The Principle of Superposition
For small amplitudes, waves will pass through each other
and emerge unchanged.
Superposition Principle: When two or more waves overlap,
the net disturbance at any point is the sum of the individual
disturbances due to each wave.
Overlapping waves algebraically add to produce a resultant wave (or
net wave).
Overlapping waves do not in any way alter the travel of each other.
Two traveling wave
pulses: left pulse
travels right; right
pulse travels left.
Superposition
y1 = A sin (k1 x - w1 t)
and
y2 = A sin (k2 x - w2 t + ϕ).
Since
sin (a) + sin (b) = 2 cos ((a - b) / 2) sin ((a + b) / 2),
we can form the sum
y=y1+y2
= 2 A cos [(Δk x - Δw t + ϕ) / 2] sin [(Σk x - Σw t + ϕ) / 2],
where Δk is k 2 - k 1, Δw is w 2 - w 1, Σk is k 1 + k 2 and Σw is w 1 + w 2
Interference and Diffraction
Two waves are considered coherent if they have the
same frequency and maintain a fixed phase relationship.
Two waves are considered incoherent if the phase
relationship between them varies randomly.
When waves are in phase, their superposition gives
constructive interference.
When waves are one-half a cycle out of phase, their
superposition gives destructive interference.
For "coherent sources", in which the two waves have
the same wave number and frequency
(k 1 = k 2, w 1 = w 2):
y(x,t) = 2 A cos (ϕ / 2) sin (k x - w t + ϕ / 2).
Amplitude
…..back to coherent waves
When two coherent waves
travel different distances to
reach the same point, the
phase difference is
determined by:
d1 − d 2
λ
phase difference
=
2π
http://www.falstad.com/ripple/
We can represent wave vectorially with a phasor.
We can use phasors to combine waves even if their amplitudes are
different.
For the horizontal components we have
For the vertical components we have
Thus, the resultant wave has an amplitude of
and a phase constant of
Beats
When two waves with the same amplitude but different
frequency are added together a phenomenon called "beating"
y2 = Asin(2πf2t)
occurs. y1 = Asin(2πf1t)
The resultant waveform can be thought of as a wave with frequency fave = (f1 +
f2)/2 which is constrained by an envelope with a frequency of fb = |f1 - f2|.
The envelope frequency is called the beat frequency.
http://library.thinkquest.org/19537/java/Beats.html
Standing Waves
Pluck a stretched string such
that y(x,t) = A sin(ωt + kx)
When the wave strikes the wall, there will be a reflected
wave that travels back along the string.
If two sinusoidal waves of the same amplitude and wavelength travel in
opposite directions along a stretched string, their interference with each
other produces a standing wave.
The reflected wave will be 180° out of phase with the wave
incident on the wall. Its form is y(x,t) = -A sin (ωt - kx).
Apply the superposition principle to the two waves on the
string:
y ( x, t ) = y1 ( x, t ) + y2 ( x, t )
= A(sin (ωt + kx ) − sin (ωt − kx ))
= (2 A cos ωt )sin kx
The previous expression is the mathematical form of a
standing wave.
A
A
A
N
N
N
N
A node (N) is a point of zero oscillation. An antinode (A) is
a point of maximum displacement. All points between nodes
oscillate up and down.
http://www.falstad.com/loadedstring/
http://www.falstad.com/membrane/
The nodes occur where y(x,t) = 0.
y ( x, t ) = 2 A cos ωt sin kx = 0
The nodes are found from the locations where sin kx=0,
which happens when kx = 0, π, 2π,…. That is when kx = nπ
where n = 0,1,2,…
The antinodes occur when sin kx=± 1; that is where
kx =
π 3π
,
,…
2 2
(
2n + 1)π
kx =
and n = 0 ,1, 2 ,…
2
If the string has a length L, and both ends are fixed, then
y(x=0,t) = 0 and y(x=L, t) = 0.
y ( x = 0, t ) ∝ sin k (0 ) = 0
y ( x = L, t ) ∝ sin kL = 0
kL = nπ
2π
λ
The wavelength of
a standing wave:
L = nπ
2L
λ=
n
where n = 1, 2, 3,…
2L
λn =
n
These are the permitted wavelengths of standing
waves on a string; no others are allowed.
The speed of the wave is:
v = λn f n
The allowed frequencies are then: f n =
v
nv
=
λn 2 L
n =1, 2, 3,…
The n=1 frequency is called the fundamental frequency.
nv
⎛ v ⎞
fn =
=
= n⎜ ⎟ = nf1
λn 2 L
⎝ 2L ⎠
v
All allowed frequencies (called harmonics) are integer
multiples of f1.
Example (text problem 11.55): A Guitar’s E-string has a length
65 cm and is stretched to a tension of 82 N. It vibrates with a
fundamental frequency of 329.63 Hz. Determine the mass
per unit length of the string.
For a wave on a string:
v=
F
μ
Solving for the linear mass density:
F
F
F
μ= 2 =
= 2
2
v
(λ1 f1 ) f1 (2 L )2
(
82 N )
=
(329.63 Hz )2 (2 * 0.65 m )2
= 4.5 × 10 − 4 kg/m