![]() ![]() Coupled to a resonator it may tend to shift toward pipe-resonance but this is small (you tune the pipe TO the bar for best output, you don t tune the bar to the pipe to reach pitch). The frequency may be slightly different in air, hardly enuff to notice. You won't hear it, but you can sense the vibration and frequency many ways. The bar's resonance is NOT (significantly) affected by air. I don't have a clue what the tempco of wood is considering the nature of wood, I bet it is not the same piece to piece, even from the same log. It comes down to a chainsaw (not the good plan) or a FedEx sticker and 2-week wait. Not the same.Īnd the bar resonance is _NOT_ easily user adjustable. ![]() The bar resonance will shift with the speed of sound in WOOD. As I understand your business, that's not your job. However the tube is user-adjustable, at home and touch-up on stage. The tube resonance will shift with the speed of sound in AIR. Since wavelenght on string is exactly the same as lenght of freely vibrating part of string, that would mean that for different tensions of vibrating object, wavelenghts for C would differ, but C will still be C. You will allways find fretting postion on fretboard where plucked string produces lets say C note, but that postion would shift up and down the fretboard depending on string tension. So, pitch is not afected by air temp.Īnd another way to look at it. That would cause tiny random vibrato, and that is real issue, but effect is so small (because it depends on derivate of air temp) that only practical effect is that impulse response reverbs dont sound like real rooms because of large distance traveled during all that bouncing around the walls (btw, this effect is real, and rooms do "breathe" - I've done some dereverberation algos, and it's big issue in such cases). What would be an issue is if air temperature would shift during wave travel to listener. That wavelenght would depend on sound velocity, and thus air temeperature, but time between same positions on waveform would be the same, and thus frequency of wave, since timing is governed by initial oscilating body. Depending on air temeperature, initial wavefront would have traveled certain distance from string, that would be wavelenght in air. After ceratin amount of time string will be (almost) back to initial position after strike, that time would be first full period of wave. When string starts to move after initial disposition it creates ripple - wavefront in sorounding air. So I am extending that to assume that the Peterson is "listening" to the wavelengths and so a shorter wavelength would appear to be a higher pitch, and a lower wavelength would appear to be a lower pitch.Ĭlick to expand.yes. ![]() So therefore, in cold weather, the frequency that is leaving the marimba key is a shorter wavelength than it is when it is hotter, right? If I want it to resonate at the normal pitch, I need to shorten the tube. Now when it gets cold outside, that resonator tube goes flat-meaning that I would have to play a lower pitch in order to achieve that resonance. The key is struck, and the frequency wave goes down the resonator and bounces off the bottom and when it reaches the top of the tube, the wavelength is at its widest point, so creates the resonance. What is throwing me off is when You tune a resonator tube to be in tune with a pitch, you make the tube to be 1/4 the wavelength of the pitch. It vibrates at the frequency that it is tuned to. So I understand that the pitch is formed inside the key when struck. ![]() I understand that I'm incorrect, and I'm still trying to wrap my head around what is wrong with my thinking, so please bare with me and point out where I am going astray: Thank you for your replies, and for helping me try to understand this dilema. ![]()
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