Talk:Acoustic resonance

I don't understand how the clarinet is closed at one end, but the saxophone and oboe aren't. A saxophone is almost the same as a clarinet in the way it is played. Any insight, anyone?--Nathan (Talk) 17:18, 9 March 2006 (UTC)[reply]

First of all, of course none of these instruments is truly closed at the end. Air has to enter, after all. But the mouthpiece is not a large opening, and it doesn't connect to the air at atmospheric pressure in the room. As a result a clarinet's acoustical behavior is a good approximation to that of an ideal closed cylindrical pipe, and -- as it says in the article -- the acoustical behavior of saxophones and oboes are good approximations to that of an ideal closed conical pipe. To be nitpicky, a conical pipe can't be open, because a cone comes to a point. But a pipe shaped like a frustum of a cone can be open or closed. In either case, provided the truncated part of the cone is short, the acoustic behavior is nearly the same as that of a complete cone, though an open frustum behaves like a cone of the same length while a closed frustum behaves like a cone with a greater length. So in fact all three instruments behave like closed pipes: cylindrical for the clarinet, (frustum of) cone for oboe and sax. And cone, too, for bassoon. Of the common orchestral woodwinds only the flute approximates an open pipe. -- Rsholmes 01:00, 21 June 2006 (UTC)[reply]

In the equation f=nv/2L, what is the length of the string expressed in? Also, the "n"=1,2,3.. What does that mean? 71.80.171.94 22:56, 26 June 2006 (UTC)[reply]

It doesn't matter what units the length is expressed in as long as all the units are consistent. For example, if the speed v is about 600 meters/second (for purposes of illustration, don't actually use that value for real calculations) and L is 1 meter, then for n = 1, f = (1 * 600 m/s) / (2 * 1 m) = 300 (1/s) or 300 hz. Likewise if v is in feet per second and L is in feet, the formula still works.

An ideal string has an infinite number of resonant frequencies corresponding to an infinite number of modes in which it can vibrate. "n=1,2,3..." means n can take any positive integer value. So using the above (fictitious) numbers again, the resonant frequencies of a 1 meter string where v is 600 m/s are 300 hz (n=1), 600 hz (n=2), 900 hz (n=3), 1200 hz (n-4), and so on. -- Rsholmes 23:45, 26 June 2006 (UTC)[reply]

Resonance within mooth as a play technique

I have removed the section of this name since it didn't make any sense. The original text follows:

A musician, forming cavity resonator within mooth, may amplify and emphasize some overtones from base tone of his/her voice or instrument. This is used for overtone singing and playing on jew's harp.

Pdefer 15:16, 28 May 2007 (UTC)[reply]

Closed Tube

f=nv/4L for a closed tube. Is this based on experimental data or is there a way to prove it? Gagueci (talk) 23:54, 23 November 2007 (UTC)[reply]