1
Traveling Waves
Having discussed simple harmonic motion, we have learned many of the concepts
associated with waves. Particles which participate in waves often undergo simple
harmonic motion.
Section 20.1, 2 Wave motion
Examples of waves: Ocean waves, waves in crops, earthquakes, speech is a wave in the
air, waves in ropes.. Electromagnetic waves travel in a vacuum and involve electric and
magnetic fields. Covered in P102. We will largely be talking about mechanical waves in
Physics 101 but there will be some discussion of light waves later.
Classic example of a wave is the wake following a boat
Waves have oscillatory behaviour and operate in both space and time. i.e. at any instant
in time, the wave has a characteristic shape and at any point in space, the wave has a time
dependence. e.g. the bobbing we experience when we are moved up and down by the
wake of as boat.
A wave is a disturbance which propagates through a system. Although individual
masses oscillate about their equilibrium positions, they do not move with the wave.
A wave transports energy not matter.
Wave variables are
1) Wavelength, λ
2) Period, T
3) Frequency, f=1/T
4) Amplitude, A
5) Velocity, v
λ
Arrangement in space
at a fixed time.
v
y
x
T
Time course at a
fixed point in
space.
y
t
y
A single wave
pulse. Boat
wake.
x or t
Wave velocity: The wave travels a distance λ in time T; then v = λ/T = λf
These are called traveling waves.
2
Transverse and Longitudinal waves
A travelling wave that causes the particles of the disturbed medium to move
perpendicular to the wave motion is called a transverse wave.
e.g. waves on a rope
A travelling wave that causes the particles of the medium to move parallel to the
direction of wave motion is called a longitudinal wave.
e.g waves in a spring, sound
Sound waves involve longitudinal movement of air particles which results in
compression and expansion and density changes.
Ocean waves are a combination of transverse and longitudinal, each particle undergoes
circular motion. There is no net displacement of molecules in an ocean wave or a sound
wave. Note: Ocean waves are surface waves.
Earthquakes have two types of waves:
1) P (primary or pressure) longitudinal waves which travel through the earth at about 7-8
km/s and
2) S (secondary or shear) transverse waves which travel along the earth's surface at 4 -5
km/s.
Transverse waves are not conducted by a fluid since a fluid cannot transmit shear forces.
This is why secondary earthquake waves are not transmitted through the earth's core.
It is possible to estimate the distance to the earthquake by recording the times of arrival
of the two types of wave.
Note that waves can be one dimensional, two dimensional or three dimensional.
Examples of
1D waves: waves on a rope, air moving along a tube
2D waves: surface waves like ocean waves, crop waves
3D waves: earthquakes, sound in the lecture theatre, light
3
Now consider a general wave or wave pulse. e. g. a one dimensional travelling wave (on
a rope),
We assume that the shape of the wave doesn't change as it moves down the rope.
y
A
v
t=0
v
t=t
P
x
vt
P
y
The -vt part of the function ensures
that the wave moves to the right with
velocity v. If you consider a point P on
the rope, it is distance x from the wave
crest at time 0 and a distance x-vt at
time t.
x
The wave function, D, represents the y coordinate (x –coordinate if we were talking about
a longitudinal wave) of any point of the medium at time t.
The wave has a velocity v.
At t= 0, D= D(x),
At t= t, D = D(x-vt) when the wave is travelling to the right, assuming that v is positive.
Example wave function: y = 2
(x-5t)2 + 2
Show plots for t=0,2,5.
What do we write if the wave pulse were traveling to the left?
What determines the velocity of a wave?
For example, suppose I send a wave pulse down a rope. What determines how fast the wave
travels to the end of the rope?
What determines how long it takes my voice to reach the back of the lecture theatre?
Is it affected by how fast I jerk the rope? Or how loud I shout?
Will a wave pulse move faster along a thin rope than along a thick rope?
4
Velocity of transverse waves
Consider a transverse travelling wave moving with velocity v in a string under tension Ts
and with mass per unit length of µ (kg/m). Consider a small arc of the traveling wave
from a reference frame which moves with the wave at velocity v to the right. The
acceleration of the string in this moving frame is v2/R (centripetal acceleration).
θ
θ
Ts
Tr
R
In this frame, the segment of string is moving to
the left with velocity v and following a curved
arc. The arc represents an angle change of 2θ
relative to the centre of curvature of the path. The
horizontal components of the tensions cancel.
The net force is down in the radial direction,
equivalent to circular motion about the centre of
the arc.
Ts
θ θ
Note that for a travelling wave the medium oscillates about a
fixed point while the wave moves at velocity v to the right.
ΣTr = 2Tssinθ, use sinθ = θ for a small section of arc.
Then set Tr = ma: 2Tsθ = mv2/R
Mass of rope segment m = ∆sµ = 2Rθµ,
Make a right angle triangle with the hypotenuse from the
centre of the axis. Then it is straightforward to work out
that the lower angle is θ.
Then 2Tsθ = 2Rθµv2/R
Large µ – thick rope; small µ – thin rope
v = √Ts/µ
The velocity of the wave increases with increasing string tension and decreases with
increasing string mass per unit length.
Velocity of longitudinal waves in a fluid (e.g sound in a liquid or gas)
Consider a wave pulse traveling in a fluid in a long tube of cross-sectional area A, with a
piston in one end. At time t=0, the piston starts to move to the right at a speed v'. In time
t, it has moved a distance v't and the longitudinal wave it produces will move at the speed
of sound, v, a distance vt.
Po, ρ
vt
v't
A
Tube area A
P0 + ∆P
v'
Sound wave front
Note that the speed of the wave,
as before, is a property of the
medium and not determined by
the mechanism which is
producing the wave.
v
We assume v’ < v;e.g. v’ < the speed of
sound
5
Net force on the compressed region of the fluid:
Fnet = (Po + ∆P)A - PoA = A∆P
The impulse given to the compressed fluid is
Fnett = ∆mv' (recall that impulse = change in momentum) Here, (ρAvt) is the mass of
fluid that was given the velocity v' when the piston moved.
A∆Pt = (ρAvt)v' => ∆P = ρvv'
The Bulk modulus of the fluid is defined as
The bulk modulus is the ratio of the change in pressure
to the fractional volume change. Note that an increase
in pressure leads to a decrease in volume. See Chapter
15 (15-6) for more details.
B = - ∆P/(∆V/Vo) = -ρvv'/(∆V/Vo)
The assignment of Vo and ∆V is
obvious from the figure.
Where Vo is the original volume, Avt, and ∆V = -Av't
So, B = -ρvv'/(-Av’t/Avt) = ρv2
And v = √B/ρ
2.00 m
Problem
θ L/4
A
L/4
T1
B
L/2
Ts
Mg
At point A:
Light string of mass 10.0g and length L = 3.00m.
Walls separated by 2.00 m
Each object has mass 2.00 Kg
How long does it take a wave to travel from A to B?
µ = m/L =.01Kg/3.00m = 0.00333 Kg/m
Sin θ =[(2.00m – L/2)/2]/L/4 = [(2m-3/2m)/2]/[3m/4] = 0.25m/0.75m θ = 19.4o
Cosθ = 0.943
∑Fy = T1cosθ-mg =0 T1 = mg/cosθ = (2.00kg)(9.81kg)/(0.943) = 20.81 N
∑Fx = T1sinθ -Ts =0 then Ts = T1sinθ = 20.81N/3 =6.935N
6
Then the speed of travel of a travelling wave in this rope is
v = √Ts/µ =√ 6.935N/0.00330Kg/m = 45.64m/s.
Then the time required for a pulse to travel from A to B is t = (3m/2)/45.64m/s = 32.9ms
Section 20.3 Sinusoidal waves
λ
Consider these two
representations of the
same travelling wave.
We want an equation
which will take into
account both the spatial
and temporal aspects of
the wave motion.
v
D
T
x
D
We use the letter D for displacement in order to have the treatment general for
both transverse and longitudinal waves.
t
Note that the sinusoidal wave repeats itself (i.e. covers 2π radians) every wavelength
spatially and every period temporally.
Then D(x,t) = DMsin [(2π/λ) x - (2π/T) t + φ ] where φ is the phase at t=0
Recall that 2π/T = ω and we now define 2π/λ = k the wave number.
Then D(x,t) = DMsin(kx - ωt + φ)
Note that here, k = the wave number, not the
spring constant, k, from Hooke’s Law.
Note that v = λ/T = λf = (2π/k)(ω/2π) = ω/k
The transverse speed of a particle in this traveling wave is given by:
vy = dD(x,t) = -ωDMcos(kx - ωt + φ)
dt
Note: vy is not the propagation velocity, v.
7
Problem from December 2004 Final
v
8
B
Zero crossings at 4, 9, 14, 19, 24, 29
D(cm)
0.05
0.10
A
0.15
0.20
0.25
x (m)
-4
-8
The plot above represents the displacement of a wave at t = 0.10 s. This wave has a
frequency of 5.0 Hz and is traveling in the negative x direction along a string
(a)
(b)
(c)
(d)
What is the speed of the wave?
Write an equation describing the wave including all constants
What is the transverse speed at point A? at point B?
Sketch the displacement at t=0
sin(π-φ) = sinφ, hence we consider π/6
and 5π/6
(a) From inspection of the plot. λ = 0.10m, then v = fλ = 5.0 Hz × 0.10m = 0.50 m/s
(b) Amplitude = 8.0 cm, k = 2π/λ = 20π m, ω = 2πf = 10π rad/s-1 At x=0, and t=0.1s,
we have -4.0cm =8.0cm sin(ωt + φ); ωt =π. Note sin (π + φ) = -sinφ, φ = sin-1 (½);
wave moving to the left, so φ = π/6 or 5π/6; Consider how the plot evolves from
x=0 at t=0.1s, it is at phase π+π/6 and going down, If we used 5π/6, the curve
would go up, then φ= π/6. Then x = 8cm sin(20πx + 10πt + π/6)
(c) at A, the transverse speed is ωA = (10π s-1)(8cm) = 80π cm/s; at B v = 0;
(d) Move the plot to the right a distance v(0.1s) = 0.05m. Now the plot starts at
displacement 4 cm and increases.
Consider a wave traveling on a rope which is made up of two sections with different
diameters. Then the speed of the wave is v1 in the left side and v2 on the right side. v1/v2
= (µ2/µ1)½ using vi = √Ts/µi.
Ts
µ1
v1
µ2
v2
Ts
t=0
t=T
λ1= v1T
λ 2 = v2 T
On the right hand side, the wave crests move at velocity v1 and wave crests travel one
wavelength in time v1T = λ1.
Actually this is a
bit more
complicated
because some of
the wave will be
reflected back at
the boundary.
8
Imagine that a wave crest reaches the junction at t=0. The crest will undergo a
discontinuous change in velocity at the junction from v1 to v2. A time T later, the junction
will undergo the next wave crest. In this time, the previous wave crest will have moved a
distance v2T = λ2. Then λ1 /λ2 = v1/v2 and the period (and frequency) is unchanged.
The period cannot change at the junction, since if it did, adjacent particles in the rope
would get out of phase with each other and the rope would break!
Problem A cord with two sections
Ts
µ1 = 0.10 kg/m
λ1
µ2 = 0.20 kg/m
Ts
λ2
Incident wave function: D = (0.050m) sin(6.0x - 12.0t)
Find (a) λ1, (b) Ts and (c) λ2
The wave function on the
other side will be different
From the wave function, D = DMsin(kx - ωt) where k=2π/λ and ω = 2π/T
(a) Then λ1 = 2π/k1 = 2π/6.0 m-1= 1.047m, k1 = 6.00 m-1
(b) v1 = ω1/k1 = √Ts/µ1 => Ts = µ1[ω1/k1]2 = (0.10 kg/m)[12.0 s-1/6.0 m-1]2 = 0.40N
(c ) Since the tensions are equal, and the period (or ω) is equal on each side, what
changes are v and λ. v2 = √Ts/µ2 = ω2/k2 = ω2λ2/2π => λ2 = 2π√(Ts/µ2)/ω2
λ2 =2π√(0.40N/0.20kg/m)/12.0 s-1 = 0.741 m
9
Section 20.5 Sound and light (Read Giancoli Chapter 16 sections 1,2,3 on sound)
Sound (light later)
Creation of a sound pulse by moving a piston in a gas chamber
High density region
v
We already derived a relation for the velocity of sound.
Regardless of how we set up a sound wave, its propagation velocity depends only upon
the medium.
For a liquid or gas or a solid volume, use v = √B/ρ where B is the bulk modulus of the
medium and ρ is the density of the medium.
See table 20-1, page 629, for velocities of sound for various materials. vs varies from
343m/s on air to 1480m/s in fresh water to > 6000 m/s in rock and aluminum.
Sounds have pitch (frequency) and loudness (intensity).
Three categories of sound waves (based on pitch):
1) Infrasonic - low frequency (< 20Hz)
2) Audible - 20 to 20,000 Hz
3) Ultrasonic - high frequency (> 20,000 Hz)
Infrasonic waves are created by earthquakes, thunder,
and vibrating heavy machinery.
Ultrasonic waves have several important applications.
E.g ultrasonic imaging and auto focus for cameras.
10
Mathematical representation of longitudinal waves
.
.
.
.
.
. . . .
.
.
.
.
.
.
.
. . . .
.
.
.
.
.
.
.
.
.
.
Note that the displacements change direction at the points of maximum and minimum displacement
The higher pressure regions are called compressions and the lower pressure regions are
called expansions. Both regions move with the speed of sound. The distance between two
successive compressions is called the wavelength λ.
This is just like a slinky; air pressure
is scaled by the brightness of blue.
Displacement D
along the wave:
Note that this displacement is along
the direction of motion of the wave.
There are displacement minima (i.e.
zero) when the pressure is minimum
or maximum.
x
Note that the pressure change is 90o
out of phase with the displacement.
Pressure difference
∆P along the wave
D = DMsin(kx - ωt + φ)
∆P = -∆PMcos(kx - ωt + φ)
As always, sin (θ + ½π) = cos(θ).
Derivation of the equation for ∆P:
Consider a volume A∆x of gas of thickness ∆x and cross-sectional area A
A
x
For each position along the tube, there is a gas displacement D
from the equilibrium position. ∆D is the change in this
displacement, across the distance ∆x.
∆D
∆x
The change in volume V (=A∆x) accompanying the pressure change is ∆V (= A∆D)
where ∆D is the difference between D at x and at x + ∆x.
Then ∆P = -B∆V/V = -BA∆D/A∆x = -B∂D
∂x
This is the definition for bulk modulus: See Section 15-6
Partial derivatives are used here we are considering
the system at a fixed time.
11
using D = DMsin(kx - ωt), we get ∆P = -BkDMcos(kx - ωt) = -∆PMcos(kx-ωt)
where ∆PM is the pressure amplitude.
Using B = ρv2 and k = v/ω, ∆PM = ρv2(ω/v)DM = ρvωDM
Section 20.6 Power and Intensity
Waves and energy are two of the key concepts in
Physics.
v
Area A, width l = vt
We draw a small volume of the medium of
area A and width vt, All the molecules in this
volume participate in the wave motion.
Consider a sinusoidal transverse (or longitudinal) wave moving at velocity v through an
elastic medium. Each particle in the medium moves with simple harmonic motion and
each particle has an energy E = ½kDm2 where Dm is the maximum amplitude of the
motion for that particle. Recall that ω2 = k/m or k =ω2m. Then E = ½ω2mDm2
For 3D waves, m = ρV where V = Al =Avt, so we can write
E = ½ω2ρAvtDm2
This simple equation shows us that the energy transported by a wave is proportional
to the square of the amplitude, the wave velocity and to the square of the frequency.
If we divide the Energy by time, we get the average power (or rate of energy transfer)
P = E/t = ½ω2ρAvDm2
Energy is like money, power is like a salary
and intensity across unit area perpendicular to the flow I = P/A = ½ω2ρvDm2
For a 3D wave in an isotropic medium (e.g. sound), known as a spherical wave, I = P/A
=P/4πr2
So I α 1/r2 the 'so called' inverse square law.
For two points as distances r1 and r2 from the source, we have I1/I2 = r22/r12
12
Note that if the Intensity decreases as 1/r2 then the amplitude, Dm must decrease as 1/r for
a spherical wave.
Intensity of Sound: Decibels (Not in Knight)
The expressions we derived above for transverse waves hold equally well for longitudinal
waves. The only difference is the direction of the displacement.
From before: I = P/A = ½ρv(ωDM)2 or ∆PM2/2ρv
The faintest sounds the ear can hear have I = 1.00x 10-12 watts/m2
Using the above equations and ρair = 1.29 kg/m3, v = 343 m/s and ω = 1.00 kHz, we find
∆PM = (2ρvI)½ = [2(1.29 kg/m3)(343m/s)(1.00x10-12watts/m2)]½ =2.97x10-5 N/m2
compared to Po of 1.013x105 N/m2
DM = ∆PM/(ρvω) = 2.97x10-5N/m2)/[(1.29kg/m3)(343m/s)(2π(1.00x103)] = 1.07 x10-11 m
= about 1/10 the size of a hydrogen atom.
The loudest sounds the ear can process have an intensity of 1.0 watts/m2. This
corresponds to a pressure ∆PM of 29.7 N.m2 and DM of 1.07x10-5m or 11 µ.
The ear is a sensitive device. The same is true of our noses and our eyes! We hear a range
of displacements spanning 106 and intensities spanning 1012!
Sound level in decibels
The ear can hear sounds with intensities as low as 1x10-12 watts/m2 and as high as about
1.0 watts/m2. Also, the ear hears logarithmically, i.e. when the sound intensity is
increased by a factor of 10, we perceive a factor of two in loudness.
Because this range is so large, a logarithmic scale has been devised. The sound level, β, is
defined by:
β = 10 log(I/Io) then I = Io10β/10
where Io is the reference intensity, taken to be 1.00x10-12 watts/m2, i.e. the threshold of
hearing. The units are decibels or dB
The threshold of hearing Io corresponds to 0 dB and the threshold of pain, 1.00 watts/m2
corresponds to 120 dB. Look at Table 16.2 to compare sound levels from different
sources. Rock concerts are too loud (120 dB) and cause ear damage.
Whisper 20 dB, Conversation 65dB, car interior at 90km/hr 75dB,
13
Problem: A loudspeaker is placed between two observers who are 110m apart along the
line connecting them. If one observer records a sound level of 60.0 dB and the other a
sound level of 80.0 dB, how far is the speaker from each observer?
I1
I2
r2
r1
Using β = 10 log(I/Io), or β/10 = log(I/Io) and I/Io = 10β/10
I1 = Io10(60.0/10) = (1.00x10-12 watts/m2)x 106 = 1.00x10-6 watts/m2
I2 = (1.00x10-12 watts/m2)x 108 = 1.00 x 10-4 watts/m2
I1/I2= r22/r12 then r1/r2 = (I2/I1)½ = [(1.00x10-4)/(1.00x10-6)]½ = 10
Then r1+ r2 = 110m and r1/r2 = 10 so r1 = 100m and r2 = 10m.
Section 20.7 The Doppler Effect
Consider an observer O and a sound source S. The air is stationary and the observer
moves with velocity, vo. The source velocity is vS. The sound has frequency fo,
wavelength λ and velocity v.
O
vO
S
λ
The observer registers sound
frequency by the number of wave
fronts which they detect per unit
time.
vS
Wave fronts move from source to observer with velocity v, the speed of sound relative to
the air which is stationary.
When vO and vS are both zero, the observer counts fo wave fronts per second. This
corresponds to the frequency of sound that the observer hears.
If vO = 0 and vS ≠ 0, the wavefronts are no longer separated by λ. When vS > 0, i.e. the
source moves towards the observer, λ' = λ - ∆λ = λ - vST = λ - vS/fo The sound velocity is unchanged
Then the frequency heard by the stationary observer is f' = v/λ' = v / (λ - vS/fo) =
v/ [ v/fo - vS/fo ]. Then f’ =
fo
= fo v
The perceived frequency increases if the
(v – vS)
(1 - vS/v)
source is moving toward the observer
14
If the source moves away from the observer, the wavelength is increased to λ + vST
giving f’ = fo
= fo v
The perceived frequency decreases if the source
(1 + vS/v) (v + vS)
is moving away from the observer.
When vS = 0 and vO is not zero, the relative velocity of wave fronts to observer is v' = vO
+ v but the wavelength λ is unchanged so the frequency, f', increases:
f' = v'/λ = (v + vO)/λ since λ = v/fo, we get
f' = (1 + vO/v)fo = fo (v + vO)/v
The perceived frequency increases if the
observer is moving toward the source
If the observer is moving away from the source, the relative speed of the sound waves
and the observer is v' = v - vO and f' = (1 - vO/v)fo
The perceived frequency decreases if the
= fo (v – vO)/v
observer is moving away from the source.
To summarise all the above derivations, we can write
Toward the sound means an increase in frequency and away from the
sound means a decrease in frequency.
f' = (v ± vO) fo
(v Ŧ vS)
where the upper signs correspond to the source and observer moving towards each other
and the lower signs correspond to motion away from each other.
Problem: A bat moving at 5.0 m/s is chasing a flying insect. If the bat emits a 40.0 kHz
chirp and receives back an echo at 40.4 kHz, at what relative speed is the bat moving
toward or away from the insect?
vB
vI
We shall assume that the insect moves away from the bat with speed vI. Then frequency
heard by the insect is
fI =
(v - vI)fo
(v - vB)
The insect is listening. The insect is moving away from the
sound, the bat is moving towards the sound.
The frequency heard back by the bat again after the sound reflects from the insect is
fB =
(v + vB)fI
(v + vI)
Now the bat is listening.
Then fB = (v - vI)(v + vB)fo
(v + vI)(v - vB)
Solve to get 40.4 = 40.0 (343 + 5.00)(343 - vI)
15
(343 - 5.00) (343 + vI)
(343 + vI) = (40.0/40.4)(348/338)(343- vI) = 1.0194(343 - vI)
vI ( 1 + 1.0194) = 343(1.0194 -1); vI = 6.651/2.0194 = 3.29 m/s
Then the bat is closing in on the insect at 5.00m/s - 3.29 m/s = 1.71 m/s
Echolocation in Bats
Bats make use of sound waves to locate insects which they eat. Bat species make use of
two approaches.
1) They emit short sound pulses of duration a one to two ms. These pulses are both
amplitude and frequency modulated. The frequency changes from 70kHz to 30 kHz
during the pulse. When searching for insects, bats emit these pulses about 4 times per
second. Once they have found a reply, the rate of sending pulses can increase to as high
as 200 per second. They maximise the reply until they can scoop up the insect.
Considerations: The frequencies of 70kHz to 30kHz correspond to λ = vs/f = 5mm to
11mm; the approximate size of the insect.
Given that the pulses are sent out 4 times /s, at what range must an insect be located in
order that its reflection of the sound gets back to the bat before the next pulse? There are
0.250s between pulses. A sound wave will travel x = vt = (343m/s)(0.250s) = 86 m. This
wave travels to the insect and back so the distance to the insect is 86m/2 = 43m.
Another factor is the amount of power reflected back from the insect compared to the
threshold of hearing on the bat. If the bat produces P = 10-5 Watts, and its hearing
threshold is 1x10-12 watts/m2 (human); what is the maximum distance from which it can
detect an echo. Assume the insect has cross sectional area 1x10-4m2 (1cmx1cm).
The P of 10-5 watts is spread over all directions in a spherical wave. The fraction which
strikes the insect is Pi = P(ainsect/4πr2) where r is the distance between the bat and the
insect. The intensity of reflected sound from the insect follows Ii = Pi./4πr2. Then the
intensity returning to the bat is IB = Painsect/(4πr2)2. If the threshold intensity is 1x1012
watts/m2;
Then r4 = ainsectP/(IB(4π)2) = (1x10-4m2)(1x10-5watts)/[(1x10-12watts/m2)(4π)2] = 6.33m4
And r = 1.6m a more realistic minimum distance from which a bat can detect an echo
from its meal. (Some bats can direct their sound waves in a narrow beam so that more of
the power gets to the insect.)
2) The second type of echolocation makes use of the Doppler effect. This species of bat
uses a long pulse of about 0.1s long and frequency 60.0 kHz and takes advantage of the
Doppler effect to detect the reflection at 61.8 kHz from the insect. By maximising the
16
intensity of the Doppler reflected sound which he can hear while sending out his pulse,
the bat can move to the location of his prey.