Standing Waves
If the same type of waves move through a common region and
their frequencies, f, are the same then so are their wavelengths, λ.
This follows from: v = λf .
Since the waves move through a common region, the material they
move through is the same, so their speeds must be same. With v
and f the same for the two waves, λ must also be the same for the
two waves.
We saw last class that in the region where such waves overlap
their superposition creates a new wave having the same speed,
frequency and wavelength, but with an amplitude that ranges
from (A1+A2) for fully constructive interference to |A1– A2| for
fully destructive interference, with Ai the respective amplitude
of each incident wave (and a phase offset of from 0 to 2π).
If the two waves are counter propagating (i.e. move in opposite
directions) a new phenomena can occur. They can generate what is
called a standing wave.
Simulation:
http://www2.biglobe.ne.jp/~norimari/science/JavaEd/e-wave4.html
To be definitive we consider transverse waves propagating in opposite
directions along a long string (though this happens in all types of
waves).
What characterizes the stationary wave is that it has positions along
the string where the amplitude is always zero (called nodes) and
positions half way between the nodes where the string oscillates
with large amplitude (called anti-nodes).
The anti-nodes often move too fast to see and are just a blur.
One way for such counter propagating waves to occur is to have a
taught string tied off to a wall at one end and oscillated at the other
end to launch a sinusoidal wave down it,
y1 (x,
=
t) A sin(ωt − kx)
x
showing just forward wave
At the tied off end the wave will reflect and invert to generate the
counter propagating wave (moving in the opposite direction).
y 2 (x, t) =− A sin(ωt + kx)
x
showing just the reflected wave
If we consider just these 2 propagating waves:
y1 (x,
=
t) A sin(ωt − kx)
y 2 (x, t) =− A sin(ωt + kx)
Then their superposition
gives
=
y(x, t)
y(x,
=
t) y1 (x, t) + y 2 (x, t)
[2A sin(kx)] cos(ωt)
(see GRR p.409)
Which is not a traveling wave. The cosine part is a harmonic
oscillation in time, but this is spatially modulated by the term
in the square brackets which is an amplitude that changes with
spatial position (since it depends on kx).
x
Showing just the original right going wave & the
envelope of the resulting standing wave
=
y(x, t)
[2A sin(kx)] cos(ωt)
The cosine term just oscillates back and forth between ±1 as time
progresses.
x
x=0 λ
4
The dashed green line shows the behavior of the sine.
2π
=
y(x) sin(kx)
= sin( x)
λ
Consider specific points:
λ
At x = :
4
λ
x
=
So at
:
4
2π
(recall k =
)
λ
2π λ
λ
π
y( ) = sin(
) = sin( ) = +1
4
λ 4
2
λ
y(
=
, t) 2A cos(ωt) oscillates between ±2A
4
giving an anti-node.
=
y(x, t)
[2A sin(kx)] cos(ωt)
x
λ λ 3λ
,
x = 0 , , ,λ
2π
4 2 4 =
y(x) sin(kx)
= sin( x)
λ
λ
At x = :
2
λ
2π λ
y( =
) sin(
) sin( π=
) 0
=
4
λ 2
λ
λ
So at x = : =
y( , t) 2A(0)cos(
=
ωt) 0 Giving a node.
2
2
3λ
2 π 3λ
3π
3λ
) = sin( ) = −1
At x = : y( ) = sin(
4
λ 4
2
4
3λ
3λ
So at x =
: y( , t) =2A( −1)cos(ωt) oscillates between 2A
4
4
giving an anti-node.
±
=
y(x, t)
[2A sin(kx)] cos(ωt)
x
λ λ 3λ
,
x = 0 , , ,λ
2π
4 2 4 =
y(x) sin(kx)
= sin( x)
λ
2π
y(
)
sin(
=
λ
=
λ) sin(2=
π) 0
At x = λ :
λ
y(λ, t) 2A(0)cos(
=
ωt) 0 Giving a node.
So at x = λ : =
So every half wavelength is a node with anti nodes in between
them, also separated by a half wavelength.
=
y(x, t)
[2A sin(kx)] cos(ωt)
x
For two counter propagating waves each of amplitude A the
amplitude of the standing wave is 2A.
But if the system has low damping (low friction) then the reflected
wave will reflect back off of the oscillated end again and then reflect
off the fixed end again and so on, until there are many more than two
waves adding to give a net amplitude for the standing wave that is
much larger than the input wave oscillation.
This is called resonance and is much like pumping your legs at just
the right time when you are on a swing. By doing that at the right
time (phase of the swing) the amplitude of the swing can get very
large even though the energy you are pumping in is relatively small.
Now such standing wave resonance can’t happen for just any
wavelength (or frequency). It can only happen if an integer number
of half wavelengths fit between the two ends of the string.
Because the input oscillation amplitude is so much smaller than the
resonant amplitude at the antinodes we can to good approximation
treat the system as if the oscillated (driven) end is also a node.
Mathematically, this condition for resonance on a string of length L
is,
λ
n   = L , n = 1, 2, 3, …
2
Solving for λ gives the wavelengths of the waves that can generate
standing waves on a string of this length,
λ=
2L
n
,
n = 1, 2, 3, …
Now since, v = λf ,
v
f=
λ
So since the wavelengths of waves that can generate standing waves
on a string of length L are,
2L
λ=
n
,
n = 1, 2, 3, …
Then the frequencies that can generate standing waves are,
v
v
nv
f n= =
=
,
2L
2L
λ
n
Where recall that for
waves on a string:
F
v=
µ
n = 1, 2, 3, …
String tension = F
linear density µ =
mass
length
The first few harmonic frequencies that can generate standing waves:
nv , n = 1, 2, 3, …
fn =
2L
Frequency Designation
v
f1 =
2L
2v
2L
2nd harmonic
3v
f3 =
2L
3rd harmonic
4v
f4 =
2L
4th harmonic
f2 =
…
A = anti-node
…
N = node
fundamental
or 1st harmonic
Example
The E string of a guitar has a length of 65 cm and is stretched
to a tension of 82 N. It vibrates at a fundamental frequency of
329.63 Hz. What is the mass of the string?
With fundamental,
nv
fn =
2L
with n = 1 for the fundamental, so,
v
f1 =
2L
So,
F
µ
f1 =
2L
But
F
v=
µ
Squaring both sides
F
µ
2
=
f1
4L2
F
µ
f12 = 2
4L
Solving for µ
F
µ= 2 2
4L f1
But
m
µ=
L
So
m
F
= 2 2
L 4L f1
F
82 N
=
m =
4Lf12 4(0.65 m)(329.63 Hz)2
m=
2.90 × 10 −4 kg =
0.29 gm