Chapter 15: Superposition and
Interference of Waves
“Real” waves are rarely purely sinusoidal
(harmonic), but they can be represented by
superpositions of harmonic waves
In this chapter we explore what happens when
harmonic waves combine to form more
complicated waves
Linearity of the wave equation
The solutions to the wave equation are linear
(
True for small displacements where
the restoring force remains linear
)
What does this mean?
If a(x, t) and b(x, t) are solutions to the wave equation,
then so is
z ( x , t ) = a ( x, t ) + b ( x , t )
Check:
∂2z 1 ∂2z
= 2 2
2
v ∂t
∂x
∂ 2 ( a + b) 1 ∂ 2 ( a + b)
= 2
2
∂x
v
∂t 2
∂ 2 a ∂ 2b 1 ∂ 2 a 1 ∂ 2b
+ 2 = 2 2 + 2 2
2
v ∂t
v ∂t
∂x
∂x
is true if
∂ 2a 1 ∂ 2a
∂ 2b 1 ∂ 2b
= 2 2 and
= 2 2
2
2
∂x
v ∂t
∂x
v ∂t
The sum (or difference) of two solutions of the wave
equation is again a solution of the wave equation
=> superposition principle
Interference
The superposition principle allows us to add waves of
different amplitudes, wavelengths, and frequencies,
each of them moving in different directions.
=> The result: Interference
Mathematically: the sums are algebraic sums
Examples:
(Red curves: algebraic sum of the blue and green curves)
Simple example for destructive interference:
z1 ( x, t ) = z0 sin(kx − ωt )
z2 ( x, t ) = z0 sin(kx − ωt + π ) = − z0 sin(kx − ωt )
⇒ z1 + z2 = 0
Simple example for constructive interference:
z1 ( x, t ) = z0 sin(kx − ωt )
z2 ( x, t ) = z0 sin(kx − ωt )
⇒ z1 + z2 = 2 z0 sin( kx − ωt )
Important trigonometric relationships for
general cases of interference:
sin(a + b) = sin(a ) cos(b) + cos(a ) sin(b)
cos(a + b) = cos(a ) cos(b) − sin(a ) sin(b)
Example:
z1 ( x, t ) = z0 sin( kx − ωt )
z2 ( x, t ) = z0 sin( kx − ωt + ϕ )
⇒ z1 + z2 = ?
z1 + z2 = z0 [sin(kx − ωt ) + sin(kx − ωt + ϕ )]
a
a+b
use:
sin(kx − ωt + ϕ ) = sin(kx − ωt ) cos(ϕ )
+ cos(kx − ωt ) sin(ϕ )
plug in:
z1 + z2 = z0 [sin(kx − ωt ) + sin(kx − ωt + ϕ )]
= z0 [sin( kx − ωt ) + sin(kx − ωt ) cos(ϕ )
+ cos(kx − ωt ) sin(ϕ )]
= z0 (1 + cos(ϕ )) sin(kx − ωt ) + cos(kx − ωt ) sin(ϕ )
=> if ϕ = 0 → z1 + z2 = 2 z0 sin(kx − ωt )
if ϕ = π → z1 + z2 = 0
Coherence: Phase is stable in time.
Standing Waves Through
Interference
Adding two oppositely traveling waves to get
a standing wave
zleft ( x, t ) = z0 sin(ωt + kx)
z right ( x, t ) = z0 sin(ωt − kx)
z ( x, t ) = z right ( x, t ) + zleft ( x, t )
z ( x, t ) = z0 [sin(ωt − kx) + sin(ωt + kx)]
sin(ωt ) cos(−kx) + sin( −kx) cos(ωt ) + 
= z0 

sin(
)
cos(
+
)
+
sin(
+
)
cos(
)
ω
t
kx
kx
ω
t


= 2 z0 sin(ωt ) cos(kx)
z ( x, t ) = 2 z0 sin(ωt ) cos(kx)
note that ω and k are same for traveling waves
and resulting standing wave
Superposition of traveling waves
Interference of incident and reflected
waves
To get node at wall, reflected wave must
have opposite sign as incident wave (but
same magnitude at wall)
Beats
Beats occur when two waves of nearly the same
frequency are superposed.
Example: Two combs with different tooth spacing
Where the two combs superpose, a new pattern
emerges (here called Moire pattern).
A similar thing happens when two waves superpose.
Two sine waves with nearly same wavelengths
superpose to a wave that has an envelope that is again
a sine wave but with longer wavelength. => beats
=> How to calculate the wavelength of the beats?
To simplify analysis, fix position x and look at wave
as function of time. (Could equivalently fix time t and look at wave as
a function of position)
z1 ( x, t ) = z0 sin(ω1t − kx ) → z0 sin(ω1t )
z2 ( x, t ) = z0 sin(ω 2t − kx ) → z0 sin(ω 2t )
Lets define two new frequencies:
1
Ω = (ω1 + ω 2 ) and δω = ω1 − ω 2
z1 (x, t) = z 0 sin(
t)
fix position x
2ω1t − k1x) → z 0 sin(ω1δω
z 2 (x, t) = z 0 sin(ω2 t − k 2 x) → z 0 sin(ω2 t)
δω0= ω1 − ω2
and
Ω=ω=
1
ω1 + ω
(ω
Ω
2)
2
2
ω
ω1
z1 (x, t) =

δω  
 δω 
 δω 
z 0 sin  Ω +
t
=
z
sin
Ω
t
cos
t
+
z
cos
Ω
t
sin
t
(
)
(
)
0
 

 0

2  
 2 
 2 

z 2 (x, t ) =

δω  
 δω 
 −δω 
z
cos
t
sin
t
z 0 sin  Ω −
t
=
z
sin
Ω
t
cos
t
+
Ω
(
)
(
)
0
 

 0
−
2  
 2 
 2 

 δω 
δω
z1 (x, t) + z1 (x, t) = 2z 0 sin ( Ωt ) cos 
t
z1 ( x, t ) + z 2 ( x, t ) = 2 z20 sin(
Ω
t
)
cos(
t)

2
average frequency:
Ω
(fast)
beat frequency: ω beat
1
= δω
2
(slow)
f beat
ω beat
1 ω1 − ω 2 1
=
=
= ( f1 − f 2 )
2π
2π
2
2
Note: since the ear is only sensitive to the amplitude,
it cannot tell the difference between
1
cos( δωt )
2
and
1
− cos( δωt )
2
⇒the perceived beating frequency is
f pulse = 2 ⋅ f beat = f1 − f 2
= pulse frequency
Demo: Beat frequency
ω1
ω2
Ω
Can hear average frequency f =
(fast)
2π
And pulse frequency
f pulse = 2 ⋅ f beat = f1 − f 2
What happens to the pulse frequency when
the two frequencies get close to each other?
What happens to the pulse frequency when
the two frequencies move apart?
Spatial Interference Phenomena
(fix time t by taking snapshot and look at
x-dependence)
In phase:
Constructive
interference
∆x = nλ
n = 0,±1,±2,…
Phase difference:
n ⋅ 2π
Out of phase:
Destructive
Interference
1
∆x = ( n + ) λ
2
n = 0,±1,±2,…
Phase difference:
1
( n + ) ⋅ 2π
2
∆L=L2-L1
minima when ∆L=(n+1/2)λ,
maxima when ∆L=nλ, n=0, ±1, ±2, …
Recall: λ
λ
2
λ
4
→ 2π
→π
→
π
2
cos( x + λ ) = cos( x )
cos( x +
λ
cos( x +
2
λ
4
) = − cos( x )
) = − sin( x )
Locations of maxima and minima:
∆L = d sinθ
for maxima:
∆L = nλ
→
nλ
sin θ =
d
for minima:
1

∆L =  n +  λ
→
2

where n = 0, ±1, ±2, ±3…
1

n + λ
2

sinθ =
d
Demo: Interference from two speakers
S1
S2
Raise your hand when an acoustic minimum
passes you.
Collision of pulses
What happens here??
Note that the center point at the dot remains
motionless.
⇒Connection to reflection of pulse on fixed end
Reflection from a fixed end
Think of this as a “real” pulse moving to the right and
an “imaginary” pulse moving to the left.
Before reflection: only “real” pulse visible
(“imaginary” one is somewhere beyond wall)
After reflection: previously “imaginary” pulse is real
now and moves to left
What happens during the reflection?
⇒superposition of left and right moving pulses
(such that rope end remains fixed, see previous slide)
Reflection from a vertically free end
What happens here?
Reflection due to change in µ (note
tension is constant!)
v∝?
Lighter to heavier medium
heavier to lighter medium
Power is conserved (power in=power out)
µ1ω A v
2
i
2
i i
=
µ1ω A v
2
r
2
r r
+
2
2
vi = vr = v1 and vt = v2
ωi = ω r = ωt
µ 2ω A v
2
t
2
2
t t
Fourier decomposition of waves
∞
f (t) = ∑ A n sin(nωt + φn )
n =1
∞
g(x) = ∑ A'n sin(nkx + θn )
n =1
This is analogous to decomposing a vector
into orthogonal direction components (x, y,
and z), except here we are decomposing a
function into orthogonal frequency
components
∞
8
1
z(t) = z 0 ∑ 2 cos(nωt)
π n =odd n
1
1


= (0.81)z 0  cos(ωt) + cos(3ωt) + cos(5ωt) + ...
9
25


Real Waves
Ch 15 Useful Equations
f beat =
1
( f1 − f 2 )
2
Interference due to two sources
∆L = d sin θ
if two sources are in phase (∆φ = 0 ):
for maxima: ∆L = nλ
sin θ =
→
1

for minima: ∆L =  n +  λ
2

where n = 0 , ±1, ±2 , ±3…
→
nλ
d
1

+
n

λ
2
sin θ = 
d
If two sources are out of phase (∆φ = π )
1

for maxima: ∆L =  n +  λ
2

for minima: ∆L = nλ
µ1ωi2 Ai2 vi
=
µ1ω r2 Ar2 vr
→
+
2
2
vi = vr = v1 and vt = v2
1

n
+

λ
2

→
sin θ = 
d
nλ
sin θ =
d
µ 2ωt2 At2 vt
2
ωi = ω r = ωt