19-1

Note 19 Traveling WavesSections Covered in the Text: Selections from Chapter 20, except 20.7

Whole courses in physics are devoted to the study ofwaves (PHYB20H3S, to name one). Waves occurthroughout nature from mechanical waves on astring, to sound waves, to heat waves, toelectromagnetic waves. Waves take many forms andmove in various kinds of media (or none, as in thecase of electromagnetic waves). It is remarkable thatthe same basic mathematical language can be used inthe description of all of them. In this note we surveythe properties of traveling waves. We shall continuein Note 20 with the study of the superposition ofwaves.

A Few Words About “Medium”For the most part, we shall think of a wave as a dis-turbance that travels through a medium. For example,a mechanical wave can travel on a stretched string—the continuum of particles making up the string is themedium. By resultant disturbance of the wave we meanthe resultant displacement of the string. In this notewe shall also apply these ideas, appropriately modi-fied, to waves on water and to sound waves in air.Eventually, we shall want to apply these ideas to elec-tromagnetic waves that travel through a vacuum. 1

A WaveA wave can be described as a disturbance or a dis-placement in a medium that changes in form or in positionor in both as time goes by. Though the medium supportsthe propagation of the disturbance from place toplace, the material of the medium itself does not movealong with the wave.

There are two basic types of waves: transverse andlongitudinal. In a transverse wave the displacement ofthe medium is at right angles to the direction of wavepropagation (Figure 19-1). In a longitudinal wave, thedisplacement of the medium is in the direction of thewave propagation (Figure 19-2). A wave may involvea combination of longitudinal and transverse motion.For example, a wave propagating on the surface ofwater moves the particles of the water in a circularmotion (Figure 19-3). We shall not be concerned withthis unusual motion here.

An important characteristic of a wave is that it canconvey energy from one point to another even thoughno net movement of matter occurs. Waves are often 1 In the case of electromagnetic waves we shall associate theresultant disturbance with the electromagnetic field and not with amedium. We shall cover this subject if time permits.

employed to transmit information from one point toanother, sound waves and radio waves being goodexamples.

Figure 19-1. Examples of a transverse wave (pulse) on astring.

Figure 19-2. Example of a longitudinal wave on a slinky.

Figure 19-3. When a wave moves along the surface of waterthe water particles move in circles.

We begin our mathematical description of waves withthe simplest kind of wave, a hypothetical transversewave on a line.

Note 19

19-2

A Transverse Wave on a LineWe shall show in due course that the most general ex-pression for a wave pulse moving with speed v alonga string or a line, say the x axis, is

D = D+ + D– = f(x–vt) + g(x+vt).

Here D represents the displacement of the line—atransverse displacement if the line is a string—and fand g are arbitrary functions of their arguments (x ± vt).What this expression means physically can be seen bystudying the two terms separately. The first term is:

D+ = f(x–vt). …[19-1]

Suppose for convenience that D represents the trans-verse displacement of a string. And suppose that at t =0 the displacement is as shown at the left of Figure 19-4. Thus the function D = f(x) has a peak when x = 0.You should be able to see that the transverse displace-ment at an arbitrary later time t is given by thefunction

D+ = f(x–vt).

Figure 19-4. A general wave pulse represented by thefunction f(x) at two clocktimes t = 0 and t.

At a later time t (at the right side of the figure), thestring looks exactly the same as at time t = 0, but thepeak has moved to the value of x for which (x–vt) = 0.In other words, the peak is now at the position x = d =vt. 2

Since the peak moves a distance d = vt in an elapsedtime t, the peak moves in the +x direction with speedv. Therefore, D+ = f(x–vt) describes a wave of arbitrary(pulse) form propagating with speed v in the positivex direction. By a similar argument the function D– =g(x+vt) describes a wave of arbitrary (pulse) formmoving with speed v in the negative x direction. Themost general form for a wave includes both compon-ents D+ and D–, though we shall usually focus on awave consisting of only one of the two components. 2 Assuming of course that no dissipative or frictional forces dampor otherwise remove energy from the disturbance.

For the moment we consider a wave consisting ofonly one component, and for convenience thecomponent moving to the right. We shall consider awave having both components in Note 20.

Speed of a Transverse Wave:Calculated From Dimensional Analysis

In a moment we shall calculate the speed of a wave ona string by applying Newton’s laws. But we can alsocalculate the speed using the method of dimensionalanalysis (introduced in Note 02) by making certainassumptions. We assume the speed v depends only onthe mass m of the string, its length l and tension T.Since we propose only three variables we can writethe speed most generally as

€

v = (const)malbTc , …[19-2]

where a, b and c are constants to be determined in theanalysis. This equation is valid only if the dimensionsof the left and right hand sides of eq[19-2] are equal.Hence we need to prove

€

[v] = [malbTc ],

or

€

L.T−1 = MaLb (MLT−2)c , …[19-3]

where square brackets indicate the operation of takingdimensions. (Tension T is different from dimensionT.) In order for eq[19-3] to hold it follows that

0 = a + c (from the powers of M on both sides)

1 = b + c (from the powers of L on both sides)

–1 = –2c (from the powers of T on both sides)

These equations have a consistent solution. It is a =–1/2, b = 1/2, c = 1/2. So eq[19-3] is true if the speedof the wave is

€

v = (const) Tlm

= (const) Tµ

, …[19-4]

where µ is the mass per unit length of the string (m/l).We shall see in a moment that a rigorous treatment ofthis subject will confirm eq[19-4] and yield a value of1 for the constant.

An especially important kind of wave is the sinusoid-al wave, which we consider next.

Note 19

19-3

A Sinusoidal WaveA sinusoidal wave is the kind of wave that would beset up on a very long string if one end of it were givena periodic displacement (moved up and down con-tinuously), or in air by a loudspeaker driven by asinusoidal voltage source. Each peak of the wavemoves to the right with a constant velocity, and thewave is periodic both in space (at a fixed time t) andin time (at a fixed position x). A snapshot graph at t =0 of such a wave is shown in Figure 19-5.

Figure 19-5. Snapshot graph of a sinusoidal wave at a time t= 0.

The transverse displacement of the wave at time t = 0is given by the function

€

D(x, t = 0) = Asin 2π xλ

+ φ0

. …[19-5]

You can see that the wave has a number of attributes.A or Dmax is the amplitude of the wave, or the max-imum transverse displacement (in m). λ (in m) is thewavelength of the wave; it is the distance betweenequivalent points on the wave. In other words, it isthe distance between adjacent peaks or the distancebetween adjacent troughs of the wave. It is also thedistance between every other zero of the wave.φ0 is called the phase constant. The phase constant sets

the initial condition of the wave, or the transverse dis-placement at x = 0. From eq[19-5] we have

€

D(0 m,0 t) = Asinφ0

so

€

φ0 = sin−1 D(0,0)A

. …[19-6]

The wave as shown in Figure 19-5 and described byeq[19-5] is “a snapshot in time” at t = 0. To get the

wave “moving” in the x-ve x direction we need tosubstitute the expression (x – vt) for x. In doing so weobtain the most general form of the displacement

€

D(x, t) = Asin 2π x − vtλ

+ φ0

= Asin 2π x

λ−tT

+ φ0

…[19-7]

A history graph of the wave at x = x1 and a snapshotgraph of the wave at t = t1 are shown in Figures 19-6.Here A, λ , and φ0 have their meanings as justdescribed. T is the period of the wave, or the time (in s)for the wave to repeat itself.

Figure 19-6. History and snapshot graphs of the wave.

The angular frequency ω (in units of rad.s–1), linearfrequency f (in units of cycles per second or Hz) andperiod T (in units of seconds) are related by 3

€

ω = 2πf =2πT

. …[19-8]

If we define the wave number k as

€

k =2πλ

, …[19-9]

3 These terms have the same meanings as defined for oscillatorymotion in Note 12.

Note 19

19-4

and use the relationship v = λf that applies for allwaves we have

€

v = λf =2πk

ω2π

=ωk

,

or

€

ω = νk . …[19-10]

Substituting eqs[19-8] through [19-10] into eq[19-7] wehave for the wave

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D(x, t) = Asin kx −ωt + φ0( ) . …[19-11]

Let us consider the analysis of an example sinusoidalwave.

Example Problem 19-1Analyzing a Sinusoidal Wave

A sinusoidal wave of amplitude 1.0 cm and frequency100 Hz travels at 200 m.s–1 in the positive x direction.At t = 0 s, the point x = 1.0 m is on a crest of the wave.(a) Determine the values of A, v, λ, k, f, ω, T, and φ0 forthis wave.(b) Write the equation for the wave’s displacement asit travels.(c) Draw a snapshot graph of the wave at t = 0 s.

Solution:(a) From the statement of the problem we have A = 1.0cm, v = 200 m.s–1, f = 100 Hz. With these values we cancalculate

λ = v/f = 200/100 = 2.00 m,

k = 2π/λ = π rad.m–1 or 3.14 rad.m–1,

ω = 2πf = 200π = 628 rad.s–1,

T = 1/f = 0.0100 s = 10.0 ms.

The phase constant φ0 is determined by the initial con-ditions. We know that a wave crest, with displace-ment D = A, is passing x0 = 1.0 m at t0 = 0 s. Eq[19-6] atx0 and t0 is

€

D x0,t0( ) = A = Asin k 1.0 m( ) + φ0[ ] .

This means that sin[k(1.0 m) + φ0] = 1, which requires

€

k(1.0 m) + φ0 =π2

rad.

Thus

€

φ0 =π2rad − (π rad.m−1)(1.0 m) = −

π2

rad.

(b) With the information from part (a) the wave’sdisplacement is

€

D(x, t) = (1.0 cm) ×

€

sin (3.14 rad.m−1)x − (628 rad.s−1)t − π2

(rad)

(c) We know that x = 1.0 cm is a wave crest at t = 0 sand that the wavelength is λ = 2.0 m. Because theorigin is λ/2 away from the crest at x = 1.0 m, weexpect to find a wave trough at x = 0. This isconfirmed by calculating D(0 m, 0 t) = –1.0 cm. Thesnapshot graph of the wave is drawn in Figure 19-7.

Figure 19-7. A snapshot graph of the wave.

Speed of a Transverse Wave:Calculated from Newton’s Laws

For a wave moving along a string the displacement Dof the string, eq[19-11], is just the y-coordinate of aparticle in the string at the position x, that is, for sucha wave eq[19-11] reduces to

€

y(x, t) = Asin kx −ωt + φ0( ) . …[19-12]

The transverse velocity of a particle in the string atposition x is the time derivative of eq[19-12]:

€

vy(t) =dy(t)dt

= −ωAcos kx −ωt + φ0( ) …[19-13]

The transverse velocity vectors of particles in the

Note 19

19-5

string are drawn in Figure 19-8. Each particle vibratesup and down with simple harmonic motion with atime dependence given by eq[19-13]. Be careful todistinguish the movement of the wave from the move-ment of the string particles. The string particles moveup and down; the wave moves to the right.

Figure 19-8. A snapshot graph of a wave on a string withvectors showing the instantaneous transverse velocityvectors of the string particles at various points.

We are now ready to derive an expression for thewave speed. Consider a small segment of the stringwith length ∆x << λ at a wave crest (Figure 19-9). Thetension in the string at either end of this segment pullsdownward on the segment tending to bring it back tothe equilibrium position. Applying Newton’s secondlaw to the segment yields

Figure 19-9. A small segment of string at the crest of a wave.

€

Fnet( )y = may = (µΔx)ay …[19-14]

where m = µ∆x and µ is the mass per unit length of thestring.

The point of maximum displacement of the string atthe peak of the crest is also the point of maximumacceleration of the string particles. This acceleration isgiven by the time derivative of eq[19-13]:

€

ay(t) =dvy (t)dt

= −ω 2Asin kx −ωt + φ0( ) .

…[19-15]

The maximum value of this acceleration is ay = –ω2A.From eq[19-10], ω = vk where k is the wavenumber.Thus the maximum acceleration is

€

ay = −ω 2A = −v 2k 2A . …[19-16]

This shows that a large wave speed v causes the stringparticles to oscillate more quickly and thus to have alarger acceleration.

The y-component of the net force acting on the seg-ment of string is

€

Fnet( )y = 2TS sinθ . …[19-17]

The angle θ is very small (∆x << λ) so we can put sinθ≈ tanθ:

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Fnet( )y ≈ 2TS tanθ . …[19-18]

At this instant of time t = 0, with the crest of the waveat x = 0, the equation of the string is given to a goodapproximation by

€

y(x) = Acos(kx) .

Now

€

tanθ =dy(x)dx at x=Δx / 2 = −kAsin(kx) at x=Δx / 2

€

= −kAsin kΔx2

€

≈ −kA kΔx2

= −

k 2AΔx2

, …[19-19]

using the small angle approximation. Substitutingeq[19-19] into eq[19-18] we get

€

Fnet( )y = −k 2ATSΔx . …[19-20]

Using eq[19-16] for ay we have

Note 19

19-6

€

Fnet( )y = −k 2ATSΔx = (µΔx)ay

€

= −v 2k 2AµΔx . …[19-21]

Cancelling –k2A∆x from both sides we are left with theexpression for the wave speed:

€

v =Tµ

. …[19-22]

Thus we confirm that the speed of the wave is asgiven in eq[19-4] with the constant equal to unity.Note that the speed depends only on the mass perunit length and the tension in the string. Specifically,it does not depend on the shape of the wave or pulse.Let us consider a typical calculation.

Example Problem 19-2The Speed of a Wave on a String

Calculate the speed of a wave on a stretched string ofmass per unit length 0.001 kg.m–1 under a tension of10.0 N.

Solution:The speed of a wave on a stretched string is given byeq[19-22]:

€

v =Tµ

=10.0 N

10−3 kg.m−1 =100 m.s–1.

Waves in Two and Three DimensionsWe have all had the experience of dropping a pebblethrough the still surface of a pond. Ripples moveaway from the point where the pebble enters thewater surface (Figure 19-10a). The ripple waves arecircular, moving in 2D space with peaks and troughsthat are readily seen, peaks alternating with troughs.The waves are transverse waves (or at least appearso). Alternate wave peaks are spaced a wavelengthapart and the waves travel with a definite speed thatcan be measured. The speed of such waves is NOTgiven by eq[19-22].

If we observe such waves a great distance from thesource, say as waves on a beach, then we see them asparallel (Figure 19-10b). We see them as parallel wave“fronts” because the radius of the wave circles is verylarge. Waves on the ocean originate in storms (or dis-

locations in the Earth’s crust) far out to sea and travelmany miles before encountering the shore.

Figure 19-10. The wave fronts of a circular or spherical wave.

In 3D space, waves move away from the source in theform of spherical waves. An example is light. If weobserve these waves a great distance from the sourcethen we see them as plane parallel waves (Figure 19-11).

Figure 19-11. A plane wave.

Note 19

19-7

Phase and Phase DifferenceThe argument of the sine function, (kx – ωt + φ0), ineq[19-12] is called the phase of the wave. It is oftendenoted just φ. We shall see that the concept of a wavephase will be important in our discussion of interfer-ence in Note 20. In fact the wave fronts illustrated inFigures 19-10 and 19-11 are surfaces of constant phase.You can see this if you write the displacement asD(x,t) = Asinφ. Because each point on a wave front hasthe same displacement, it follows that φ must have thesame value at every point on a wave front.

Let us now examine the notion of phase difference ∆φbetween two different points on a sinusoidal wave.Two points are shown on the wave in Figure 19-12.The phase difference between them is

€

Δφ = φ2 −φ1 = (kx2 −ωt + φ0) − (kx1 −ωt + φ0)

€

= k x2 − x1( ) = kΔx = 2π Δxλ

. …[19-23]

Figure 19-12. The phase difference between two points on asinusoidal wave.

Thus the phase difference between any two points ona wave depends on the ratio of their separation ∆x tothe wavelength λ. For example, two points on a waveseparated by ∆x = λ/2 have a phase difference ∆φ = π.The phase difference between any two adjacent wavefronts (∆x = λ) is ∆φ = 2π. Let us consider a numericalexample of phase difference.

Example Problem 19-3The Phase Difference between Two Points on a SoundWave

A 100 Hz sound wave travels with a wave speed of343 m.s–1 (20 ˚C).(a) What is the phase difference between two points60 cm apart along the direction the wave is traveling?(b) How far apart are two points whose phase differsby 90˚?

Solution:(a) The phase difference between two points atpositions x1 and x2 is given by eq[19-23]:

€

Δφ = 2π Δxλ

.

Solving for the wavelength λ with ∆x = 60 cm = 0.60 mwe have

€

λ =vf

=343 m.s−1

100 Hz= 3.43 m.

Thus

€

Δφ = 2π 0.60 m3.43 m

= 0.350π rad = 63.0˚ .

(b) A phase difference of ∆φ = 90˚ is π/2 radians. Thussolving for ∆x:

€

Δx =Δφ2π

λ =π /22π

(3.43 m) = 0.858 m.

Notice that the length of the sound wave (of a rela-tively low frequency) is of the order of a dimension ofan average-sized living room.

Sound WavesIf you examine the cone of a loudspeaker closelyenough while the speaker is reproducing a low freq-uency sound you can actually see the speaker conemove in and out continuously. As the cone moves outit compresses the air molecules in front of it. As thecone moves in it causes the air molecules in front of itto become less dense. These regions of condensationand rarefaction of the air molecules move away fromthe speaker as a longitudinal wave (Figure 19-13). Thedistance between alternate condensations or betweenalternate rarefactions is the wavelength λ.

Sound can travel in gases, liquids and in solids. The

Note 19

19-8

speed of sound depends on the properties of the med-ium. For air at room temperature (20 ˚C) the speed ofsound is about 343 m.s–1. In liquids and solids thespeed of sound is generally greater than this. Thehuman ear is capable of detecting sound with freq-uencies between about 20 Hz and 20 kHz. Ultrasonicdevices operate at frequencies of 40 kHz and above. 4

Figure 19-13. A representation of a sound wave.

Electromagnetic WavesWhat is loosely referred to as light are waves that canbe seen by the human eye. Light is an electromagneticwave, an oscillation of the electromagnetic field.Radio waves and microwaves are examples ofelectromagnetic waves that happen, because of theirfrequencies, to be invisible to us. A section of theelectromagnetic spectrum is shown in Figure 19-14.Visible light occupies a thin slice of spectrum.

Figure 19-14. The electromagnetic spectrum from 106 Hz to1018 Hz.

All electromagnetic waves, of whatever type, travelthrough a vacuum with the same speed. This speed,denoted c, has been measured to have the value

c = 299,792,458 m.s–1.

4 As does the ultrasonic motion detector used in the first yearphysics lab in the experiment “Linear Motion”.

In most calculations we shall use c = 3.00 x 108 m.s–1.Notice from Figure 19-14 that visible light has wave-

lengths in the range 700 nm to 400 nm (1 nm = 10–9 m).The colors of the corresponding light range from redto blue, respectively. What is referred to as whitelight, like the light from the sun, was first shown byIsaac Newton to be a superposition of light of all ofthese colors.

Example Problem 19-4An Example of an Electromagnetic Wave

A satellite orbiting Jupiter transmits data to the Earthas a radio wave with a frequency of 200 MHz.(a) What is the wavelength of the electromagneticwave?(b) How long does it take the signal to travel the 800million kilometers from Jupiter to the Earth?

Solution:Solving for the wavelength we have

€

λ =cf

=3.00x108 m.s−1

2.00x108 Hz=1.5 m.

Solving for the elapsed time we have

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Δt =Δxc

=8.0 x 1011m

3.00 x 108m.s−1 = 2700 s = 45 min.

This wavelength, 1.5 m, is about the same as thewavelength used in taxicabs and police radios.

Power and IntensityAll forms of traveling waves, be they mechanical,sound or electromagnetic, transfer energy from onepoint to another. The power of a wave is the rate, injoules per second, at which the wave transfers energy.The intensity I of a wave of power P is defined

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I =Pa

, …[19-24]

where a is the area on which the wave impinges(Figure 19-15). The unit of intensity is W.m–2.Obviously, a wave focused into a small area will havea greater intensity than a wave of equal power that isspread out over a large area. For example, sunlight ona summer day can be focussed by a lens onto a

Note 19

19-9

pinpoint on a piece of paper to burn a hole in it.According to eq[19-24] if a source of spherical waves

(Figure 19-16) of power P radiates uniformly in alldirections then the power at distance r is spreaduniformly over the surface of a sphere of radius r.Thus for a spherical wave

Figure 19-15. Plane waves of power P impinge on area a withintensity I = P/a.

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I(r) =Psource4πr2

. …[19-25]

Thus the intensity decreases with the square of thedistance. This is referred to as rectilinear propagation.

Figure 19-16. A source emitting uniform spherical waves.Shown here is a 2D slice.

Eq[19-25] means that for a source of a certain powerthe intensity I1 at a distance r1 and intensity I2 at adistance r2 are related by

€

I1I2

=r22

r12 …[19-26]

For any wave type of amplitude A, the intensity canbe written

€

I = CA2, …[19-27]

where C is a constant of the particular wave type.

An early researcher into the physics of sound wasnone other than the inventor of the telephone,Alexander Graham Bell (1847-1922). In researchingthe physiology of hearing Bell found that the humanear responds to sound intensity in a logarithmic,rather than linear, fashion. Bell concluded that soundintensity I is most logically expressed as the logarithmof an intensity ratio, that is, as

# in db =

€

10log10II0

, …[19-28]

where “db” stands for decibel (named after Bell, thebasic unit being the bel) and Io = 1.00 × 10–12 W.m–2 isthe approximate intensity of the weakest sound thatcan be heard by the human ear.

The normal ear can distinguish sounds that differ inintensity by 1 db. An intensity of 120 db causes severediscomfort.

This concludes our survey of wave attributes. Weshall continue in Note 20 with the effects when wavescoexist at the same point in space.

Note 19

19-10

To Be Mastered

• Definitions: medium, wave, transverse wave, longitudinal wave• General expression for: speed of a wave on a string• General expression for: sinusoidal wave• Definitions: period T, amplitude A, wavelength λ, frequency f, angular frequency ω, wave number k, phase constant φ0,

phase φ, phase difference ∆φ

• Calculation of: speed of a wave on a stretched string:

€

v =Tµ

• Calculation of: phase φ, phase difference ∆φ• Description of: sound waves, electromagnetic waves• Definitions: power P, intensity I, rectilinear propagation

Typical Quiz/Test/Exam Questions

1. Explain the difference between a longitudinal and a transverse wave.

2. Give one example of each of the above, and also mention one kind of wave that is neither longitudinal nortransverse.