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Since I was a kid I was always wondering why is it that when I sing in a small room (i.e. bathroom!), whenever I touch a certain frequency, the whole room vibrates sympathetically.

What are the parameters that make an acoustic body (be it a room, cell or whatever) resonate to a specific note?

Is there a way to reproduce the resonance in other acoustic bodies? I'm wondering if we can encapsulate an instrument in a box that changes its size according to the played pitch to amplify it and make it resonate in this wild manner.

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How funny, I was just doing this a few days ago at the bathroom at my work: humming a low frequency that was resonating strongly. Since I was a kid, too, I had noticed that –  Michael Martinez May 1 at 19:23

4 Answers 4

up vote 19 down vote accepted

Because of dynamics called room modes.

Room modes are the collection of resonances that exist in a room when the room is excited by an acoustic source such as a loudspeaker. (...) each frequency being related to one or more of the room's dimension's or a divisor thereof.

To keep things simple, we will assume the room has 6 parallel walls (right prism or cube) and will just focus on one type of room mode: axial (because it's the simplest, and it is predominant over other modes). The calculations of the room modes of more complex shapes can get very complex, very fast.

The relevant parameters are the dimensions and shape of the room, and the wavelength of the sound. When the wavelength of the sound and the dimensions of the room coincide, there will be resonance. Similar to how a string will resonate to specific frequencies depending on its length (and tension, and materials, etc).

To make a room resonate:

  1. Measure one dimension of the room: either height, length, or width. For the formula we are using we will measure it in meters.

  2. Find the frequency that has a wavelength that is double the length of one of the three dimensions of your room. It's easier than it sounds, just use this formula:

    Frequency = 1/2 x 343 m/s / the dimension of the room that you just measured

    Where 343 m/s is the speed of sound in meters per second (an approximation, since it is not constant and depends on many factors) If you want to use other length unit, you can convert the speed of sound to your unit of choice, and use that unit in the measurement of the room and calculations.

  3. Playback (or sing, if it's within your vocal range) that frequency inside that room.

For example, suppose there is a distance of 4 meters between the walls of your bedroom. The frequency we want would be 1/2 x 343 m/s / 4m = 42 Hz (that's between a E and an F).

Smaller rooms resonate to higher frequencies. That's why you noticed this in the bathroom using your voice, since a small room might have a room mode in the frequency range of the human voice. With one meter between walls: 1/2 x 343 m/s / 1 m = 171.5 Hz (very close to an F).

A given room also has higher resonating frequencies. If 42 Hz is a resonating frequency of your bedroom: then 42 Hz * 2 = 84 Hz, 42 Hz * 3 = 126 Hz and so on also are resonating frequencies of the room.1 Because the lower resonating frequencies of a room are usually very low, these might be easier to make resonate.


  1. Actually, waves which “bounces” against perpendicular walls can also resonate, but the formula which corresponds to them is slightly more complicated. If l, w and h are the length, width and height of a room (in metres), and for any integers (whole numbers) n, p and q, the frequency F given by the following formula resonates:

F = 1/2 * 343 m/s * square_root ( n^2 / l^2 + p^2 / w^2 + q^2 / h^2 ).

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Am I right that different positions within the room will resonate differently? –  Bob Broadley Apr 29 at 12:39
    
@BobBroadley no, but when you stand in a corner, you get a bunch of echoes w/ very small delay, leading to the impression of resonance. –  Carl Witthoft Apr 29 at 12:45
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I also wanted to add that singing the lowest resonating frequency of a room is unrealistic except for the smallest rooms. 42 Hz is nearly one octave below what is generally required of a bass in a choir. –  Édouard Apr 29 at 13:06
    
@Édouard That's a very important note. It is addressed at the end of the answer, but I might have needed to be clearer about it. Will add more about it to the answer. –  JCPedroza Apr 29 at 13:07
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@Tim You can! But you might need more than a few measurements. Bob McCarthy's Sound System Design and Optimization is a good place to start: amzn.com/B00BMS2QLA –  JCPedroza Apr 29 at 15:02

JCPedroza's answer is correct for a square room, but I think it's worth pointing out that the shape of the room is not just it's dimensions. For example, a square room with an open window will act different than if the window is shut. In acoustics, we often model the response of a room as a circuit. If you break up the space into pieces, each piece can be represented as a circuit element. There are multiple analogies you can make, but as one example, a tube of air acts like an acoustic inductor, and a box of air acts like an acoustic capacitor. If you draw an equivalent representative circuit for the geometry you're interested in and work out the circuit analysis, you can figure out exactly what will happen.

So how does a resonance happen? Well, a big empty room is going to have a large acoustic capacitance, but it will also have a small acoustic inductance. When a capacitor and inductor are connected to each other, they begin to slosh energy back and forth between each other. There's one frequency where the sloshing reaches a maximum, and that's the resonance frequency. So that's why you will hear the resonance frequency of the room louder than you normally would (at omega = 1/sqrt(LC)). There are equations you can use to measure the acoustic inductance and capacitance of a room, but it's quite a lot of work.

Two rooms can have the same resonant frequency, but one may seem to ring longer than another. In circuit analysis, we call that the Q (quality) factor of a resonator, and it depends on the resistance in the circuit. You can add more resistance by putting a bunch of furniture or carpet or sound panels in the room, which will cause the ringing to stop much sooner, since they'll dissipate energy and prevent it from being sloshed back into the capacitor or inductor. In an electric circuit, that's dissipated as heat, and in a room with a sound wave, it's also dissipated as heat!

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This is a very interesting answer. I don't understand it, but it's still a very interesting answer. I'll have to study it awhile. –  BobRodes Apr 29 at 18:57
    
I think we have to invent an expandable acoustic box that customizes its size/shape according to the played frequency. It will give a very warm sound I'm sure. Question is if the size must be a room or we can trick it with different means like tubes etc. LOL. –  Shimmy Apr 30 at 21:44
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Shimmy, you've just described a speaker! Of course, the box doesn't change shape according to the frequency, but instead we just make a couple different "rooms" that are good at certain frequency ranges. –  blueintegral May 1 at 2:38

One form of the phenomenon is called the "standing wave". As it applies to rooms or acoustic spaces, it is sometimes refered to as "resonant room mode".

Here is a link to an article in Audioholics Magazine that discusses how these phenomena affect room acoustics.

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I'm wondering if we can make an instrument that changes its size according to the played tone to amplify it and resonate it in this wild manner.

Well, most wind instruments work in that manner. There are rather few, like fanfares, that are mostly indeterminate regarding their resonances.

And natural trumpets don't have valves or holes, either, but still resonances.

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Trombone comes to mind as an instrument that obviously changes in size to effect changes in the resonant frequencies. –  Dave Sep 30 at 21:16
    
Of course, with valved brass, the valves select the length of tubing to use. With plain woodwinds (e.g. recorders, flutes, oboes, clarinets, and bassoons) opening holes in the bore approximates the effect of shortening the tube length. –  Caleb Hines Sep 30 at 22:39
    
No. In wind instruments, the size change of the vessel affects the frequency thus the pitch. I'm talking about an instrument that should sympathetically vibrate according to the current frequency. Try going to a small bathroom and hum a chromatic range from the lowest tone you can. If that tone is in your vocal diapason, you'll suddenly find one pitch (sometimes has to be refined microtonally) that makes the whole room fully resonate. –  Shimmy Sep 30 at 23:33

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