vwave

0θ

y = sin (θ + φ)

(θ + φ) = 0

–φ

vwave

Figure 12Adding a phase angle φφ shiftsthe wave back a distance φφ.


STANDING WAVESIn addition to the waves traveling down a rope, anotherkind of wave pattern that is easy to achieve are thoseshown in Figure (13). All you have to do is shake theend of the rope at the right frequency and one of thesewaves will appear. Change the frequency and you canchange to one of the other patterns.

The waves in Figure (13) are called standing wavesbecause the pattern does not move along the rope. Thepoints of zero amplitude, the points called nodes of thewave, stay at fixed positions while the rope betweennodes oscillates back and forth.

The difference between the wave patterns is character-ized by the number of nodes. In Figure (13) all thewaves have nodes at the ends, and there are zero, one,two and three nodes in between as we go from the leftto the right pattern.

The two kinds of waves on a rope, the traveling waveof Figures (8) and (9) and the standing wave of Figure(13) are closely related to each other. A carefuldemonstration shows how traveling waves can turninto standing waves.

If you start shaking a rope a traveling wave starts downthe rope as shown in Figure (8). After a while the wavereaches the other end of the rope, is reflected, and startsmoving back the other way. This reflection is mosteasily seen if you send a single pulse down the rope sothat you can see it bounce off the fixed end and comeback to you.

If you send a series of pulses down the rope, if youcreate a traveling sine wave, then the reflected pulseshave to move back through the pulses that are stillcoming in. You now have the superposition of twotraveling waves moving through each other in oppositedirections.

Figure 13Standing waves on a rope.

15-19

You might think that the sum of two traveling wavesmoving through each other could lead to a complexpattern, and when you try it in a demonstration it oftenlooks that way. The problem with a demonstration isthat the reflected wave reaching your hand can reflectagain and you begin to build up a mixture of manywaves.

If you do the demonstration carefully, however, youcan observe a simple result. The sum of the travelingwave and the reflected wave, moving through eachother, is a standing wave. The addition of two travel-ing waves of equal amplitude and wavelength movingin opposite directions through each other is illustratedin Figure (14). The two traveling waves are shown onlines (a) and (b) at five different times t = 0, T/4, T/2,3T/4 and T, where T is the period of the waves. On thebottom line (c) we have added the amplitudes of the twotraveling waves to get a picture of the amplitude of theresulting wave.

In the first frame t = 0, the two waves match exactly,producing a sum that has twice the amplitude of eithertraveling wave. A quarter of a period later, at t = T/4,the traveling waves are precisely opposite each other.And the sum is zero all along the wave. This neverhappens in a traveling wave. A traveling wave is nevercompletely flat, there is always a crest moving along.

At time t = T/2, half a period later, the traveling wavesagain line up producing a wave of twice the amplitude.Note now that the points in the sum wave that werebelow the axis at t = 0 are now above the axis att = T/2, and vice versa. At time t = 3T/4 the travelingwaves are again out of phase and add up to zero. Att = T, we are back to where we started.

From line (c) of Figure (14), we see that the nodes of thesum wave remain stationary and the rope between thenodes oscillates up and down. This is exactly what wesee in the photographs of the standing rope waves inFigure (13).

t = 0 t = T/4 t = T/2 t = 3T/4 t = T

(a)

(b)

(c)

Figure 14Making a standing wave out of two traveling waves on a rope. In the top line (a) we show a travelingwave moving to the right. A section of rope is shown at times t = 0, T/4, T/2, 3T/4 and T, where T is theperiod of the wave. In (b) we have, in the same section of rope, five views of a traveling wave moving tothe left. In (c), we have added the two traveling waves and get a standing wave with stationary nodes. Todemonstrate the addition, at the center of each time frame we have drawn arrows to show the height ofthe wave at that point. The length of the bottom arrow is the sum of the lengths of the upper two arrows.To add two waves, you add up the heights at each point along the wave.


Using a trigonometric identity, we can show math-ematically that the sum of two traveling waves movingthrough each other creates a standing wave. Theformula for a sine wave traveling to the right is fromEquation 26.

ymoving right = A sin kx – ωt (26)

If you worked Exercise 6, you found that the formulafor a similar wave moving left is

ymoving left = A sin kx + ωt (33)

The formula for the sum of the two waves is

ysum wave = A sin kx – ωt + A sin kx + ωt

The trigonometric identity we will use is

sin a ± b = sin a cos b ± cos a sin b (34)

This gives

sin kx – ωt = sin kx cos ωt – cos kx sin ωt

sin kx + ωt = sin kx cos ωt + cos kx sin ωt

Add these two sine waves, the cos kx sin ωt termscancel and we are left with

ysum wave = 2A sin kx cos ωt (35)

To interpret Equation 35, write it in the form

ysum wave = A x cos ωt (36a)

where the x dependent amplitude A x is

A x = 2A sin kx (36b)

Equation 36a tells us that the entire wave is oscillatingin time as cos ωt . However the amplitude of theoscillation depends upon the position x long the wave.Equation (36b) tells us how the amplitude varies withposition. It varies sinusoidally as sin kx, with nodespermanently located at the points where sin kx = 0. Thissinusoidal variation in the amplitude along the wave isclearly seen in the photographs of the standing waveson the rope, Figure (13).

WAVES ON A GUITAR STRINGPerhaps the clearest example of standing waves are thewaves on the strings of a stringed instrument such as theguitar. The advantage of working with these waves isthat you get to both see the shape of the wave and hearits frequency.

The shape of guitar string waves are the same as thestanding rope waves in Figure (13). In Figure (15), wehave sketched the allowed standing wave patterns on astring of length L. Because the string is fixed at theends, we can only have waves with nodes at the ends.

bridge string nut

L

λ = 2L1

first harmonic or fundamental

second harmonic

third harmonic

fourth harmonic

fifth harmonic

n th harmonic

1

λ = 2L22

λ = 2L33

λ = 2L44

λ = 2L55

λ = 2Lnn

Figure 15Allowed standing waves on a guitar string. Theformula for the wavelength of the nth harmonicis seen to be λλ n = 2 L / n .

15-21

The wave with no nodes between the ends is called thefundamental or first harmonic. Its wavelength is 2L,twice the length of the string. In the second harmonic,with one node in the middle, a full wave fits on thestring at one time and we have λ2 = L . Each time weadd a node we go up to one higher harmonic. (Thesecond harmonic is also called the first overtone, etc.)

The general formula for the wavelength of the nthharmonic can be seen is we write the progression ofwavelengths in the form

λ1 =2L1

; λ2 =2L2

; λ3 =2L3

, etc.

It is clear that the formula for λ n is simply

λ n =

2Ln

wavelength ofthe nth harmonic

(37)

The easiest way to remember Equation 36 is draw asketch of the allowed standing waves and write downthe progression λ1 = 2L/1, λ2 = 2L/2, etc.

Frequency of Guitar String WavesWhat notes do you hear when you pluck a guitar string?That depends very much upon how you pluck it.Usually when you pluck the string, you create a numberof the standing wave patterns at one time, and the noteyou hear is a rich mixture of the frequencies of theindividual waves.

With care, however, you can pluck the string so thatmost of the vibration is in one of the harmonics. Agentle pluck at the center of the string will excite mostlythe fundamental or first harmonic. Pluck the string 1 41 4of the way from one end and briefly place your fingerat the center of the string to create a node there. Thisway you can excite mostly the second harmonic. Youwill then notice that the sound of the note is one octaveabove the sound of the fundamental. Accomplishedguitar players can selectively excite still higher har-monics.

What are the frequencies of oscillations of these vari-ous standing wave patterns? We can answer thisquestion because of our knowledge that standing waves

can be made from two traveling waves moving througheach other. The resulting standing wave has the samefrequency and wavelength as the traveling wave, thuswe can use the traveling wave formulas to determinethe frequency of the standing wave.

Using dimensions, we see that a traveling wave ofwavelength λ cm/cycle , traveling at a speed v cm/sec,has a frequency f cycles/second given by

fcycles

sec = v metersec × 1

λ meter/cycle

= vλ

cyclessec (38)

The speed v of a transverse wave on a string, that hasa mass per unit length µ, and tension T, is from Equation5

vwave =

T

µ(5)

Using Equation 5 in Equation 38 gives us as theformula for the frequency of the traveling wave

f =

1

λT

µ(39)

The same Equation 39 must also apply to the standingwaves on the guitar string. We get for the frequency fnof the nth harmonic, which has a wavelength fn ,

fn =

1λn

Tµ

frequency of thenth harmonic

(40)

For anyone who has tuned a guitar, Equation 40 makesa lot of sense. First note that when you go from the firstharmonic λ1 = 2L, to the second harmonic λ2 = L, thewavelength is cut in half and the frequency doubles. Adoubling of the frequency of a note corresponds togoing up one octave.

When you are tuning the guitar, you raise the frequencyof a string by tightening it and increasing the tension.That is also predicted by Equation 40.


On a guitar or most stringed instruments the low notesare played on fat wires that have a greater mass per unitlength µ than the skinny wires used for the high notes.The reason for using the fat wire is that you can increasethe tension and still keep the frequency down. Themore tension in the wire and the more mass in the wire,the more energy you can store in the wire and the louderthe sound you can produce. It is hard to get as muchsound out of the low frequency strings and you need allthe help you can get.

Exercise 8You have a guitar string of length L, with a tension T anda mass per unit length µ. (L is the distance from the nutto the bridge.)

(a) What is the f frequency of the fundamental mode ofvibration? Express your answer in terms of L, T, and µ.

(b) Show that the nth harmonic has a frequency n timesas great as the fundamental.

Exercise 9One end of a wire is attached to a post as shown inFigure (16). The wire is then run over a pulley where amass m is hung on the other end. The distance d fromthe post to the pulley is 1 meter and the mass of onemeter length of the wire is 5 grams.

(a) How big a mass m must be hung on the wire in orderto get the wire to vibrate in its fundamental mode at afrequency of 440 cycles/second, which is middle A?(Answer: 395.10 kg).

(b) Describe four distinct ways one could double thefrequency of oscillation of the wire.

(c) How much mass would you have hung on the wirein Figure (16) to get the wire to oscillate in its fundamen-tal mode at a frequency two octaves above middle A?(Answer 6321.63 kg.)

Sound Produced by a Guitar StringWhen you pluck a guitar string, the standing wave onthe string produces a traveling sound wave in the air.This is analogous to plucking the end of a rope toproduce a traveling wave along the rope as illustratedin Figure (8b). The wavelength of the sound wave isdetermined by the frequency of oscillation of the stringand the speed of sound. (It is not the same as thewavelength of standing waves on the string.)

Exercise 10A guitar string is tuned to oscillate at a frequency of 440cycles/second in its fundamental mode. What are thewavelengths of the sound waves produced by the firstthree harmonics

(a) in air at 20° C

(b) in helium at 20° C

m

d

Figure 16An easy way to adjust thetension in a vibrating string

ch 15 1d waves (bad?) - physics2000 - relativity first · chapter 15 one dimensional ... had an...

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