General Physics 3: Physics 207
Waves, Optics and
Modern Physics
Dr. Cerne (pronounced chair-nay)
jcerne@buffalo.edu
Office Hours:
128 Fronczak Hall
How about:
Mon 2-3 pm?
Wed 3-4:pm?
No Recitations this week. Start next week.
Physics 207 Homepage!!!:
http://www.physics.buffalo.edu/phy207
Recipe for Success in Physics 207
1. Prepare for lectures
• attempt assigned reading before class
• be active in class (paying attention to
lecture is more important than taking
exacting notes)
2. Come to lecture
• ¼ of class time spent solving problems
3. Come to recitation section
4. Do homework
• best way to prepare for exam is by doing
problems
5. Don’t fall behind! Cover 1 Chap/week!
6. If you have questions, get them answered
7. Discuss and explore material with
classmates
8. Come to office hours when you need help
Course Outline
Mechanical Waves
14.Waves
15. Superposition and Interference of Waves
Electro-Magnetic Waves
34.Maxwell’s Equations
35.Light
36.Mirrors and Lenses and their uses
37.Interference
38.Diffraction
Particle Waves
40. Quantization of Angular Momentum and
Energy Values
41.Quantum Physics
42.Quantum Effects in Large Systems of
Fermions and Bosons
Motivation
Why learn about waves?
1. Fundamental, quantitative, and creative
thinking
→ what you learn Physics 207 will make
you a better scientist or engineer
2. Fundamental to classical and modern
physics
• How does your TV/radio function?
• How do telescopes and microscopes
work?
• Why don’t hydrogen atoms and neutron
stars implode?
3. Material is interesting and exciting (of
course…!)!
z(x,t)
x or t
A wave is an organized oscillation causing a
displacement in z (position, pressure,
temperature, electric field, density, etc) as a
function of x or t
Waves:
• mechanical (sound, water, etc.)
• non-mechanical (EM)
continuation of Physics 108
• optics
• quantum mechanics: electron,
photon, hydrogen atom
• Schroedinger’s wave equation
• Pauli principle
• Periodic table
• Optical properties of materials
• Lasers
Mechanical Waves
medium: material that carries the wave, e.g.,
air, water, solid. Can be modeled as masses
connected by springs
Transverse waves
Longitudinal waves
Longitudinal and transverse waves
a+
b
=d
c
Water is incompressible
Traveling wave (speed of wave determined by
mass density and tension)
The Wave Equation
Look at a short segment of a string:
length: ∆x
mass: m = µ∆x
String is under tension T
We look at a transverse wave => look at the
transverse net force on the segment
For small angles:
And we also know:
T sin θ ≈ T tan θ
∂
tan θ =
z ( x, t )
∂x
With this, we can calculate the net force on the
String segment
∂
FL = −T sin θ L ≈ −T
z ( x, t )
∂x
and
∂
FL = T sin θ R ≈ T
z ( x + ∆x, t )
R
∂x
The combined force on the string segment is
∂
∂
F = FR + FL = T [ z ( x + ∆x, t ) − z ( x, t )]
∂x
∂x
Recall: the definition of the derivative is
d
f ( x + ∆x ) − f ( x )
f ( x) =
dx
∆x
And we get
∂ ∂z
∂2 z
F = T ( ∆x ) ( ) = T ( ∆x ) 2
∂x ∂x
∂x
Note that curvature, not displacement determines F!
And together with Newton’s 2nd law
∂2 z
ma = µ ( ∆x ) 2
∂t
we get
∂2 z
∂2 z
F = ma = µ ( ∆x ) 2 = T ( ∆x ) 2
∂x
∂t
And after canceling the ∆x:
∂2 z
∂2 z
µ 2 =T 2
∂t
∂x
This is the wave equation for a string
The wave equation applies to standing waves
and traveling waves! (We assumed neither one
during derivation!)
In its more general form, the wave equation
describes not just harmonic waves on a string,
but all forms of harmonic waves
Wave equation for a string:
∂2z µ ∂2z
=
2
T ∂t 2
∂x
Dimension of
µ
:
T
2
1
 µ  [M / L]  T 
=
=
=
 T  ML / T 2  L2  velocity2
 
[
]
[
Dimension suggests that
We define
µ
T
v≡
]
plays the role of a speed
T
µ
We will see that v is the speed with which a wave
(or disturbance) propagates
∂2z 1 ∂2z
= 2 2
2
v ∂t
∂x
General form of the
wave equation
∂2z 1 ∂2z
= 2 2
2
∂x
v ∂t
The wave equation applies not just to strings with
mass densities µ and tension T, but to many
other waves (like sound, light, radio waves, …)
For the string, the speed of waves was v =
T
µ
For mechanical waves, it is in general true that:
v=
restoring force factor
mass factor
Demo: speed of wave on latex tube
Wave speed depends on medium
[ tension ]
kg ⋅ m/s 2 m 2
=
= 2 = v 2 
[lin. mass density] kg/m s
restoring force factor
v=
mass factor
v↑
if restoring force ↑ (quicker snap back)
if mass factor ↓ (less inertia)
Traveling wave latex tube demo:
Filled with air
Filled with water
How else can we change the wave speed?
Traveling Waves
Consider a disturbance moving along the string:
Note that the shape of the wave does not change!
What is the most general way to describe this
behavior?
z ( x, t ) = f ( x − vt )
z
v
t0=0
0
x
z = f ( x)
z
0
v
0
→ x2 = x1 − vt1 + vt2
t2=2T
x2
z = f ( x − 2vT )
x2 = x1 + v ( t2 − t1 )
x → moving to right,
x must increase as
t3=3T t increases to keep
z
0
0
x x2 − vt2 = x1 − vt1
x1
z = f ( x − vT )
z
z
z ( x,t ) = f ( x − vt )
t1=T
x3
x
z = f ( x − 3vT )
v
t4=4t
x4
x
z = f ( x − 4vT )
( x - vt )
constant
Simple Harmonic Motion
Use sine or cosine as function:
f ( x − vt )
⇒
sin[k ( x − vt )]
k, the wave number, makes the
argument unitless.
z ( x, t ) = z0 sin[k ( x − vt )]
= z0 sin[kx − kvt ]
= z0 sin[kx − ωt ]
where
ω = kv
z ( x, t ) = z0 sin( kx − ωt ) Travels into the
positive x direction
z ( x, t ) = z0 sin(kx + ωt ) Travels into the
negative x direction
Periodic Waves
v
Wavelength:
P
λ : Length after which the wave repeats itself
2π (spatial
Wavenumber:
k=
frequency)
λ
Frequency:
number of crests passing point p (temporal
f:
frequency)
1 second
Angular frequency: ω = 2π ⋅ f
Period:
τ : Time between crests passing point P
1 2π
τ= =
f
ω
Speed of a Wave
2π
λ=
k
As we watch a wave moving through, we measure
the time ∆T it takes for one wavelength λ to pass.
λ is a distance and the wave moved this distance
in the time ∆T
=> The speed of the wave is: v =
λ
∆T
The time for one wave to pass is called the period τ
and we have
λ
v = = λ ⋅ f because f = 1
τ
τ
Example:
Every 3 seconds a 2 m long wave passes you
λ 2m 2 m
=> v = =
=
τ 3s 3 s
Note that:
ω
f =
2π
and
2π
λ=
k
and therefore
2π ω ω
⋅
=
v = λf =
k 2π k
If we plug in a simple sinusoidal wave into the
wave equation:
z ( x, t ) = z0 sin( kx − ωt )
1 ∂ z ∂ z
= 2
2
2
v ∂t
∂x
2
2
1
∂
∂
(− ω cos(kx − ωt ) ) = z0 (k cos(kx − ωt ) )
z
2 0
v
∂x
∂t
1
2
2
z
ω
sin(
kx
−
ω
t
)
=
z
k
sin(kx − ωt )
0
2 0
v
1 2
2
=
k
ω
v2
z ( x, t ) = z0 sin(kx − ωt ) satisfies the wave equation
if
v=
ω
k
Traveling wave f ( x − vt ) satisfies the wave eq.
y ( x,t ) = x − vton
f ( x,t ) = f ( x − vt ) = f ( y )
single variable y replaces x and t
want to show that
∂2 f ( y ) 1 ∂2 f ( y )
= 2
2
∂x
∂t 2
v
recall the chain rule for differentiation:
∂f ∂f ∂y
∂f ∂f ∂y
=
and
=
∂x ∂y ∂x
∂t ∂y ∂t
∂f ( y ) df ( y ) ∂y
→
=
=
dy ∂x
∂x
∂y
=
∂x
HW Problem # 15
17
∂y
=
∂t
Power delivered by a traveling
wave
z
T
θ
Tsin θ
Tcos θ
1D system
x
O
Power moving past point x:
x
P = Ftransverse ⋅ vtransverse
The transverse force is (for small θ )
Ftransverse
∂z
= −T sin θ ≈ −T tan θ = −T
∂x
and the transverse velocity of the string is just
vtransverse
∂z
=
∂t
For a harmonic wave
z ( x, t ) = z0 sin(kx − ωt )
the power moving past point x is then:
∂z ∂z
P = Ftransverse ⋅ vtransverse = −T
∂t ∂x
= Tkω z02 cos2 ( kx − ωt )
= µvω z cos ( kx − ωt )
2 2
0
=>
2
P = µvω z cos ( kx − ωt )
2 2
0
2
and the time average of this is
1
2 2
P = µvω z0
2
(the average over cosine squared over long time is ½)
Energy Transport and Energy Density
We did not consider any friction or other
mechanism for loss of mechanical energy
of the wave.
⇒Power supplied by “emitter” is transmitted
with the wave
⇒Power is a rate of energy transport
Vibrating string has an energy density µ
(energy per unit length)
∆l = v ⋅ ∆t
P(x,t)
v
x
x0
Energy passing through point x0 in time ∆t is
∆E = P ⋅ ∆t
The energy density (energy per unit length) is then
Length ∆l = v∆t moves past point x in time ∆t
∆E P∆t P
u( x,∆t E
) == P∆t =moves =past point x in time ∆t
Energy
∆l
v∆t v
energy density per2unit
u ( x,t )
2 length
2
= µω z0 cos ( kx − ωt )
∆E P∆t P
u ( x,t ) =
=
=
∆l energy
v∆t perv unit length is
The average
= µω 2 z02 cos12 ( kx2− ω
2 t)
E =
µω z0
average energy per2unit length E
1
E = µω 2 z02 → contains both kinetic and
2
potential energy.
For a simple harmonic oscillator
E = K +U
1
K = E
2
1
U = E
2
28
See problem # 32
Standing Waves
wavelength determined by boundary conditions
z
x
Mathematical description of standing
waves
z ( x,t ) = f ( x ) g ( t )
g ( t ) = cos (ω t + φ ) where is ω the frequency
of simple harmonic motion and φ is the phase.
Initial conditions determine φ
e.g., if g ( t = 0 ) = 1 → φ = 0
if g ( t = 0 ) = 0 → φ =
?
→ Note that all parts of the wave have the same
time-dependence
f ( x ) = z0 sin ( kx + δ )
k ≡ wavenumber is the spatial frequency (analogous
to temporal frequency ω )
k=
2π
Which has a larger k?
λ
x
Rad
→ number of oscillations*2π per meter
[k ] =
m
z
Z0
z(x,t)=g(x)*f(t)=Z0sin(kx)*cos(ωt)
t = 0 → cos(0) = 1
x
-Z0
Increasing time
z
Z0
ωt = π/4 → cos(π/4) = 0.71
x
-Z0
z
Z0
ωt = π/2 → cos(π/2) = 0
x
-Z0
z
x
L
Nodes at ends (let δ = 0)
sin ( k × 0 ) = 0
and
sin ( k × L ) = 0
→ k × L = nπ for n = 1, 2, 3, 4, …
nπ
→ kn =
L
2π 2 L
→ λn =
=
kn
n
→ λ is quantized!
⇒Have found the permitted wave numbers
kn =
2π
λ
=
nπ
L
⇒This tells us how many half wavelengths
fit within the boundaries
2L
λn =
n
There are three half wavelengths => third harmonic
⇒There are TWO nodes ( )
There are always n − 1 nodes in a standing
wave of the nth harmonic mode
Sill needed: how fast does this wave swing?
Found the permitted wave numbers k or the
permitted wavelengths n of the standing wave
with boundary conditions
Now: want to find the angular frequency ω
,
and thus, how fast the standing wave oscillates.
Reminder: Angular frequency
ω = 2πf
ω is determined by “dynamics”, that is by
Newton’s 2nd law:
F = ma
How to apply this to get ω ??????
Frequency of Standing Waves
Substitute our standing wave into the
wave equation:
z ( x, t ) = z0 sin(kx + δ ) cos(ωt + ϕ )
∂2z
2
=
z
sin(
kx
+
δ
)(
−
ω
) cos(ωt + ϕ )
0
2
∂t
=>
∂2 z
2
=
z
(
−
k
) sin( kx + δ ) cos(ωt + ϕ )
0
2
∂x
∂2 z
∂2 z
and therefore µ 2 = T 2 becomes:
∂t
∂x
µω 2 = Tk 2
or
ω =k
T
µ
=
2π
T
λ
µ
for strings in general,
and for standing waves on a string:
ωn =
nπ
L
T
µ
Demo: Standing wave on a string
 nπ 
z ( x) = A sin ( kn x ) = A sin 
x  at fixed time
 L
z
x
L
n=
fn =
n=
fn =

n=
fn =
n=
fn =
Do our results fit to theory?
n T
fn =
2L µ
Summary
Standing Waves: Longitudinal and Transverse
z ( x,t ) = f ( x ) g ( t )
= z0 sin ( kx + δ ) cos (ω t + φ )
boundary conditions → kn and λn (discrete)
properties of medium → ω and τ
Traveling Waves:
z ( x,t ) = f ( x − vt )
e.g., z0 sin ( kx − ω t )
Both satisfy the wave equation:
∂2 z 1 ∂2 z
= 2 2
2
∂x
v ∂t
Sound Waves
No net flow of molecules in direction of wave!
∆ρ = Aρ sin ( kx − ω t )
Why no transverse waves?
Speed of sound
in air:
γ 0 P0
v=
ρ0
γ 0 = 1.4
,
v = 330 m/s = 740 mi/hr at STP
in solids:
v=
Y
ρ
,
Y = Young's modulus and ρ = density
Y =?
F
A
Y]=?
[
∆L
L
F
stiff material Y ↑
Y=
soft material Y ↓
∆L
A
L
Speed of sound waves in various media
Y
v=
ρ
Medium
Density
Hydrogen
Air
Liquid mercury
(20oC)
Methyl alcohol
Water
Polyethylene
Lead
Silver
Aluminum
Beryllium
0.09 g/l
1.3 g/l
11.3 g/cm3
10.5 g/cm3
2.7 g/cm3
1.85 g/cm3
Speed of
sound (m/s)
1284
330
1450
1189
1402
920
1210
2700
5000
12870
Pressure Wave
Both ends open
Displacement
Amplitude
z(x,t)
Pressure
Amplitude
(also called
∆ρ(x,t))
Standing sound waves
Boundary condition: no displacement on closed
ends and maximum displacement on open ends
displacement
v=λf
v = 330 m/s (fixed by P0 and ρ0 )
2L
λn =
n
v
n
v
fn =
=v
= nf1 where f1 =
λn
2L
2L
For L = 4.0 m
330 m/s
f1 =
= 41 Hz
2 ( 4.0 m )
ff32 =
= 23 ff11 ==83
124HzHz
f
3f
124 H
f 2 = 2 f1 = 83 Hz
z=0
x
f1
∂z
=0
∂x
f2=3f1
f3=5f1
∂z
=0
∂x
f1
x
∂z
=0
∂x
f2=2f1
f3=3f1
Demo: The trumpet
mouthpiece
Mouthpiece + trumpet (cavity)
1. Mouthpiece: frequencies and volume?
2. Mouthpiece+Trumpet: frequencies and
volume?
3. Changes in cavity length: what happens to
frequency?
a) Pressing a valve
b) Pulling slide
c) Changing bell shape with hand
Intensity
P
I=
A
Ptotal
[I ] =
W/m
2
PA
Want to keep track of I over many orders of magnitude
→ use logarithmic scale. This scale deals with the
exponent of 10 required to obtain I relative to a very
small reference intensity I0.
I
β ≡ 10 log10 → 10 times the number of orders of
I0
[ β ] = dB
magnitude that I is larger than I 0
or decibels, I 0 = 10−12 W/m 2
I = I0 → β = 0
0 orders of magnitude
I = 10 I 0 → β = 10
1 order of magnitude
I = 100 I 0 → β = 20
2 orders of magnitude
I = 105 I 0 → β = 50
5 orders of magnitude
human ear can hear 0-120 dB range (10-12 − 1 W/m 2 )
range for human eye is similar (though I0 is different)
The Doppler Effect
Moving source, stationary observer
v
vs
λ=
f0
v
v
A
Source moving away
from observer A
lengthened λ
Source moving towards
v
v + vs
λ' = λ + s =
f0
f0
vs v − vs
λ' = λ − =
f0
f0
observer not moving,
sees waves travel at v
observer not moving,
sees waves travel at v
 v 
v
f'=
= f0 

v
v
λ'
+
s 

f ' < f0
 v 
v
= f0 
f'=

−
λ'
v
v
s 

f ' > f0
B
observer B
shortened λ
source moving away from
source moving towards
observer (v opposite sign of vs ) observer (v same sign of vs )
Demo: Doppler effect due to moving
source and stationary observer
vs
Observer
Doppler effect: stationary source,
moving observer
Since source is not moving, wavelength in medium is
unchanged → λ’=λ
v
vr
v
v
v
However, the effective velocity of the sound wave
changes because the observer is moving with respect
to the medium!
v
vr
v
v
v
λ' = λ
λ' = λ
observer moving away from observer moving towards
source, sees waves travel at v' source, sees waves travel at v'
v' = v + vr
v' = v − vr
f'=
v'
λ
=
v − vr
λ
f'=
v'
λ
=
v + vr
λ
 v 
f ' = f 1 − r 
v

f ' < f0
 v 
f ' = f 1 + r 
v

f ' > f0
observer moving away
from source (v same
direction as vr )
observer moving
towardssource (v
opposite direction as vr )
Doppler effect: both source and observer
moving
1. Start with source, which is moving with respect
to the medium at a speed vs
λ' =
λ'
v − vs
→ wavelength in medium
f0
vs
v
λ'
v
2. Observer moving at speed vr with respect
to medium, sees λ ' and new wave speed v'
v' = v − vr (no longer need source)
λ'
v
λ'
 v − vr 
v'
v − vr
=
=
f'=
 f0
λ ' ( v − vs ) f 0  v − vs 
vr
v
Note that f’ is not symmetric!
Receiver moving at 5 km/hr towards a
stationary source is NOT the same as having
the source moving at 5 km/hr towards a
stationary observer!
Sign conventions for Doppler calculation
Level 1:
a) If vs (vr) is in the same direction as v, use
positive value for vs (vr) .
b) If vs (vr) is in the opposite direction as v, use
negative value for vs (vr) . Plug into
formula
 v − vr 
f' =
 f0
 v − vs 
Level 2:
Choose signs so that :
a) source moving towards observer increases
f’
b) observer moving towards source increases
f’
Level 3:
a) Start at source and calculate λ’
b) Look at receiver to get relative wave speed
v'
v’
f'=
λ'
Traveling waves generated at a fixed
point (X), in a moving medium
X
vm
Strea
m flo
vs=?
vr=?
w
v=speed of sound
90o-θ1
not θ1
vt2
v
=
cos θ 0 =
vs t2 vs
v ( t2 − t1 ) v
=
cos θ1 =
vs ( t2 − t1 ) vs
vs
Mach number ≡
v
Large Mach number : θ1 → 90"
vs < v
vs > v
Shock wave produced by supersonic
bullet passing over a candle
Charged particle moving through water
at a speed greater than the speed of light
in water → Cerenkov radiation
Ch 14 Useful Equations
Standing wave:
z ( x,t ) = sin ( kx + δ ) cos (ω t + φ )
Transverse wave on string:
∂2 z
∂2 z
T 2 =µ 2
∂x
∂t
2π T
ω=
λ
µ
Wave equation:
∂2 z 1 ∂2 z
=
∂x 2 v 2 ∂t 2
Sinusoidal traveling wave in x-direction:
z ( x,t ) = z0 sin ( kx − ω t )
v=λf =
λ ω
T
= =
τ k
µ
Energy density:
dE
= µω 2 z02 cos 2 ( kx − ω t )
dx
Average power delivered by wave:
1
P = µω 2 z02 v
2
γ P0
vsound =
ρ0
 I 

 I0 
β = 10 log10 
λ' =
v − vs
f0
f'=
v '  v − vr 
=
 f0
λ '  v − vs 
where I 0 = 10−12 W/m 2
v' = v − v r