WavesPart 1: Travelling Waves

Last modified: 15/05/2018

Contents Links Contents

Travelling WavesHarmonic Waves

WavelengthPeriod & FrequencySummary

Example 1Example 2Example 3Example 4

Transverse & Longitudinal WavesWaves on a String

Transverse VelocityTransverse AccelerationWave SpeedPower Transmitted

ExampleSummary

Travelling Waves Contents

A travelling (or progressive) wave is a disturbance in a medium thatmoves while maintaining its shape over time.

This could be an ocean wave, a wave on a string, a sound wave etc.While the details of these differen types of wave vary, the basic propertiesare the same and will be discussed in this lecture.

The wave disturbance can have any shape, but the condition that thisshape doesn’t change puts some limits on the mathematical descriptionof the wave.

A one-dimensional wave, moving in the x -direction can be described by awave function y(x , t) which describes the magnitude of the disturbanceat position x and time t. For a wave on a string y would be the distanceaway from the straight line of the string that the wave has moved thestring at that time.

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v

y(x , t)

x

If a wave is to maintain its shape after moving a distance ∆x in time ∆t,then we must have:

y(x , t) = y(x + ∆x , t + ∆t) = y(x + v∆t, t + ∆t)

v

y(x + ∆x , t + ∆t)∆x = v∆t

This can only be true if:

y(x , t) = F (x − vt)

where F can be any function at all. (F will determine the exact shape ofthe wave.

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We have assumed here that the wave is travelling in the +x -direction - ast increases, so does x . If instead the wave is travelling in the−x -direction, we use similar logic to find:

y(x , t) = y(x −∆x , t + ∆t) = y(x − v∆t, t + ∆t)

which requires:y(x , t) = F (x + vt)

where again F can be any function.

I A travelling wave in the +x direction is described by:y(x , t) = F (x − vt)

I A travelling wave in the −x direction is described by:y(x , t) = F (x + vt)

where in each case, the speed v is a positive number.

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If we have two waves travelling in the same medium at the same speedand direction:

y1(x , t) = F1(x − vt) and y2(x , y) = F2(x − vt)

Then the disturbances caused by the two waves can be added:

y(x , t) = y1(x , t) + y2(x , t) = F1(x − vt) + F2(x − vt) = G(x − vt)

y is again a function of x − vt, so is also a travelling wave, called thesuperposition of the two waves. The same principle will apply to three,four, five etc waves.

A very useful technique is to do this procedure in reverse - expressing awave as the superposition of harmonic waves, where F is the sinfunction.

Because of this, we will focus on the properties of harmonic waves in thefollowing.

Harmonic Waves Contents

The general equation for a harmonic wave is:

y(x , t) = A sin (k(x − vt) + φ)

where A, k and φ are constants. This is often written using ω = kv as:

y(x , t) = A sin (kx − ωt + φ)

At t = 0 this looks like:

x

y

A

A

φk

Since the maximum value of ‘sin’ is1, A represents the maximum valueof the disturbance, or the amplitudeof the wave.

φ is the starting phase - it indicatesthe starting point of the wave. Of-ten this is unimportant and can beassumed to be zero.

Wavelength Contents

An important property of a wave is the wavelength, λ, which measuresthe distance between two consecutive peaks of the wave (or actuallybetween any point, and the next repeat of that point).

x

y

x + λx

λ

x

yλλ

λ

λ

Mathematically, this means:

y(x , t) = y(x + λ, t)

A sin(kx − ωt + φ) = A sin(k(x + λ)− ωt + φ)

= A sin((kx − ωt + φ) + kλ)

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Clearly, we must have:kλ = 2π

And so, the wavelength λ is connected to the wave number k by:

kλ = 2π ⇒ λ = 2πk

The term ‘wave number’ is used for historical reasons, don’t stress aboutwhat it means.

The SI unit of wavelength, λ is of course metres (m)

The SI unit of wave number, k is metres−1 (m−1)

Period & Frequency Contents

Another important property of a wave is the period, T , which measuresthe time taken for a point on the string to complete a full up and downoscillation.

t = 0

t = 18 T

t = 14 T

t = 38 T

t = 12 T

t = 58 T

t = 34 T

t = 78 T

t = T

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Mathematically:

y(x , t) = y(x , t + T )

A sin(kx − ωt + φ) = A sin(kx − ω(t + T ) + φ)

= A sin((kx − ωt + φ)− ωT )

⇒ ωT = 2π

and the angular frequency, ω is connected to the period:

ω = 2πT = 2πf

where f = 1T is the frequency of the wave.

The SI unit of period, T is of course seconds (s)

The SI unit of angular frequency ω is Hertz (Hz) (1 Hz = 1 s−1)Frequency f is also measured in Hertz.

Harmonic Waves Summary Contents

The general equation for the displacement caused by a harmonicwave is:

y(x , t) = A sin (±kx ± ωt + φ)

where:

A is the amplitude of the wavek is the wave number (always positive),I the wavelength λ is calculated from k: λ = 2π

kω is the angular frequency (also always positive),

I the period T and frequency f are calculated from ω:T = 2π

ω f = 1T = ω

2πφ is the starting phase of the waveThe speed v of the wave is:

v = f λ = ωk

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NOTE: both k and ω are positive, the signs in front of thesenumbers indicate the direction of the wave:

I if the signs are opposite (i.e -,+ or +,-) then the wave istravelling in the +x direction

I if the signs are the same ( +,+ or -,-) then the wave istravelling in the −x direction

From maths you should remember cos(X ) = sin(X + π2 ), so:

cos (kx − ωt + φ) = sin(

kx − ωt + (φ+ π

2 ))

It is therefore possible to rewrite a wave equation, using ‘cos’ in-stead of ‘sin’, provided that we also change the phase φ. As alreadymentioned, the exact value of φ is usually not very important.

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A travelling wave is described by the following equation:

y(x , t) = 0.25 sin(0.5πx − 4πt)

For this wave find the:(a) amplitude(b) wave number(c) angular frequency(d) wavelength(e) frequency

(f) wave speed

(g) direction of travel

(h) displacement at x = 1.3 mand t = 4.2 s

Compare the given equation with the general form:

y = 0.25 sin(0.5πx − 4πt)y = A sin(kx − ωt + φ)

(a) A = 0.25 m(b) k = 0.5π = 1.57 m−1

(c) ω = 4π = 12.57 Hz

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(d) λ = 2πk = 2π

0.5π = 4.0 m

(e) f = ω2π = 4π

2π = 2.0 Hz

(f) v = ωk = 4π

0.5π = 8.0 m/s or v = f λ = (2)(4) = 8 m/s

(g) There are opposite signs in front of k and ω, so the wave istravelling in the +x direction

(h) y(x = 1.3, t = 4.2) = 0.25 sin(0.5π × 1.3− 4π × 4.2)

= 0.25 sin(−50.74 RADIANS)

= −0.11 m

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A travelling wave is described by the following equation:

y(x , t) = 0.25 sinπ(0.5x − 4t)

For this wave find the:(a) amplitude(b) wave number(c) angular frequency(d) wavelength(e) frequency

(f) wave speed

(g) direction of travel

(h) displacement at x = 1.3 mand t = 4.2 s

y(x , t) = 0.25 sinπ(0.5x − 4t) = 0.25 sin(0.5πx − 4πt)

This is the SAME equation as we just saw! Don’t be fooled byour pitiful attempts to confuse you!!!Obviously all answers are the same!

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A travelling wave is described by the following equation:

y(x , t) = 0.25 cosπ(0.5x − 4t)

For this wave find the:(a) amplitude(b) wave number(c) angular frequency(d) wavelength(e) frequency

(f) wave speed

(g) direction of travel

(h) displacement at x = 1.3 mand t = 4.2 s

This is the same equation as the previous example except with ‘cos’instead of ‘sin’.The answers (a) - (g) are unchanged, only (h) requires re-doing:

y(x = 2, t = 4) = 0.25 cosπ(0.5× 1.3− 4× 4.2)= 0.25 cos(−50.74) again RADIANS= 0.22 m

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A travelling wave is described by the following equation:

y(x , t) = 0.25 cosπ(4t − 0.5x)

For this wave find the:(a) amplitude(b) wave number(c) angular frequency(d) wavelength(e) frequency

(f) wave speed

(g) direction of travel

(h) displacement at x = 1.3 mand t = 4.2 s

Remember cos(−X ) = cos X , so:

y(x , t) = 0.25 cosπ(4t − 0.5x) = 0.25 cosπ(0.5x − 4t)

This is the same equation as the previous example and is anotherattempt to confuse you. The ‘x’ term does not have to come first!

Transverse & Longitudinal Waves Contents

There are two common types of travelling waves, differing in the exactmotions they cause:

I Longitudinal waves have the wave vibrations in the same directionas the direction of the wave. Sound is a longitudinal wave,consisting of pressure variations in the air.

I Transverse waves are more common, including waves on a string,ocean waves and light. In these waves, the direction of the mediumvibrations is transverse to (across, or perpendicular) the wavevelocity.

In both cases, the average motion of the medium is zero - particlesoscillate about their original position. The definitions of wavelength,frequency apply equally to both.

We’ll now look a little more closely at one example of a transverse wave -waves on a string.

Wave on String: Transverse Velocity Contents

Imagine an ant holding on to a string at position x as a harmonic wavepasses by. The transverse displacement (i.e. the displacement in adirection transverse to the waves direction of motion - up/down) of theant’s point on the string at time t is:

y(x , t) = A sin (kx − ωt + φ)

We know that the wave speed is v = ωk , but what is the speed of the

ant?

The ant is moving up and down with variable transverse velocity, vtwhere:

vt(x , t) = ∂y∂t = −ωA cos (kx − ωt + φ)

(where we use ∂∂t - the partial derivative with respect to time, which

requires x to be treated as a constant)

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Remember, this speed is unrelated to the speed of the wave. It is thespeed of the up/down motion of the string (and therefore the ant) at x .

Notes:I The maximum transverse speed is ωA, which occurs when

cos(. . .) = ±1, i.e. when sin(. . .) = 0 - the ‘undisturbed’ position ofthe string.

I The minimum transverse speed is 0, which occurs when cos(. . .) = 0,i.e. when sin(. . .) = ±1 - the maximum amplitude of the movement.

I This transverse velocity is different in magnitude and direction tothe wave velocity.

I The ant (and each piece of the string) is in simple harmonicmotion. ( Remember y = A sin(ωt + constant) ) These motions areall slightly out of sync, as the value of the constant changes.

Wave on String: Transverse Acceleration Contents

Clearly, the transverse velocity is changing, so we know that there mustalso be a transverse acceleration.

Of course the transverse acceleration is the (partial) time derivative ofthe transverse velocity:

at(x , t) = ∂vt∂t = −ω2A sin (kx − ωt + φ) = −ω2y

Notes:I The maximum transverse acceleration is ω2A, which occurs when y

is also maximum. (just like simple harmonic motion)I The minimum transverse acceleration is 0, when y = 0.

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A travelling wave is described by the following equation:

y(x , t) = 0.25 sinπ(0.5x − 4t)

For this wave find the maximum

(a) transverse speed (b) transverse acceleration

(a) max. transverse speed is ωA = 4π × 0.25 = 3.1 m/s.

We previously found the speed of this wave to be 8 m/s, sothe wave is faster than the transverse motion of the string.For a different wave, this could be the other way around.

(b) max. transverse acceleration is

ω2A = (4π)2 × 0.25 = 39.5 m/s2 ≈ 4g

Wave on String: Wave Speed Contents

Two important properties of a stretched string are:I the mass per unit length (or linear mass density) : µI the tension T in the string

We should expect the values of these properties to affect the propertiesof a wave on the string.

To investigate this connection, let’s consider wave travelling on a stringat speed v to the right. Now, we zoom in on a small segment which ismomentarily at rest, at the peak of the wave.

v

A

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If we now consider ourselves to be in a frame of reference moving withthe same velocity as the wave i.e. v to the right, then the velocity of thesmall segment relative to us is v to the left.

The acceleration of the segment is the trans-verse acceleration seen earlier, directeddown.

We can draw the free body diagram of theforces acing on the segment as seen at right,where T1 and T2 are tensions, and m is themass of the segment.

amg

v

T2T2θθ

Assuming that the weight of the segment is negligible compared to thetensions, then T1 = T2 = T .

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Applying Newton’s Second Law:∑down

F = ma = 2T sin θ

The acceleration of the segment is perpendic-ular to the velocity, so the motion can be con-sidered to be circular, with a connected to thespeed v by a = v2

R for an unknown radius R.

2T sin θ = ma = mv2

R

�2T sin θ = (µ�2�Rθ)v2

�R(m = µ× segment length)

�θT = µ�θv2 (sin θ ≈ θ for small angles)

⇒ v =

√Tµ

a

v

R

TTθθ

θθ

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Any wave of any frequency travelling on this string must have thisspeed.

Other waves also have this property, for instance sound waves of allfrequencies will travel at the same speed in air. This speed is determinedby the physical properties of the air - pressure, density etc.

For some waves, the speed in a material will depend on the frequency.Probably the best-known example of this light travelling through glass orwater. The differing speed of different wavelengths (i.e. colors) causes arainbow formed by a prism, or by moisture in the air.

A material like this, where speed depends on frequency of the wave isknown as a dispersive medium and won’t be discussed any further inthis course.

Waves on String: Power Transmitted Contents

The particles of the string do not move in the direction of the wave, butenergy is transmitted along the string by the wave. The rate at whichthis occurs is the power carried by the wave.

Let’s look again at a small segment of the string we’ve just dealt with.As it moves up and down, the energy of this small segment of the stringis converted between potential and kinetic energies (just like a mass on aspring). When the displacement is zero, then the energy is all kineticenergy, where the speed is the maximum transverse speed vt = ωA.

v

vt

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If the length of our small segment is taken to be ∆x then its mass will beµ∆x and so its kinetic energy is:

KE = 12mv2

t = 12 (µ∆x)(ωA)2

If the time taken for this energy to be transfered is ∆t, then the powerP carried by the wave is:

P = ∆E∆t = 1

2µ(

∆x∆t

)ω2A2 = 1

2µvω2A2

This particular formula only applies to waves on a string, but similarformulas can be found for other waves, all with the common factor that:

P ∝ ω2A2

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A string with linear mass density 5.0 × 10−2 kg/m is stretchedwith a tension of 80 N. A harmonic wave with amplitude 6 cm andfrequency 60 Hz travels along the string. Find:(a) the speed of the wave(b) the power transferred by the wave down the string.

(a) v =√Tµ =

√80

5.0×10−2 = 40 m/s

(b) P = 12µvω2A2

= 12 (5.0× 10−2)(40)(2π × 60)2(6× 10−2)2

= 512 W

Waves on a String Summary Contents

As well as the definitions common to all waves for wavelength,frequency etc., there are some extra equations that apply only fora wave on a string:

I The speed v of the wave is determined by the physicalproperties of the string:

v =

√Tµ

where T is the tension in the string, and µ is the linearmass density or mass per length of the string.

I The power P, carried by the wave is:

P = 12µvω2A2