# Standing Waves

1. Jan 31, 2014

### tyneoh

Do all frequencies produce standing waves or just the harmonics? My physics textbook stated that standing waves are form when two wave trains with equal amplitude and frequency meet each other in opposite directions. Does the common frequency of the wave trains have to be one of the harmonics? Suppose we have two speakers (two sources) facing each other, both connected to the same signal generator. Will a standing wave form for all frequencies? In another situation where we have a single source, i.e. transverse waves on a string reflected from a fixed point, does the same principle apply?

2. Jan 31, 2014

### Meizirkki

In your example, two speakers facing each other, there will always be a point right in the middle of the two where air moves into neither direction no matter what kind of signal is input. So the answer is yes.

3. Jan 31, 2014

### Staff: Mentor

Without reflections, there are no harmonics at all - if the signal is a sine wave and if we restrict the waves to a single dimension (plus some other idealizations), you get standing waves everywhere. If the signal is more complicated, but constant in time, you can still talk about standing waves, but the waves will look more complicated. And if the signal varies with time, you can get arbitrary pattersn - just directly in the middle, you will never get moving air due to symmetry, as mentioned by Meizirkki.

4. Jan 31, 2014

### sophiecentaur

Do not confuse standing waves with resonances. You will always get a standing wave when there is a reflection. To get a resonance, in which overtones are reinforced, you need multiple reflections (as at either end of a string or pipe) which allow energy to build up when an exciting signal of the appropriate frequency is introduced.

5. Jan 31, 2014

### tyneoh

If so then standing waves are just resonance in work?

6. Jan 31, 2014

### BruceW

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.

7. Jan 31, 2014

### tyneoh

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?

8. Jan 31, 2014

### BruceW

the boundary conditions give you the equation for the harmonics. If there is only one solid boundary, you can choose any frequency you like, and you still get a standing wave. If you have two solid boundaries, then the frequency of the wave does have to be one of the harmonics, so that the boundary conditions are satisfied.

9. Jan 31, 2014

### tyneoh

So to answer my question, standing waves on a string attached to a fixed point do not require harmonics whereas standing waves between two speakers do?

10. Jan 31, 2014

### BruceW

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

11. Jan 31, 2014

### tyneoh

Also, are standing waves examples of resonance?

12. Jan 31, 2014

### AlephZero

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.

13. Feb 1, 2014

### tyneoh

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 travelling in opposite directions? For resonance only then you need harmonics?

14. Feb 1, 2014

### BruceW

yes, you can form a standing wave at any frequency as long as you have two waves of equal amplitude and frequency, travelling in opposite directions. Therefore, standing waves are (generally) not a kind of resonance, because they happen at any frequency. On the other hand, if you have a solid boundary at both sides of the string, then only the harmonic frequencies are allowed. This is a kind of resonance because only certain frequencies will appear.

15. Feb 1, 2014

### tyneoh

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?

16. Feb 1, 2014

### BruceW

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.

Last edited: Feb 1, 2014
17. Feb 1, 2014

### sophiecentaur

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.

18. Feb 1, 2014

### BruceW

In the strict sense of the word "fundamental frequency", the instrument can only produce frequencies that are integer multiples of the fundamental frequency. And I'm guessing this is the definition that other members of PF have heard of. So if someone says the fundamental frequency is $f_0$ but the instrument can produce a frequency $1.9f_0$ then they are not using the 'strict' definition. Instead, they might be using some definition, like if the frequencies are approximately integer multiples of some frequency, then we choose that frequency as an approximate fundamental frequency (in a certain sense). but this is not really a 'true' fundamental frequency.

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.

19. Feb 1, 2014

### sophiecentaur

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 overtone crystal. That means it is intended for a circuit designed to produce its specified frequency when operated in that mode. If you make it operate in its fundamental and then select the fifth harmonic, you will get the wrong frequency. You can say it is being overly picky of me but there have been threads on PF which have argued long and loud about far less significant differences.

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.

20. Feb 1, 2014

### AlephZero

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 approximately have 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 do not necessarily 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.

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