Wave Phenomena
Physics 15c
Lecture 9
Wave Reflection
Standing Waves
What We Did Last Time
Energy and momentum in LC transmission lines
1
1 V0I0
  Transfer rates for normal modes:
V0I0 and
2
2 cw
The energy is carried by the EM field around the wires
  The wires guide the waves
  Poynting vector gives the power density S = E × H
 
Studied the wave reflection
 
Determined by the impedance matching
R−Z
V− =
V+
R+Z
 
R−Z
I− = −
I+
R+Z
Power is reflected or absorbed according to
Preflected
(R − Z)2
=
P
2 input
(R + Z)
Pabsorbed
4RZ
=
P
2 input
(R + Z)
Goals for Today
Reflection of mechanical waves
We expect reflection in mechanical waves as well as in
electromagnetic waves
  Where the sound hit a solid wall
  Where the string is tied to the door knob
  We expect “open” and “shorted” ends
  Can we define “impedance” for mechanical waves?
 
Standing waves
Created when sinusoidal waves are reflected
  Basic principle for (almost) all musical instruments
 
Transverse Waves on String
T
T
ξ(x)
Transverse wave ξ(x – cwt) is traveling on a string
 
The string has tension T and linear mass density ρl
From Lecture #6:
Wave equation
∂ 2ξ(x,t)
∂ 2ξ(x,t)
ρl
=T
2
∂t
∂x 2
Wave velocity
ω
cw = =
k
T
ρl
x
Reflection at a Fixed End
x=L
ξ(x)
Suppose one end of the string is tied to a fixed point
 
This end cannot move  ξ(L) = 0
Reflection must occur to satisfy this condition.
 
Let’s call the original and reflected waves as
ξ+(x − cwt) and ξ−(x + cwt)
At x = L, ξ+ (L − cw t) + ξ− (L + cw t) = 0
Reflected waves are negative of the incoming waves
 
Similar to the shorted end of an LC transmission line, where
V+ (L − cw t) +V− (L + cw t) = 0
Reflection at a Free End
Imagine that an end of the string is “free”
 
Let the string move vertically without any force
x=L
Not easy to
make this
ξ(x)
 
Vertical component of the tension is given by
∂ξ(x,t)
F = T sin θ ≈ T
∂x
 
F
T
θ
The “free” end at x = L means
⎛ ∂ξ ⎞
F(x = L) = T ⎜ ⎟
⎝ ∂x ⎠
=0
x =L
i.e. the string is horizontal
Reflection at a Free End
⎛ ∂ξ+ (x − cw t) ∂ξ− (x + cw t) ⎞
The boundary condition is ⎜
+
=0
⎟
∂x
∂x
⎝
⎠ x =L
 
It takes a little trick from here
⎛ ∂ξ+ (x − cw t) ⎞
⎛ ∂ξ− (x + cw t) ⎞
= −⎜
⎜
⎟
⎟
∂x
∂x
⎝
⎠ x =L
⎝
⎠ x =L
1
−
cw
⎛ ∂ξ+ (x − cw t) ⎞
1
=−
⎜
⎟
∂t
cw
⎝
⎠ x =L
⎛ ∂ξ− (x + cw t) ⎞
⎜
⎟
∂t
⎝
⎠ x =L
d ξ+ (L − cw t) d ξ− (L + cw t)
=
dt
dt
  Integrate with time, and assume ξ+ = ξ− = 0 at t = −∞
ξ+ (L − cw t) = ξ− (L + cw t)
Reflected waves are the same as the incoming waves
Mechanical Impedance
Reflection of electromagnetic waves on an LC transmission
line was determined by the impedance matching
 
We calculated general solutions using R and Z
To analyze reflection of mechanical waves, we need
mechanical impedance
Consider transverse mechanical waves ξ(x,t) = f (x ± cw t)
∂ξ
  Vertical force is F = T
= Tf ′(x ± cw t)
∂x
proportional to each other
∂ξ
= ±cw f ′(x ± cw t)
  Vertical velocity is v =
∂t
F
T
This is a constant:
Consider the ratio:
=
= ± T ρl
Mechanical impedance
v ±cw
Mechanical Impedance
Mechanical impedance of transverse waves: Z = T ρl
  For longitudinal waves, Z = K ρ
l
m
L
  For LC transmission lines Z =
1/ kS
C
The force and the velocity are related by
+ for forward-going waves
F = ±Zv
− for backward-going waves
 
Unit: [force]/[velocity]  N/(m/s) = kg/s
Let’s see how this helps our calculation
Connection Between Two Media
ρl1
ρl2
Consider waves traversing a boundary between two media
Heavy string to light string
  Sound transmission between different materials
 
glass
copper
At the connection between two media,
The force F must be equal on both sides
  The wave amplitude ξ(x, t) must be equal on both sides
  This can be achieved if the velocity v is equal
(Assuming it was OK to at the beginning of time)
 
Connection Between Two Media
Define forces and velocities for the incoming, reflected,
transmitted waves
F1(x − c1t) v1(x − c1t)
F2 (x − c 2t) v 2 (x − c 2t)
Z1
Z2
Fr (x + c1t) v r (x + c1t)
x=0
 
Boundary conditions are
v1(−c1t) + v r (c1t) = v 2 (−c 2t)
F1(−c1t) + Fr (c1t) = F2 (−c 2t)
Z1v1(−c1t) − Z1v r (c1t) = Z2v 2 (−c 2t)
Connection Between Two Media
v1(−c1t) + v r (c1t) = v 2 (−c 2t)
Z1v1(−c1t) − Z1v r (c1t) = Z2v 2 (−c 2t)
Z1 − Z2
v r (c1t) =
v1(−c1t)
Z1 + Z2
2Z1
v 2 (−c 2t) =
v1(−c1t)
Z1 + Z2
Z1 − Z2
Fr (c1t) = −
F1(−c1t)
Z1 + Z2
2Z 2
F2 (−c 2t) =
F1(−c1t)
Z1 + Z2
Reflected and transmitted power are
2
⎛ Z1 − Z2 ⎞
Pr = − ⎜
P1
⎟
⎝ Z1 + Z2 ⎠
P2 =
4Z1Z2
(Z
1
+ Z2
)
2
P1
Add up to P1
Air  Water
Sound waves travel from air into water inside a pipe
 
Mechanical impedance: Z = K ρl = A MB ρv
Zair
γ air Pair ρvair
1.4 ⋅ 9.8 × 10 4 [N/m2 ] ⋅1.3[kg/m3 ]
−4
=
=
=
2.9
×
10
Z water
MBwater ρvwater
2.1× 109 [N/m2 ] ⋅103 [kg/m3 ]
4Zair Z water
Fraction of power transmitted
= 0.0012
2
(Zair + Z water )
  ~99.9% is reflected
 
Engineering problem for the middle ear 
Air in the ear canal  water in the inner ear
  Solution 1: Ear drum >> oval window
  Solution 2: Use levers (ossicles) to reduce velocity
and increase force
 
Standing Waves
Suppose a sinusoidal wave train ξ = ξ0cos(kx – ωt) is being
reflected at either a fixed or a free end
 
Let’s define the position of the end as x = 0 for simplicity
x=0
Reflected waves are ±ξ0 cos(kx + ω t)
  The sum of the incoming and reflected waves is
 
⎧ 2ξ cos kx cos ω t
⎪
0
ξ0 cos(kx − ω t) ± ξ0 cos(kx + ω t) = ⎨
⎪⎩ 2ξ0 sin kx sin ω t
free end
fixed end
Standing Waves
⎧ ξ cos kx cos ω t
⎪ 0
⎨
⎪⎩ ξ0 sin kx sin ω t
fixed end
free end
λ = 2π / k
λ = 2π / k
node
node
node
node
node
antinode
antinode antinode
antinode
antinode
Standing waves have nodes and antinodes
 
There are two nodes (antinodes) in one wavelength
Standing Waves on String
What happens when both ends of a transmission line are
fixed, free, or a combination of both?
Example: a string stretched between two fixed points
L
 
Stringed musical instruments (piano, guitar, violin)
The standing waves must have nodes on both ends
 
The wavelength must have particular values that match the length L
of the string
Frequencies and Wavelengths
Standing wave for one fixed end at x = 0 is
ξ(x,t) = ξ0 sin kx sin ω t
 
To satisfy the other fixed end at x = L,
ξ(L,t) = ξ0 sin kL sin ω t = 0
nπ
k=
L
2π 2L
λ=
=
k
n
kL = nπ
nπ cw
ω = cw k =
L
Only particular, discrete, set of frequencies
and wavelengths are allowed
any integer
ω ncw
ν=
=
2π
2L
Greek nu =
frequency!
The lowest-frequency mode is called the fundamental mode
  Others are called harmonics
 
Fundamental and Harmonics
Fundamental
n =1
2nd harmonic
n=2
3rd harmonic
n=3
4th harmonic
n=4
cw
λ = 2L ν 0 =
2L
λ=L
2ν 0
2
λ= L
3
3ν 0
1
λ= L
2
4ν 0
Stringed Instruments
Strings on piano, guitar, etc. oscillates at the fundamental
frequency plus its harmonics: ν0, 2ν0, 3ν0, 4ν0, …
Fundamental freq. is determined by L and cw
  cw is determined by T and ρl
 
cw
1
ν0 =
=
2L 2L
You hear the sum of the fundamental + harmonics, and
recognize the whole sound as having a distinct pitch
 
String instruments adjust the pitch of a note by L, T, and ρl
Relative amplitudes of higher harmonics determine timbre,
i.e., the character (piano-like, guitar-like) of the sound
 
Without harmonics, everything would sound like a tuning fork
T
ρl
Plucking a String
Plucking a string sets up
the initial condition
ξ(x,t = 0)
One can break it into a
Fourier series
  Each term of the Fourier
series is a harmonic
  Where you pluck in x determines the mixture of the harmonics
 You hear different timbre
  As the oscillation decay with time, higher harmonics disappear faster
 Timbre changes with time
 
Wind Instruments
An open end of a pipe acts as a free end for sound
Pressure of the air = 1 atm where the pipe ends
  This is equivalent to having no force  a free end
 
Wind instruments (pipe organ, woodwinds, etc.) are basically
a pipe with at least one open end
 
The other end is usually closed
Wind Instruments
Standing wave for one free end at x = 0 is
ξ(x,t) = ξ0 cos kx cos ω t
 
To satisfy the other fixed end at x = L,
ξ(L,t) = ξ0 cos kL cos ω t = 0
(2n − 1)π
k=
2L
2π
4L
λ=
=
k
2n − 1
2n − 1
kL =
π
2
(2n − 1)π cw
ω (2n − 1)cw
ω = cw k =
ν=
=
2L
2π
4L
Wavelength of the fundamental is 4L
  Only the odd-multiples of the fundamental exist
 
any integer
Wind Instruments
Fundamental
n =1
3rd harmonic
n=2
5th harmonic
n=3
7th harmonic
n=4
cw
λ = 4L ν 0 =
4L
4
λ= L
3
3ν 0
4
λ= L
5
5ν 0
4
λ= L
7
7ν 0
Wind Instruments
Air in wind instruments oscillates at the fundamental
frequency plus odd harmonics: n0, 3n0, 5n0, 7n0, …
Fundamental frequency is determined by L and cw
  cw is pretty much constant at normal temperature
  Pipe organ must be that big to cover the lowest audible
frequency
 
cw
ν0 =
4L
cw
330
ν0 =
= 20 → L =
= 4.125 m
4L
4 ⋅ 20
 
Lack of even harmonics gives wind instruments their characteristic
sound
Inharmonicity
Sound that contains frequencies other than integer
multiples are said to be inharmonic
Inharmonic sound does not have a recognizable pitch
  Drums and other percussion instruments are inharmonic
 
String/wind instruments do have small inharmonicity
Higher harmonics are not exactly multiples of the fundamental due
to small dispersion, i.e., the wave velocity varying with the frequency
  Example: Strings are not perfectly flexible and resist bending
 Harmonics have higher frequencies
  This has interesting effects on tuning
  Our ears are accustomed to small inharmonicity in music
 
Summary
Reflection of mechanical waves
Similar to reflection of electromagnetic waves
  Mechanical impedance is defined by F = ±Zv
 
For transverse/longitudinal waves: Z =
  Useful in analyzing reflection
 
[T or K ]ρl
Standing waves
Created by reflecting sinusoidal waves
  Oscillation pattern has nodes and antinodes
  Musical instruments use standing waves to produce their distinct
sound
