Adding Waves 2

Beginning of Unit II

Superposition of two tones leading to resonance, beats, disonance, etc.

Standing Waves on a String

Resonance occurs in a confined space when traveling waves are reflected so as to add or interfer constructively.  Consider the following sum of two waves on a string of length L =  0.5 m, wavelength λ = 1.0 m, and traveling with velocity v = 275. m/s.

t=0.; 
x=0.;
L=0.5;
n=1; 
λ=2L/n;
v=275.;
Animate[Plot[{0.5Sin[2Pi/λ(x-v t)],0.5Sin[2Pi/λ(x+v t)],0.5Sin[2Pi/λ(x-v t)]+0.5Sin[2Pi/λ(x+v t)]},{x,0.,0.5},PlotRange->{-1.0,1.0},PlotStyle->{RGBColor[1,0,0],RGBColor[0,1,0],RGBColor[0,0,1]}],{t,0,.01,.00005}, DefaultDuration->20]

Actual resonance depends on the correct matching of the wavelength to the string L

Because of the requirement that the ends of the string must remain fixed (we call these boundary conditions), λ = 2L/n where n is a positive integer starting with one.  For n = 1 there is a single lobe, n = 2 has two lobes, n = 3 has three lobes and  so on.  The fixed points in the standing wave pattern are called nodes.  The points where the maximum excursions occur, either postive or negative, are called antinodes.  In this case the number of nodes is n + 1 and the number of antinodes is given by n.  Adjust n to different integer values in the input cell above and observe the  different standing  wave patterns (blue curve).  The standing wave patterns are a linerar superposition of two traveling waves, one traveling in the positive x direction and one in the negative direction.  Can you tell which is which in terms of the analytical expressions in the input cell?

The different standing wave patterns represent the different harmonics of a bowed violin string, for example.  n = 1 is the  fundamental, n = 2 is the second harmonic and corresponds to a pitch an octave above the fundamental at twice the frequency, n = 3 is a twelfth above the fundamental or a fifth above the second harmonic.  n = 4 is two octave above the fundamental or a fourth above the  fifth. n=5 is a major 3rd above the fourth, and n = 6 is  a minor 3rd above the major 3rd.

Organ pipe open at both ends

The organ pipe open at both ends is an other example of resonance with the same spectrum of harmonics as the string fixed at both ends.  However, there are important differences.  In this case the antinodes are at the ends and for n = 1 there is a single node in the middle.  This is a resonance in the air of the pipe rather than in the wire of the vibrating string.  String waves are transverse waves, i.e., the direction of the motion is perpendicular to the direction of propagation of the traveling waves.  In the case of a pressure wave inside of an organ pipe, the waves are longitudinal, i.e., the variation of the pressure is in the same direction as the traveling sound waves.

In the example below, we introduce a phase factor of π radians between the two traveling waves.  This results in a single node in the middle with the antinodes at the ends for the case of n =1.  For larger values of n, the ends remain antinodes and additional nodes and antinodes are introduced as n increases.  Now the number of antinodes = n + 1, and the number of nodes = n.

This is a useful model for sound production in a flute.

This example can also be used to study the effect of varying the phase between the two traveling waves (change α), and by changing λ.  Try setting α = 0 and varying λ,  This gets us back to the string case.  Can you see why the resonance is spoiled if λ is not equal to 2L/n?

t=0.; 
x=0.;
L=0.5;
n=1; 
λ=2L/n;
v=275.;
α=1;
Animate[Plot[{0.5Sin[2Pi/λ(x-v t)],0.5Sin[2Pi/λ(x+v t)+α π],0.5Sin[2Pi/λ(x-v t)]+0.5Sin[2Pi/λ(x+v t)+α π]},{x,0.,0.5},PlotRange->{-1.0,1.0},PlotStyle->{RGBColor[1,0,0],RGBColor[0,1,0],RGBColor[0,0,1]}],{t,0,.01,.00005}, DefaultDuration->20]

Organ pipe closed at one end

Here the boundary conditions require a node at the closed end with a pressure antinode at the open end.  Since the separation between a node and an antinode in a standing wave pattern is 1/4 of a wavelength, the longest wavelenth the organ pipe can support is now λ = 4L.

t=0.; 
x=0.;
L=0.5;
n=1; 
λ1=4L/n;
v=275.;
Animate[Plot[{0.5Sin[2Pi/λ1(x-v t)],0.5Sin[2Pi/λ1(x+v t)],0.5Sin[2Pi/λ1(x-v t)]+0.5Sin[2Pi/λ1(x+v t)]},{x,0.,0.5},PlotRange->{-1.0,1.0},PlotStyle->{RGBColor[1,0,0],RGBColor[0,1,0],RGBColor[0,0,1]}],{t,0,.01,.00005}, DefaultDuration->20]

Setting n = 1 with λ = 4L in the example above, gives us exactly what we want namely a node at one end and an antinode at the other.  However, setting n =2 gets us back to the case of nodes at both ends so this is not allowed!  n = 3 again produces a node at one end and an antinode at the other with an additional node and antinode inside of the organ pipe.  Apparently the even harmonics are not allowed and n must be odd.  So our boundary conditions require that only odd harmonics can be resonant.  In other words λ = 4L/n where n = 1, 3, 5, 7, ...  Experiment with different values of n to convince yourself that only the odd harmonics can work.

This is a useful model for sound production in the clarinet.

The phenomenon of beats

If we add two sinusoidally varying pressure waves with vey nearly the same frequency at a point in space we will hear a phenomenon known as beats.  We will start by exploring the situation analytically.

t=0.; f=440.;Δf=40.;
Plot[{Sin[2Pi f t],Sin[2Pi (f+Δf) t],Sin[2Pi f t]+Sin[2Pi (f+Δf) t]},{t,0,.055},PlotRange->{-2.0,2.0},PlotStyle->{RGBColor[1,0,0],RGBColor[0,1,0],RGBColor[0,0,1]}]

In this example the differency in frequency is Δf = 40 Hz and the envelope pattern reaches a maximum every 0.025 s.  Hence the beat frequency  = 1 / 0.025 s = 40 Hz.  However, the underlying frequency is 1/2 this or 20 HZ.  The superposition has a frequency which is half way between the two frequencies or 460 Hz.  This is called a fused tone.

The above beat frequency is too large to be heard.  The sound would just sound very rough or disonant.  Beat frequencies of up to 10 Hz can be heard by the ear and are useful in fixing the intonation between two instruments.

Making the beat frequency larger

t=0.; f=440.;Δf=88.0;
Plot[{Sin[2Pi f t],Sin[2Pi (f+Δf) t],Sin[2Pi f t]+Sin[2Pi (f+Δf) t]},{t,0,.055},PlotRange->{-2.0,2.0},PlotStyle->{RGBColor[1,0,0],RGBColor[0,1,0],RGBColor[0,0,1]}]
1/88.

Perception of two tones

Notice the  three regions.  Region I is the beat region with a fused tone.  Region II still has a fused tone, but beats can no longer be distinctly hear--the sound is hear as rough.  Region III is where we can start to perceive two different pitches, but the sound is still rough or disonant.  Region IV start the region where the two tone are no longer perceived as rough, but rather as consonant.