# The Naked Scientists Forum

### Author Topic: What is the shape of a vibrating string?  (Read 11872 times)

#### JP

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##### What is the shape of a vibrating string?
« Reply #25 on: 22/08/2011 03:24:53 »
And it's not quite correct to say it's A sine wave for the fundamental. It's actually two added together:

sin(x-wt) + sin(x+wt)

I'm probably thick, but I don't see why it necessarily follows that the shape of the string is sinusoidal at the fundamental (although I suspect it is).

The thing that is sinusoidal is the orthogonal displacement of any element of the string (assuming, of course, that the oscillation of the string never decays) and it is that displacement, or really those displacements, that produce the note.

Geezer, it's a sinusoid because it's defined that way.  If you solve for the physics of a string, the displacement of the string is a wave equation.  If you solve the wave equation subject to the conditions that the ends of the string are fixed, you can of course get a bunch of funky shapes that aren't sinusoids.  But these shapes aren't modes, either.  The modes are defined to be sinusoids, and the fundamental mode is the lowest frequency sinusoid allowed by physics (aside from an unmoving string, which is technically a sinusoid of infinitely long wavelength :p ).

The reason why we bother defining modes at all has to do with the way the physics can be described mathematically.  No matter what shape the vibrating string takes, it should be the sum of a bunch of modes.  To specify the more complex shape, you only need to specify the coefficients of the modes, which is a set of numbers.  It's often easier to deal with a set of numbers than to try to find an equation for an arbitrary wave.  This is especially true if you're representing waves in a computer, where you can figure out how many modes actually are important, and truncate your set of numbers there, making a very efficient way of storing that wave's complex shape.

#### Geezer

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##### What is the shape of a vibrating string?
« Reply #26 on: 22/08/2011 07:53:23 »
Geezer, it's a sinusoid because it's defined that way.  If you solve for the physics of a string, the displacement of the string is a wave equation.  If you solve the wave equation subject to the conditions that the ends of the string are fixed, you can of course get a bunch of funky shapes that aren't sinusoids.  But these shapes aren't modes, either.  The modes are defined to be sinusoids, and the fundamental mode is the lowest frequency sinusoid allowed by physics (aside from an unmoving string, which is technically a sinusoid of infinitely long wavelength :p ).

The reason why we bother defining modes at all has to do with the way the physics can be described mathematically.  No matter what shape the vibrating string takes, it should be the sum of a bunch of modes.  To specify the more complex shape, you only need to specify the coefficients of the modes, which is a set of numbers.  It's often easier to deal with a set of numbers than to try to find an equation for an arbitrary wave.  This is especially true if you're representing waves in a computer, where you can figure out how many modes actually are important, and truncate your set of numbers there, making a very efficient way of storing that wave's complex shape.

Pardon me JP, but you have obviously mistaken me for someone who might have a clue about the math involved

#### wolfekeeper

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##### What is the shape of a vibrating string?
« Reply #27 on: 22/08/2011 13:55:05 »
Geezer, it's a sinusoid because it's defined that way.
No, that's not quite right I think. Sinusoids aren't defined in terms of this, nor vice versa.

It's pretty much an accident of physics.

I suppose it's because there's two modes of energy storage here: potential energy (energy stored in the tension) and kinetic energy (energy of motion), and the energy oscillates between the two in a similar way that distances change along the x and y axes as you spin around a circle.
« Last Edit: 22/08/2011 13:56:52 by wolfekeeper »

#### JP

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##### What is the shape of a vibrating string?
« Reply #28 on: 22/08/2011 14:15:28 »
Geezer, it's a sinusoid because it's defined that way.
No, that's not quite right I think. Sinusoids aren't defined in terms of this, nor vice versa.

It's pretty much an accident of physics.

I suppose it's because there's two modes of energy storage here: potential energy (energy stored in the tension) and kinetic energy (energy of motion), and the energy oscillates between the two in a similar way that distances change along the x and y axes as you spin around a circle.

It's not defined to be a sinusoid, but it is defined in terms of a certain mathematical construction.  For a vibrating string, fixed at both ends, this definition requires it to be a sinusoid.

If you go through the derivation of waves on a string, you get the wave equation: http://en.wikipedia.org/wiki/Vibrating_string#Derivation

Modes are defined here: http://en.wikipedia.org/wiki/Normal_mode:  "A normal mode is a mode of a linear field among a chosen set of orthogonal modes."  That's fancy mathematical language which places requirements on what a mode is.  This, along with the underlying physics (the wave equation + both ends of the string fixed), require each mode to be a sinusoid in the case of a vibrating string.

More accurately, it's a sinusoid in position along the string, and it's amplitude also oscillates sinusoidally in time.

#### wolfekeeper

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##### What is the shape of a vibrating string?
« Reply #29 on: 22/08/2011 15:48:19 »
No. Nature doesn't use maths. Humans use maths to predict what nature does. Maths is just a collection of patterns that humans have noticed about the world and about patterns in patterns. And we give these patterns symbols and we write them down, and we call that maths.

But nothing, ever, in nature is due to our human definitions.

The fact that the fundamental must be a sine wave is because of the relationship between the energy modes and the x/y dimensions of something spinning around a circle. It's the same pattern. That's why it's sinusoidal. There's probably other ways to see it as well.

Your argument that you just have to solve the wave equation doesn't really solve anything, since ANY waveform can travel obeying the wave equation. Square, triangular, anything. You need to show why sinusoids are so common.

#### JP

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##### What is the shape of a vibrating string?
« Reply #30 on: 22/08/2011 17:59:48 »
No. Nature doesn't use maths. Humans use maths to predict what nature does. Maths is just a collection of patterns that humans have noticed about the world and about patterns in patterns. And we give these patterns symbols and we write them down, and we call that maths.

But nothing, ever, in nature is due to our human definitions.
I completely agree, and I wasn't arguing otherwise.  But modes are a human definition.  What we do in physics is to look at the string and try to model it by mathematics.  Of all the waves that can be described on the string by our model, some have a special property, and we call those modes.  That property is that the waves form a complete, normal set.  That's math-speak to say that we can mathematically describe, with our model, any wave that exists on the string by adding up a bunch of these modes, and also that any one mode cannot be written in terms of any other modes.

The real power is what I've mentioned twice above: that you can mathematically describe a wave on a string in terms of just a list of the coefficients of these modes, i.e. a list of numbers rather than a very complicated equation.

Quote
The fact that the fundamental must be a sine wave is because of the relationship between the energy modes and the x/y dimensions of something spinning around a circle. It's the same pattern. That's why it's sinusoidal. There's probably other ways to see it as well.
I'm not sure what you're getting at here, but the fundamental mode is a sine wave because all modes are sine waves for the string.  There aren't really other ways of getting at this, since modes are a mathematical definition within our model, and you only get them by enforcing completeness and normality.

Quote
Your argument that you just have to solve the wave equation doesn't really solve anything, since ANY waveform can travel obeying the wave equation. Square, triangular, anything. You need to show why sinusoids are so common.
Actually it does solve everything.  Modes are a particular wave within our mathematical model, so it makes sense we have to solve the equation on which we base our model.  As I mentioned, these more complex waves don't meet the requirements of what we call modes, but within our model, we can mathematically express them as a sum of modes.

#### wolfekeeper

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##### What is the shape of a vibrating string?
« Reply #31 on: 22/08/2011 18:23:10 »
No, sorry. Point of fact you can analyse vibrations using many different bases, including various orthogonal square waves. You're essentially implicitly using fourier analysis, but you absolutely don't have to.

In fact, for example, in image processing, sine waves are less commonly used since sine waves are mathematically cumbersome.

What you're saying actually doesn't explain it at all; it's more of an 'when you mathematically do this, it works' argument, which is 'it just happens to be true' rather than a 'it must be a sine wave because...' argument which is what the original question was.

#### Geezer

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##### What is the shape of a vibrating string?
« Reply #32 on: 22/08/2011 19:26:31 »
Oh dear! It seemed like an innocent enough question, but now I'm not so sure It will take me a considerable amount of time to try to catch up here.

#### JP

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##### What is the shape of a vibrating string?
« Reply #33 on: 23/08/2011 16:07:10 »
No, sorry. Point of fact you can analyse vibrations using many different bases, including various orthogonal square waves. You're essentially implicitly using fourier analysis, but you absolutely don't have to.

You have a good point.  I was misled by talking about modes, rather than harmonics.  We're actually talking about the harmonics here.  By definition, harmonics are solutions to the Helmholtz equation.  What that physically means for this string is that they're waves whose vibration in time is of a single frequency (in the Fourier sense).  Because of the physics of having both ends of the string tied down, these waves bounce back and forth and form standing waves.

These harmonics are also members of a complete, normal basis, but you're right--they certainly aren't the only possible basis.  They are the only one that is made up of harmonics.

I can see why the argument that they're special because they're harmonics or standing waves makes sense, and I agree those are special properties.  But I still think their use is primarily mathematical.  If you pluck a string, you don't get just one mode.  You get a traveling wave of a complex shape (probably a triangle).  The use of modes in this case is to describe the wave in terms of it's harmonic solutions.

The harmonics might also be useful because as the wave dies out, the higher order modes, which vibrate faster, will probably tend to die out sooner, due to air resistance.  As long as the original wave had a lot of the fundamental present, the wave will probably look a lot like the fundamental as the higher frequency harmonics die out.

#### Geezer

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##### What is the shape of a vibrating string?
« Reply #34 on: 23/08/2011 18:41:41 »
Er, well, what I was getting at is, in the case where there is only the first harmonic, you can model the string as a large number of masses, each moving with simple harmonic motion orthogonal to the "center line" of the string. If there is no damping, there never will be any other harmonics and each mass will oscillate indefinitely.

The envelope formed by the amplitudes of the oscillations of the masses determines the shape of the string. (Now for the tricky bit.)

Why does that insist that the envelope is sinusoidal? I'm pretty sure it is, but it's not obvious to me why it has to be. If it turned out to be triangular, as long as all the masses oscillated with the same frequency and phase, there would still only be a single harmonic.

The answer, I think, is because the tension in the string varies sinusoidally due to the sinusoidal orthogonal displacement.
« Last Edit: 23/08/2011 18:43:25 by Geezer »

#### JP

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##### What is the shape of a vibrating string?
« Reply #35 on: 23/08/2011 19:37:35 »
The answer, I think, is because the tension in the string varies sinusoidally due to the sinusoidal orthogonal displacement.

Yeah, essentially.  Of course, your initial displacement has to be sinusoidal for this to work.  If you pluck it as a triangle shape you don't get a nicely behaved sine wave.  I believe that if you let the vibration go on for a while, the higher frequency harmonics that make up that triangle will be damped, so you will end up with something that looks like the sine wave, since that's the most resistant to air resistance.

#### Geezer

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##### What is the shape of a vibrating string?
« Reply #36 on: 23/08/2011 21:16:53 »
If you pluck it as a triangle shape you don't get a nicely behaved sine wave.  I believe that if you let the vibration go on for a while, the higher frequency harmonics that make up that triangle will be damped, so you will end up with something that looks like the sine wave, since that's the most resistant to air resistance.

Yes. I think so. Isn't there also less energy in the higher harmonics to start with?

#### JP

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##### What is the shape of a vibrating string?
« Reply #37 on: 23/08/2011 21:24:58 »
Higher order harmonics tend to be important if there's a lot of fine structure in the wave.  (Which makes sense from the Fourier standpoint.)  If you pluck it as a fairly smooth wave to begin with, there won't be a lot of energy in the higher order harmonics from the start.

Even if you somehow put a lot of energy into the high order harmonics, they're higher temporal frequency, so they oscillate faster than the low order ones, so I would expect them to lose that energy to air resistance (and sound waves) quickly.

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##### What is the shape of a vibrating string?
« Reply #37 on: 23/08/2011 21:24:58 »