Harmonic Superposition and Standing Waves
11 January 2016
PHYC 1290 Department of Physics and Atmospheric Science
Superposition of Harmonic Waves
y1 y2
y1 = A sin(kx - ωt) y2 = A sin(kx + ωt)⊖ for wave moving right ⊕ for wave moving left
Let them superpose:
➞ y = y1 + y2 = A sin(kx - ωt) + A sin(kx + ωt)
Consider 2 waves with same amplitude, wavelength and period:
A
-A0
t = π/ωt = 2π/3ω
t = π/3ωt = 0
Use the trig identity
sin B + sin C = 2 sin�
12(B + C)
⇥cos
�12(B � C)
⇥
with B = kx - ωt and C = kx + ωt .
➞ y = 2A sin(kx)cos(-ωt)
But cos(-θ) = cos(θ).
➞ y = 2A sin(kx)cos(ωt)
It doesn’t move left or right!
It is a “standing wave” of amplitude 2A.
y
xπ/k 2π/k2A
-2A0
Nodes Anti-nodes
Simulation
How to make a standing wave
fixed end
drivesinusoidally
When the wave is reflected from the fixed end, it will be inverted but will have the same k, ω and A.
➞ y = ... = 2A cos(kx) sin(ωt) (a standing wave)
➞ y = A sin(kx - ωt) - A sin(kx + ωt)
inverted moving leftmoving right
DemoStanding wave on a spring
By shaking the spring at different frequencies, we can see different “harmonics” m:
m = 1
m = 2
m = 3
m = 4
DemoStanding longitudinal wave on an aluminum rod
Running fingers along an aluminum rod (using lots of resin) excites longitudinal modes that we can hear.
LEnds can move(free, not fixed)
Hold herefor m=1
Hold here for m=2
How Does it Work?
yx
Node
m=1:
x
Node Node
y
m=2:
Standing wave modes:
L
m = 1
m = 2
m = 3
m = 4
Fixed-fixed ends
�2 =2L2
�3 =2L3
�4 =2L4
Harmonic numbers:
f2 = 2v
2L
f3 = 3v
2L
f4 = 4v
2L
�1 =2L1
Wavelengths:Frequencies
(using v = fλ):
f1 = 1v
2L
GeneralizationFor standing waves with fixed ends, the wavelength for
harmonic (or mode) number m is
�m =2Lm
The lowest frequency (f1) is called the “fundamental harmonic”:
f1 =v
2Lwhich allows us to write
fm = mf1
and the frequency is
fm = mv
2L
with m = 1, 2, 3, ...
Standing wave modes
L
m = 1
Free-free ends
Harmonic numbers:
m = 2
m = 3
The generalizations are the same as before:
�m =2Lm
f1 =v
2L andfm = mf1, with m = 1, 2, 3, ...
Standing wave modes
L
m = 1
m = 3
m = 5
Fixed-free ends.
Harmonic numbers:
The generalizations for fixed-free ends are different:
�m =4Lm
f1 =v
4L andfm = mf1, with m = 1, 3, 5, ...
Extra Material
Standing wave modes
L
m = 1 �1 =2L1
Free-free ends
Harmonic numbers:
m = 2 �2 =2L2
f2 = 2v
2L
m = 3 �3 =2L3
f3 = 3v
2L
Wavelengths:Frequencies
(using v = fλ):
f1 = 1v
2L
Conclusion: We can use the same generalizations as before.
�m =2Lm
f1 =v
2L andfm = mf1, with m = 1, 2, 3, ...
Standing wave modes
L
m = 1
m = 3
m = 5
Fixed-free ends
Harmonic numbers: Wavelengths:
Frequencies (using v = fλ):
�1 =4L1
�3 =4L3
�5 =4L5
f1 = 1v
4L
f3 = 3v
4L
f5 = 5v
4L
Conclusion: The generalizations for fixed-free ends are different.
�m =4Lm
f1 =v
4L andfm = mf1, with m = 1, 3, 5, ...
Standing wave on a stringHere is a 60 Hz standing wave (m=2) on a string. Part way through a strobe light is used to isolate the string’s motion.
http://youtu.be/rkeIubVQ0N8
http://youtu.be/rkeIubVQ0N8
ExampleMiddle-C (C4) on a piano has fundamental frequency of 262 Hz.
Q: What are the frequencies of the Harmonics?
A: The string is fixed at both ends, so fm = mf1.
m=1: f1 = 262 Hz (given)
... and so on.
m=2: f2 = 2×f1 = 524 Hzm=3: f3 = 3×f1 = 786 Hz
Q: If the string is 62.5 cm long, what are the wavelengths?
A: For fixed ends, λm = 2L/m , som=1: λ1 = 2L/1 = 125 cmm=2: λ2 = 2L/2 = 62.5 cmm=3: λ3 = 2L/3 = 41.67 cm
Q: What are the wave speeds on the string?
A: Using, v = fλ,m=1: v1 = f1×λ1 = 32750 cm/sm=2: v2 = f2×λ2 = 32750 cm/s m=3: v3 = f3×λ3 = 32750 cm/s
The speeds are all the same! This is expected because the wave speed v depends only on the medium’s properties.
Q: What tension does the middle-C string have?
A: The wave speed on a string is given by
v =
�T
µ� T = µv2
We need to first determine the linear density μ=m/L.
➞ μ = m/L = density × πr 2 = 0.00614 kg/m
➞ T = 659 N
v =
�T
µ� T = µv2
Piano wire is made of steel which has density 7820 kg/m3. The middle-C string is 1 mm in diameter.
➞ m = density × volume = density × πr 2L
659 N is a lot of tension for one string! This is equivalent to having the tension on one string
drawn by a 67 kg (148 lb) mass.
There are normally about 230 strings in a piano (although there are 88 keys, some keys have
multiple strings).
This means there is a combined tension of over 150,000 N (15,000 kg or 30,000 lbs) in a piano!!
For this reason, at the heart of a piano is a cast-iron frame (or harp, pictured left) needed to
sustain the massive tension.
Image souce: http://commons.wikimedia.org/wiki/File:Fluegel-Rahmen.jpg
http://commons.wikimedia.org/wiki/File:Fluegel-Rahmen.jpg
Here is a view inside a grand piano. You can see the strings drawn across the cast-iron frame.
Image source: http://commons.wikimedia.org/wiki/File:Bosendorfer_185.JPG
http://commons.wikimedia.org/wiki/File:Bosendorfer_185.JPGTop Related