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Resonance in Musical Instruments and with Tuning Forks

Hi,

I’m confused about the relationship between wavelength and the length of the tube in which the standing wave propagates. If you increase the length of the tube (say it’s closed on one end), will the length of the fundamental wavelength increase?

For example, if you hold a vibrating tuning fork that vibrates specifically at middle A (440Hz) over a closed pipe and you hear resonance, then you know there’s an antinode at the open end, over which you have your tuning fork hanging. But (correct me if I’m wrong) if you increase the length of the tube, the resonance gradually fades, and eventually you will hear no resonance due to a node at that end? Yet the length of the wave and frequency doesn’t change? It’s just that with a longer tube, you can incorporate more parts of the wave? (e.g. Fundamental → 1/4 of a wave; say you increase the tube so it can contain 1/2 of the wave.)

So from that, I would assume that the frequency and wavelength depends solely on the source—the A440 tuning fork, designed to vibrate at that fixed frequency and wavelength. Lengthening the pipe will only result in either resonance, or no resonance, affecting amplitude.

Yet in musical instruments, such as a clarinet, you press down on keys to lengthen the tube, generating a lower pitch. Yet the source of vibration, the reed, vibrates just the same as when a low G is played and middle G is played. So this beckons me to believe that changes in the length of the pipe wherein the standing wave propagates does change the wavelength and frequency of the wave, the vibrator’s frequency notwithstanding?

Thanks.
The difference between your first example (using a fixed frequency driver / tuning fork) and a wind instrument such as a clarinet/flute/oboe is that in the case of the instrument, the driver has no specific frequency. It is more or less "white" noise, containing a large spectrum of frequencies. The tube of the instrument is tuned to a specific frequency by use of the position of the holes. The instrument tube then resonates at its own specific frequency, taking the energy from that same frequency existing within the spectrum provided by the driver.

In other words, the frequency you get from the instrument is controlled by the tube, not the reed. The reed has no particular frequency of its own.
(edited 9 years ago)
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velocitous
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But in the tuning fork, the length of the tube would not affect the frequency?
Original post by velocitous
But in the tuning fork, the length of the tube would not affect the frequency?


In the case of the tuning fork, you get resonance from the tube when the length of that tube is such that one of the natural frequencies of vibration of the tube (fundamental or overtones) matches that of the fork.

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