Announcements
• First Quiz: Feb 10, Friday, 2:30-3:20pm
• Location:
– Anderson 210 (last names A-L).
– Anderson 250 (last names M-Z).
• Free to use:
– Non-programmable calculator
– 1 sheet (2 pages) of notes
• On several Fridays, we will have lectures in 175 Willey Hall, which is
easier to use for demonstrations. Specifically, we will meet in 175
Willey on 02/17, 03/24, 04/07, and 04/28.
Waves
• In the previous courses, studied oscillations in an isolated system.
– Example: pendulum.
• Waves are oscillations that extend in space and time.
• Many examples:
– Waves on a string or wire
– Waves on springs
– Waves on water (gravity waves)
– Waves through air (sound waves)
– Electromagnetic waves (optical, microwaves, etc)
– Waves through solids
– Gravitational waves (general relativity, stretching distances
between free masses)
Classification
• Depending on the type of motion:
– Transverse: induce motion
perpendicular to the wave propagation.
– Longitudinal: induce motion in
the direction of wave propagation.
• Wave speed may depend on the medium:
– Non-dispersive: all waves have the same speed.
– Dispersive: speed depends on the frequency of the wave.
Demos
• Demo: 3B10.10 Pulse on a Rope
• Demo: 3B10.20 Slinky on Table
• Demo: 3B10.30 Bell Labs Wave Model
Pulse in a Non-dispersive Medium
• The pulse shape at some initial time
t=0 is described by some function of
the position y = f(x).
• If the pulse propagates at speed v,
after some time t the pulse will have
the same shape, but in a reference
frame x’ = x – vt.
• So, the shape can then be described
as y(t) = f(x – vt).
• Similarly, if the wave is propagating in
the –x direction,
y(t) = f(x + vt).
Wave Speed
• Determined by the tension
forces and density of the
medium.
Speed of Sound Waves
• Recall: Bulk modulus:
• Recall: speed of sound:
∆P
B=−
∆V / V
B
v=
ρ
Wave Equation
Wave Equation
• General wave equation:
2
∂ 2ψ
∂
ψ
2
=v
2
∂t
∂x 2
• Solution ψ(x) is the wavefunction.
• General solutions are superpositions of right-going and left-going
waves.
ψ ( x, t ) = Aψ R ( x − vt ) + Bψ L ( x + vt )
• The speed is determined by the restoring force and the density.
Harmonic Waves
• One solution is sinusoidal:
• Insert it into the wave
equation:
• Wave speed:
ψ ( x, t ) = A sin (kx − ωt )
∂ 2ψ
2
=
−
ω
A
sin (kx − ωt )
2
∂t
2
∂ψ
2
= − Ak sin (kx − ωt )
2
∂x
∂ 2ψ ω 2 ∂ 2ψ
= 2
2
2
∂t
k ∂x
v=
ω
k
Harmonic Waves
• Meaning of wave-number k:
– If x increases by one
wavelength, displacement
should return to the original
value.
ψ ( x, t ) = A sin kx − ωt
• Denote wavelength with λ.
 2πx 2πt 
– k λ = 2π  k = 2π / λ.
−
ψ ( x, t ) = A sin

T 
• Meaning of angular frequency ω:
 λ
– If time increases by one period T, the displacement should
return to the original value.
– ωT = 2π  ω = 2π / T = 2πf
– f = 1/T is known as the frequency of the wave.
(
v=
ω
k
=
λ
T
= λf
)
Energy Transfer by
Waves
Waves in 2D or 3D
• Waves on the surface of the water
(2D).
• Waves through the air (3D).
• Described by wavefronts propagating
in 2D or 3D.
• For a point source, the wavefronts are
circles or spheres.
– If far away, could approximate them
as plane waves.
• A planar source would also produce a
plane wave.
• Demo: 3B10.63 Ripple Tank
Example: Sound Waves
• Longitudinal, so displacement is in the direction of wave
propagation. Can denote it by s(x,t): s ( x , t ) = s sin
0
∆Eavg =
• In 1D, the average energy would be:
• In 3D, µ∆x  ρ∆V:
• Define energy density:
(kx − ωt )
1
µω 2 s02 ∆x
2
1
ρω 2 s02 ∆V
2
∆Eavg
1
= ηavg = ρω 2 s02
∆V
2
∆Eavg =
• Define intensity as power per unit area:
∆Eavg
P ∆Eavg ∆Eavg
1
I= =
v=
v = ηavg v = ρω 2 s02v
=
A
A∆t
Av∆t
∆V
2
Intensity
β = 10 log
I
I0
• For sound waves often describe intensity in decibels:
– The reference intensity usually chosen to be the barely audible
one: I0=10-12 W/m2.
Reflection
• Suppose an incident wave comes in from the
left and is reflected. The total wave is then:
ψ = ψ 0 [cos(kx − ωt ) + r cos(kx + ωt )]
• Require continuity at the reflection point (say
x=0):
ψ = ψ 0 [1 + r ]cos(ωt ) = 0
r = −1
• The wave flips sign when it reflects.
Reflection
• If the end of the string is free, then the
slope of the string is zero at the end.
∂ψ
= −kψ 0 [sin( kx − ωt ) + r sin( kx + ωt )]
∂x
∂ψ
( x = 0) = −kψ 0 [− 1 + r ]sin(ωt )
∂x
r = +1
• The reflected wave is of the same sign as
the incident wave.
Reflection and Transmission
• In general, at the boundary between two media, both reflection and
transmission are possible.
• Have to match both the wavefunction and the derivative of the
wavefunction at the boundary.
– These two conditions specify the reflection and transmission
coefficients, r and t.
ψ inc = ψ 0 cos kx − ωt
• Incident wave:
• Reflected wave:
• Transmitted wave:
v2 − v1
r=
v2 + v1
(
)
ψ r = rψ 0 cos(kx + ωt )
ψ t = tψ 0 cos(kx − ωt )
2v2
t=
v2 + v1
v1 and v2 are wave
speeds in the two media.
Reflection and Transmission
• t > 0: transmitted wave is always of the same sign as the incident
wave.
• r > 0 only if v2 > v1 , i.e. if the wave speed is larger in the second
medium than in the first one.
• Recall that energy (or power or intensity) is proportional to the
square of the amplitude.
– Energy in the incident wave ~ v1 ψ02
– Energy in the reflected wave ~ v1 r2 ψ02
– Energy in the transmitted wave ~ v2 t2 ψ02
– Conservation of energy:
v1
r +t
=1
v2
2
2
Reflection and Transmission
• In 3D, boundary between two media is a surface.
• Waves change direction (and speed) as it moves from one medium
to another: refraction.
• Incident wave is reflected and refracted (transmitted).
• Total internal reflection is possible too!
Doppler Shift
• If the source and the receiver of a wave are moving towards each
other, the wave peaks will pass by the receiver more frequently than
of they are stationary relative to each other.
• If they are stationary, the time between two peaks (as measured by
receiver) is:
distance between peaks λ
time =
wave speed
=
v
• If receiver moves at speed u toward the source, then:
distance between peaks
λ
time =
=
wave speed + receiver speed v + u
Similar applies if the source moves
toward the receiver.
Doppler Shift
• If the source and the receiver move toward each other, the receiver
observes higher wave frequency then emitted by the source.
• In general:
v ± ur
fs
fr =
v ± us
• Demo: 3B40.60 Ripple Tank - Doppler Effect
• Demo: 3B40.10 Doppler Buzzer
Shock Waves
• If the source moves faster than the wave speed, there will be no
waves in front of the source.
– The waves pile up behind the source.
– This is known as the shock wave.
– Examples: supersonic airplanes, charged particle emitting
radiation in a medium…
• Mach angle:
vt v
sin θ =
ut
• Demo: 3B45.15
Ripple Tank - Shock Wave
=
u
Superposition of Waves
• Principle of wave superposition:
– When two or more waves overlap,
the resultant wave is the algebraic
sum of individual waves.
• True for any waves, including
transient pulses and permanent
waves.
• For example, if we have two pulses
on the string, the total wave function
is the sum of the individual wave
functions.
Superposition of Waves
• The origin for this is the wave
equation.
• Suppose y1 and y2 are two
solutions to the wave equation.
• Then the sum, y3 = y1 + y2, is also
a solution of the wave equation.
• Similar proof holds for any linear
combination y = Ay1 + By2.
d 2 y1, 2
2
1 d y1, 2
= 2
2
dx
v dt 2
d 2 y3 d 2 y1 d 2 y2
=
+
2
2
dx
dx
dx 2
1 d 2 y1 1 d 2 y2
= 2
+ 2
2
v dt
v dt 2
1  d 2 y1 d 2 y2 
= 2  2 + 2 
v  dt
dt 
1 d 2 y3
= 2
v dt 2
Question
• Suppose the two pulses on the
string are identical, except one
is inverted with respect to the
other. What will be the end
outcome of the superposition?
1) The two waves bounce off of
each other.
2) The two waves cancel when
they hit each other, after which
there is no wave on the string.
3) The two waves cancel when
they hit each other, but then a
moment later reappear on
opposite sides.
A
B
A
B
A+B=0 at all times after “collision”
A
B
Answer
• Indeed, the two pulses will pass
through each other unaffected!
• How is this possible? That is, if
at some time there is no pulse
on the string, how can the
pulses reappear?
A
B
A+B=0
A
B
Answer
• Indeed, the two pulses will pass
through each other unaffected!
• How is this possible? That is, if
at some time there is no pulse
on the string, how can the
pulses reappear?
• Answer: while the displacement
at each point on the string may
be zero, the velocity is not! That
is, various points on the string
are in motion as they pass
through the zero displacement.
A
B
A+B=0
A
B
Interference of Harmonic Waves
• Suppose we have two harmonic (sinusoidal) waves, of same
frequency, wavelength, and amplitude, but not phase:
y1 = A sin( kx − ωt )
y2 = A sin( kx − ωt + δ )
• Their sum is given by:
y1 + y2 = A sin(kx − ωt ) + A sin(kx − ωt + δ )
• Recall:
sin θ1 + sin θ 2 = 2 cos
δ
θ1 − θ 2
2
sin
θ1 + θ 2
2
δ
y1 + y2 = 2 A cos sin( kx − ωt + )
2
2
• The resulting wave has the same wavelength and frequency.
• Note, however, that the phase offset between the two waves affects
both the amplitude and the phase of the resulting (summed) wave.
• If δ=0 (i.e. the two waves are
not offset in phase, they are
identical), the two waves
simply add.
– Constructive interference.
Amplitude
Interference of Harmonic Waves
y1 = A sin( kx − ωt )
y2 = A sin( kx − ωt + 0)
0
0
y1 + y2 = 2 A cos sin( kx − ωt + )
2
2
= 2 A sin( kx − ωt )
• If δ=π (i.e. the two waves are
exactly out of phase), the two
waves simply subtract,
resulting in zero displacement
everywhere and at all times.
– Destructive interference.
Amplitude
Interference of Harmonic Waves
y1 = A sin( kx − ωt )
y2 = A sin( kx − ωt + δ )
y1 + y2 = 2 A cos
π
π
sin( kx − ωt + )
2
2
π
= 2 A × 0 × sin( kx − ωt + ) = 0
2
Beats
• What if the two waves have different frequencies (and wavelengths)?
y1 + y2 = A sin( k1 x − ω1t ) + A sin( k 2 x − ω2t )
ω1
= 2 A cos (k1 − k 2 ) x −2(ω1 −ω2 )t sin (k1 + k 2 ) x −2(ω1 +ω2 )t
k1
=
ω2
k2
• At a given point in space, the k-part only contributes a phase that is
constant in time. If we ignore it:
y1 + y2 = 2 A cos
(ω1 −ω2 )t
2
sin
(ω1 +ω2 )t
2
=v
 ∆ω 
t  sin(ωavg t )
= 2 A cos
 2 
• That is, we get a product of two sinusoidal functions!
Beats
•
•
•
One of the sinusoidal functions has
frequency half-way between the
original two.
But the amplitude is modulated at a
frequency equal to the half of the
difference of the original two.
– Some define the beat frequency
to be Δf/2.
– However, often one observes
intensity and not amplitude.
– Intensity is proportional to the
square of the amplitude, so it
oscillates twice as fast as the
amplitude.
– Beat frequency = Δf!
Physical origin of the beats:
sometimes the two original waves
add constructively and sometimes
destructively.
Destructive
superposition
Constructive
superposition
 ∆ω 
y1 + y2 = 2 A cos
t  sin(ωavg t )
 2 
∆f =
∆ω
2π
Sound Beats
• Most common manifestation of beats is associated with sound
waves.
• Example: when a 440 Hz tuning fork is hit simultaneously with the Astring of a guitar, 3 beats per second are heard. The guitar string is
tightened slightly, which increases the beat frequency. What was the
original frequency of the guitar string?
Sound Beats
• Most common manifestation of beats is associated with sound
waves.
• Example: when a 440 Hz tuning fork is hit simultaneously with the Astring of a guitar, 3 beats per second are heard. The guitar string is
tightened slightly, which increases the beat frequency. What was the
original frequency of the guitar string?
• Solution: We are dealing with intensity, so fbeat = Δf , so the original
frequency of the guitar string could only be 443 Hz or 437 Hz.
– When the string is tightened, the tension in the string increases,
and so does the wave speed and its frequency. Since this further
increases the frequency difference between the guitar string and
the tuning fork, the guitar string frequency must have already
been larger than that of the fork.
– So, the answer is 443 Hz.
Demos
• 3B60.10 - Beat Forks
• 3B60.18 - Organ Pipe Beats
Spatial Interference
• Consider superposition of two waves of the same frequency but
originating from different locations.
y1 + y2 = A sin( kx1 − ωt ) + A sin( kx2 − ωt )
= 2 A cos k ( x12− x2 ) sin k ( x1 + x22 )− 2ωt
• Again, consider intensity. The phase
difference is
δ = k ( x1 − x2 ) = k∆x
• Depending on the position with
respect to the two sources, the
phase could be 0 or π, or anything
else. We get a complex interference
pattern.
Spatial Interference
• We can rewrite
δ = k∆x = 2π
∆x
λ
• Hence, if the path difference is equal to an integer number of
wavelengths, the phase would be ~2πn, and we get constructive
interference.
• If the path difference is equal to a half-odd-integer number of
wavelength, the phase would be ~2πn+π, and we get destructive
interference.
Speakers on a Rod
• Consider two speakers mounted at the ends of a 6-foot rod,
generating sounds waves at 3 kHz (in phase). By how much would
you have to rotate the bar, in order to hear the
constructive/destructive interference?
• 3B55.10 - Speaker Bar
Interference of Speakers
• Consider two speakers initially located next to each other,
generating 3 kHz sound waves out of phase (i.e. phase difference is
π). By how much should one speaker be moved so that an observer
5 m away hears constructive interference?
• 3B55.47 - Interference of Speakers
Demos
• 3B55.40 - Trombone
• 3B50.40 - Moire Pattern
Transparencies
• Ripple tank with two sources
Coherence
• We saw that two sources need not be in phase to produce a
constructive interference.
– It depends on where you are measuring the interference.
• Two sources that remain in phase or maintain constant phase
difference are called coherent.
– Otherwise, they are incoherent.
• Most sources are incoherent – e.g. two musicians producing the
same note on different instruments.
– The interference quickly switches between constructive and
destructive and no stable interference pattern can be observed.
Double Slit Experiment
• A plane wave is incident on a
screen with two slits.
• The slits behave as coherent
wave sources.
• Behind the screen, the two
waves superpose, and create
interference pattern.
• Works for many situations:
– Water in tank.
– Light.
– Even matter! (wave-particle
duality)
• Demo 3B50.25 - Ripple Tank Double Slit
Standing Waves
Antinode
• Consider a string with fixed ends.
– It is possible to have a
stationary wave pattern:
L
standing wave.
• Similar patterns exist whenever a Node
Node
wave is confined in space:
• Intuition: at the nodes, the string is
– Transverse waves on a piano
not moving, so there is no
string.
interaction with the “walls”.
– Sound in a pipe.
• No energy losses, wave pattern
– Light in optical cavities.
lasts for a long time.
• Condition: need to have nodes of • If the nodes are not at the “walls”,
the waveform at the ends.
the wave will try to “push” at the
“walls”.
• Large energy losses, wave
pattern decays away quickly.
String Fixed at Both Ends
• There are (infinitely) many
waveforms that have nodes at
the ends.
– The length of the string
must be equal to an integer
number of half-wavelengths.
L=n
λn
n = 1,2,3,…
2
2L
λn =
n
v
nv
=
fn =
λn 2 L
- Each point either oscillates or does not.
- Points that oscillate are in phase or exactly
180 degrees out of phase.
String Fixed at Both Ends
• Each of these standing waves is called a
mode of vibration.
• Frequencies corresponding to standing waves
are called resonance (or natural)
frequencies.
• The lowest frequency (f1 = v/2L) is the
fundamental frequency.
– It corresponds to the fundamental mode,
or the first harmonic.
• All of the other modes are multiples of the
fundamental frequency.
– They are called second harmonic, third
harmonic…
– Or, first overtone, second overtone…
• Together, these frequencies form the resonant
frequency spectrum.
v
nv
fn =
=
λn 2 L
v
= nf1
fn = n
2L
Demos
• 3B22.10 - Melde's Standing Wave Machine
• 3B22.50 - Slinky Standing Waves
Testing Piano Wire
• A 3m long wire of linear mass density 2.5 g/m has adjacent
frequencies at 252 Hz and 336 Hz. What is the fundamental
frequency of the wire? Is this a good wire to use in piano strings, if
the wire can handle up to 700 N of tension?
String Attached at One End
• Now suppose one end of the string
is free to move - eg it is attached to
a ring that is free to slide up and
down a pole.
• In the limit when the ring mass is
taken to zero, any vertical force on
it (by the string) would give the ring
infinite acceleration!
– So, we must have an antinode
at the free end.
– Tangent of the string at the end
is then horizontal.
• New condition: need an odd
number or quarter-wavelengths to
fit on the string.
L=n
λn
; n = 1,3,5,...
4
nv
4L
; fn =
λn =
n
4L
Sound Standing Waves
• Consider a pipe closed on one end.
• At the closed end, the air displacement must be zero, so the
pressure variation is large.
• At the open end, the pressure remains at the atmospheric level, so it
does not change much, but the air displacement is the largest.
Sound Standing Waves
• But, can also have pipes open on both ends, or closed at both ends.
• This configuration is similar to string being fixed at both ends.
– The fundamental mode has wavelength twice the length of the
pipe.
• Note: in practice, the pressure node is not exactly at the open end of
the pipe, but slightly outside of it (the wave is not purely onedimensional, there are edge effects etc).
Tuning Fork and Air Column
• When a tuning fork of 500 Hz is brought close to a tube partly filled
with water, resonances are found when the water level is 16.0, 50.5,
85.0 and 119.5 cm below the top of the tube. What is the speed of
sound in air? How far from the open end of the tube is the
displacement antinode?
Optical Cavity
• Light can form a standing wave between two partly transmitting
mirrors.
eiθ = cos θ + i sin θ
e −iθ = cos θ − i sin θ
e
iθ 2
= cos 2 θ + sin 2 θ = 1
Reflection Coefficient: r

Transmission Coefficient: t 
(
)
(
)
1 iθ
cos θ = e + e −iθ
2
1 iθ
sin θ =
e − e − iθ
2i
Reflectivity R = r2
Transmissivity T = t2
(Same for both
mirrors)
Power in cavity:
Peaks when L/λ is halfinteger.
Peak height and width
depend on the reflectivity
of the mirrors
Transmissivity of the cavity:
Peaks when L/λ is halfinteger.
Peak width depends of mirror
reflectivity. Note it never
surpasses 1.
Harmonic (Fourier) Analysis
• Any function over [0,2π) can be written as a sum of sines and
cosines:
a0
a0
f (θ ) = + ∑ an cos(nθ + δ n ) = + ∑ ( An cos(nθ ) + Bn sin( nθ ) )
2
2
n
n
• Apply this concept to a wave defined over some length L, separating
the time and space components:
a0
y ( x, t ) = + ∑ an cos(kn x − ωn t + δ n )
2
n
a0
 π

= + ∑ an cos n ( x − vt ) + δ n 
2
 L

n
a0

 π

 π

= + ∑  An sin n ( x − vt )  + Bn sin n ( x − vt ) + δ n  
2
 L

 L

n 
where kn = 2π/λn = nπ/L and ωn = 2πfn.
Harmonic (Fourier) Analysis
• That is, any wave can be written as a superposition (linear
combination) of harmonic waves.
• But what are the relevant coefficients?
• Let us ignore the space part – i.e. suppose we observe the wave at
a specific location x=0, over some time period T.
ωn = 2πf n =
2πn
T
a0
y (t ) = + ∑ ( An sin(ωnt ) + Bn cos(ωnt ) )
2
n
2
 2πnt 
An = ∫ y (t ) sin 
dt
T 0
 T 
T
2
 2πnt 
Bn = ∫ y (t ) cos
dt
T 0
 T 
T
Square Wave Decomposition
• Determine the harmonic spectrum of a square wave.
Square Wave Decomposition
Differences in Musical Instruments
Different instruments may produce
waves of the same frequency, but
they typically mix in higher
harmonics.
This gives rise to different sounds of
different instruments.
Wave Packets
• A wave packet is a localized wave, or a pulse.
• It is not periodic, so its frequency composition can be complex, not
limited to harmonics of a specific frequency.
• Fourier series  Fourier transform.
• Pulse in time domain corresponds to a “pulse” in frequency domain.
– The widths of the pulses are closely related.
Δω Δt ~ 1
Δt
Δω
Time
Frequency