Resonance in Musical Instruments and with Tuning ForksWatch
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?
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.