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tyneoh

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tyneoh

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Meizirkki

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mfb

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sophiecentaur

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tyneoh

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If so then standing waves are just resonance in work?

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BruceW

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the allowed frequencies are restricted by the boundary conditions. So for transverse waves which are reflected, there is only one boundary, and you can choose any frequency you like, to create a standing wave.

with the two speakers, if there is a pure frequency from each, and the air inbetween the speakers is really a superposition of the two waves, then it is similar to having a string inbetween two boundaries. So in this case, only certain frequencies are allowed.

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tyneoh

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the allowed frequencies are restricted by the boundary conditions. So for transverse waves which are reflected, there is only one boundary, and you can choose any frequency you like, to create a standing wave.

with the two speakers, if there is a pure frequency from each, and the air inbetween the speakers is really a superposition of the two waves, then it is similar to having a string inbetween two boundaries. So in this case, only certain frequencies are allowed.

For transverse waves on a spring, doesn't the applied frequency have to be one of the harmonics to form standing waves? By boundary conditions, I assume those are the variables in the fundamental frequency formula?

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BruceW

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tyneoh

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BruceW

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yep. sounds about right. unless the string is attached to two fixed points.

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tyneoh

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Also, are standing waves examples of resonance?

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AlephZero

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Also, are standing waves examples of resonance?

I think that is the wrong way round. A resonance is a good example of a standing wave, but you can have standing waves without resonance.

Note, that statement is true for the type of vibrating systems you will study in a first physics course, but it is not true for all vibrating systems. Some examples of resonances without standing waves, are propeller whirl in aircraft, brake squeal, or oil whip in the bearings of rotating machinery. Those all involve devices that are rotating, but the vibrations are not "standing waves" either when viewed either from a position fixed to the ground, or a position fixed to the rotating structure. The "resonance" rotates around the axis of rotation, but not at the same angular velocity as the rotating machine.

You probably won't meet any of those unless you do a grad-school-level course in rotating machinery and/or nonlinear dynamics, so don't worry about them - but they are all important "real world" engineering problems.

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tyneoh

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I am a little bit confused though, do standing waves form at all frequencies as long as you have two waves of similar amplitude and frequency traveling in opposite directions? For resonance only then you need harmonics?I think that is the wrong way round. A resonance is a good example of a standing wave, but you can have standing waves without resonance.

Note, that statement is true for the type of vibrating systems you will study in a first physics course, but it is not true for all vibrating systems. Some examples of resonances without standing waves, are propeller whirl in aircraft, brake squeal, or oil whip in the bearings of rotating machinery. Those all involve devices that are rotating, but the vibrations are not "standing waves" either when viewed either from a position fixed to the ground, or a position fixed to the rotating structure. The "resonance" rotates around the axis of rotation, but not at the same angular velocity as the rotating machine.

You probably won't meet any of those unless you do a grad-school-level course in rotating machinery and/or nonlinear dynamics, so don't worry about them - but they are all important "real world" engineering problems.

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BruceW

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tyneoh

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Okay I am starting to see the gist of it, but could you care to explain why harmonics are required for solid boundaries but not for open-ended conditions? Do sound waves in a pipe count as open-ended or bounded?

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BruceW

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harmonics are required for both open-ended boundary and closed-ended boundary. In either case, they give a boundary condition. But if you have only one closed end, and no open-end or closed-end on the other side, then harmonics are not required. I think it is important to keep in mind that an open-ended boundary is a specific boundary condition. It is possible to have neither open-ended, nor closed-ended. And it is in this case, that harmonics are not required.

really, I was saying "solid boundary" before, just to relate it to your example of two speakers facing each other. They are an example of closed-ended boundary, thinking of the sound wave in terms of displacement of molecules from equilibrium. Interestingly, we can instead think of the sound wave in terms of pressure. If we think in this way, then the two speakers are both open-ended boundary, in terms of pressure. And of course, we will still require harmonics in either case.

edit: durr, actually I think it is the other way around for the speakers. I think they are an open-ended boundary in terms of the displacement of molecules from equilibrium. And so, they are a closed-ended boundary in terms of the pressure.

really, I was saying "solid boundary" before, just to relate it to your example of two speakers facing each other. They are an example of closed-ended boundary, thinking of the sound wave in terms of displacement of molecules from equilibrium. Interestingly, we can instead think of the sound wave in terms of pressure. If we think in this way, then the two speakers are both open-ended boundary, in terms of pressure. And of course, we will still require harmonics in either case.

edit: durr, actually I think it is the other way around for the speakers. I think they are an open-ended boundary in terms of the displacement of molecules from equilibrium. And so, they are a closed-ended boundary in terms of the pressure.

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sophiecentaur

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Okay I am starting to see the gist of it, but could you care to explain why harmonics are required for solid boundaries but not for open-ended conditions? Do sound waves in a pipe count as open-ended or bounded?

For a resonating systems, the amplitudes of the standing waves are much greater - because energy gets stored in the resonance mechanism. If you have two sources of amplitude A, firing towards each other, the maximum amplitude of any standing wave will just be 2A, if the waves go off to infinity. If, instead, you have a single source and two reflecting boundaries, the waves will travel up and down the space between and, when the frequency of the source corresponds to one of the overtones of the resonator, you will get standing waves with amplitudes of nA, where the proportion of energy lost per cycle is 1/n. (n is a measure of the Q of the resonator). When the input is not at one of the resonant frequencies, the energy is lost and never builds up - hence there are no identifiable nodes and antinodes.

I use the word "overtone" despite the fact that the rest of PF seems to be blind to it (sob sob) because it is only in very simple resonators that the higher order resonances are actually at harmonics of the fundamental. It may be 'near enough' for some stringed instruments to talk about harmonics but most other instruments have a very weird set of overtones that do not coincide with harmonics unless the player does some 'bending' and fudging. That is one reason why good performers sound better than beginners.

Open ended or closed resonators still have a boundary and, hence, reflections. Simple standing waves will occur when there is no boundary on the other end. i.e. a perfectly matched termination, corresponding to all the energy escaping. The actual position of the 'open end' of a pipe is difficult to locate because of the 'end effect'. The closed end is easier to define. It's easy to see why strings are the most common resonators to be used as examples for getting a basic understanding.

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BruceW

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Anyway, sorry I'm kinda taking things off-topic... I was just interested if you would agree with that or not? I would guess that is what people mean. But I'm not sure.

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sophiecentaur

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Anyway, sorry I'm kinda taking things off-topic... I was just interested if you would agree with that or not? I would guess that is what people mean. But I'm not sure.

I know that 'everyone' in everyday life talks about harmonics and that many musical instruments do produce higher frequencies that are pretty much integer multiples of fundamental frequencies. I also know that there are a number of examples in everyday life that the high order resonances are significantly different from harmonics. Buy an oscillator crystal and it may well be marked as a 5th

I don't understand why a definition which is ' good enough' for your average user is good enough when we are trying to treat things in the depth that is normal for PF. Overtone is a well know and accepted term and it's well defined so why not encourage its appropriate use?

But you are right, it would probably be better to continue on a new thread because it's not quite on - topic when dealing with ideal, one dimensional linear resonators.

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AlephZero

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In the strict sense of the word "fundamental frequency", the instrument can only produce frequencies that are integer multiples of the fundamental frequency.

That is not true, in the real world. You might be confusing two different things. One is "free vibration" where an external force is applied to start the system vibrating, and no force is applied after that time. Musical examples would be plucking a guitar string, or striking a drum or a bell. In general the different modes of vibration do not have any simple relation to the fundamental frequency. Even for a stretched string, they only

The other situation is "forced vibration" where you apply a continuous oscillating force to make the system vibrate. Musical examples are wind instruments, or a violin played using a bow. If the sound wave produced is periodic (but not necessarily a pure sine wave), you can do a Fourier analysis and represent it as a set harmonics which are at exact integer multiples of the fundamental frequencies, but those harmonics do

These two things are easily confused in a first course on sound and vibration, because for the two systems that are usually considered (vibrating strings, and pipes with constant cross section area) the natural vibration frequencies are in fact close to integer multiples of the fundamental.

The nearest you get to seeing the difference in a typical first course is the "end correction" for the resonant frequency of pipes, but that is usually just mentioned as a factoid to be used in solving textbook questions, not as something with any deep significance.

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sophiecentaur

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That is not true, in the real world. You might be confusing two different things. One is "free vibration" where an external force is applied to start the system vibrating, and no force is applied after that time. Musical examples would be plucking a guitar string, or striking a drum or a bell. In general the different modes of vibration do not have any simple relation to the fundamental frequency. Even for a stretched string, they onlyapproximatelyhave frequencies that are integer multiples of the fundamental.

The other situation is "forced vibration" where you apply a continuous oscillating force to make the system vibrate. Musical examples are wind instruments, or a violin played using a bow. If the sound wave produced is periodic (but not necessarily a pure sine wave), you can do a Fourier analysis and represent it as a set harmonics which are at exact integer multiples of the fundamental frequencies, but those harmonics donotnecessarily correspond to the natural vibration frequencies of the instrument.

These two things are easily confused in a first course on sound and vibration, because for the two systems that are usually considered (vibrating strings, and pipes with constant cross section area) the natural vibration frequencies are in fact close to integer multiples of the fundamental.

The nearest you get to seeing the difference in a typical first course is the "end correction" for the resonant frequency of pipes, but that is usually just mentioned as a factoid to be used in solving textbook questions, not as something with any deep significance.

You put that rather nicely, A0. It's a topic (like colour) with which everyone has familiarity but it can be over-familiarity when it comes to the details.

That's an interesting comment and I wonder how that actually applies to a real instrument. The bit I wonder about is the "If the sound wave produced is periodic" bit. That sort of assumes that a bow or breath cannot be exciting overtones as well - unless you mean a very long periodic. I know this is hardly relevant for a string but I don't see why the overtone (non harmonic) could't be present. In fact, would actual harmonics only be generated if there were nonlinearities?

No wonder this stuff isn't considered in a 'first course".

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AlephZero

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The bit I wonder about is the "If the sound wave produced is periodic" bit. That sort of assumes that a bow or breath cannot be exciting overtones as well - unless you mean a very long periodic. I know this is hardly relevant for a string but I don't see why the overtone (non harmonic) could't be present. In fact, would actual harmonics only be generated if there were nonlinearities?

Yup, the "if" is an assumption (and probably never true for a physical instrument, if you look in enough detail).

If everything is linear (another assumption!) and the excitation force is periodic, you will only excite the (non harmonic) modes when the excitations starts and stops. Actually the transient sounds at the start are often quite important in terms of humans identifying one instrument from another.

If there are nonlinearities, you can excite all sorts of "hoots and whistles" that are not necessarily related to anything in an obvious way. This can be intentional (e.g. the "growls" that jazz musicians produce) or accidental (listen to any beginner trying to learn the violin, for example)

Real instruments often produce a lot of broadband noise, both from the initial transient, and continuously, as well as the nominally periodic sound. For wind instruments, this is often described as a "breathy" sound.

In fact some of it is still being actively researched. Measurement techniques have come a long way since the first attempts to make scientific sense of all this back in the 19th century!No wonder this stuff isn't considered in a 'first course".

And some of the early attempts were wrong. For example the Helmholtz (who was nobody's fool as a scientist, in several branches of physics) sent the development of brass instruments up a blind alley for about 50 years, before people got brave enough to say the boundary conditions in his models were completely wrong, but he had made two big mistakes that nearly canceled each other out!

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BruceW

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we are using different definitions of "fundamental frequency". To be honest, your definition is probably more useful. (and I'd guess your definition is the standard definition, too).That is not true, in the real world. You might be confusing two different things. One is "free vibration" where an external force is applied to start the system vibrating, and no force is applied after that time. Musical examples would be plucking a guitar string, or striking a drum or a bell. In general the different modes of vibration do not have any simple relation to the fundamental frequency. Even for a stretched string, they onlyapproximatelyhave frequencies that are integer multiples of the fundamental.

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sophiecentaur

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we are using different definitions of "fundamental frequency". To be honest, your definition is probably more useful. (and I'd guess your definition is the standard definition, too).

The fundamental frequency of a resonator is the same surely, whether responding to an impulse or a continuous wave. It's the higher orders where frequencies may (or may not) be different.

I am feeling more an more confident that it's the overtones that give instrumental colour - despite what the old electronic synthesiser makers would claim. The shortfall in their performance, compared with sampling types is likely to be due to the 'harmonic synthesis' approach as much as any other reason.

Whilst I am about it, I feel I must put the boot in against the Discrete Fourier Transform as the universal tool for the analysis of sound. The term "instantaneous frequency" is bandied about with no regard to its real meaning and context. A DFT of the waveform from many musical instruments could give some very suspect answers about the 'real' structure of its spectrum. (I have to admit that I don't know of a 'better hole to go to', though.)

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BruceW

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AlephZero

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To repeat, I think you are getting confused between the modes of (free) vibration of an object, and the components of a Fourier series for a periodic motion.

If you personally want to define "fundamental frequency" your way, that's fine - so long as ypu realize that by your definition, a guitar and a piano are NOT "musical instruments", to give just two examples.

If you disagree about the physics, sorry, but what you learned about vibrating strings in a high school level course on sound isn't the whole truth.

See http://www.simonhendry.co.uk/wp/wp-content/uploads/2012/08/inharmonicity.pdf for some real world data, on pianos.

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BruceW

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[tex]T\frac{\partial^2 y}{\partial x^2} - QSK^2 \frac{\partial^4 y}{\partial x^4} = \sigma \frac{\partial^2 y}{\partial x^2}[/tex]

and he defines the fundamental frequency as ##1/(2L) \sqrt{T/\sigma}##. And due to the higher terms in the equation, the frequencies will not necessarily be integer multiples of the fundamental frequency.

So in this definition, is the fundamental frequency useful at all? I guess maybe it is, in the case that the higher-order terms are only small, so we can make a perturbation of some kind. OK, so I have changed my mind a bit. But I would still argue that if the instrument is not approximately harmonic, then the 'fundamental frequency' as he defines it, has no meaning.

- #28

sophiecentaur

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The problem only arises when one refers to those " partials" in the link as 'harmonics'. The word 'fundamental' is common to both, IMO.

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