Simple Harmonic Motion/Fundamental Frequency

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Iman06
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Homework Statement


A tuba is a instrument that can be modeled after a closed tube and has a length of 4.9m. A frequency of 122.5hz produces resonance in the Tuba. Is this the fundamental frequency of the instrument? If not, what harmonic is it?

Homework Equations


f=λv
4l=λ(open closed tube)
v= 343m/s

The Attempt at a Solution


So for this problem, I used the equation 4l=λ to find the necessary wavelength for the length which I got as .089m. I then compared it to the actual(?) wavelength of the instrument based on the frequency and velocity, which I got as .3571m. The question is, what am I actually doing? I have no idea. Can someone at least clear up what exactly to do first?
 
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Hi iman06 and welcome to PF.

Iman06 said:
4l=λ(open closed tube)
This equation doesn't say much. There is a more complete equation that gives the fundamental and higher harmonics for a tube closed at one end. Figure out (or look up) what it is, then calculate a few frequencies for the length that is given to you and see if there is a match to the given 122.5 Hz.
 
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Iman06 said:

Homework Equations


f=λv
This equation is wrong. By that I mean if 'f' is frequency, 'λ' is wavelength, and 'v' is the speed of sound, then the above equation is wrong.

4l=λ(open closed tube)
v= 343m/s

The Attempt at a Solution


So for this problem, I used the equation 4l=λ to find the necessary wavelength for the length which I got as .089m.
Okay, using "4l=λ" is a way to find the tube's fundamental wavelength, but you didn't do that quite right. I have no idea where the 0.089 m comes from.
 
collinsmark said:
This equation is wrong. By that I mean if 'f' is frequency, 'λ' is wavelength, and 'v' is the speed of sound, then the above equation is wrong.
I missed that one. Must've been tired. Just look at the units or dimensions to figure what the equation should be.

Ill take this opportunity to mention my Insights article.https://www.physicsforums.com/insights/make-units-work/
 
scottdave said:
I missed that one. Must've been tired. Just look at the units or dimensions to figure what the equation should be.
I missed it also. I looked at the right side first (as I always do when I read an equation) and thought that I was looking at λν (lambda nu). At that point the "f" on the left side didn't register in my mind. The LaTeX nu (##\nu##) looks closer to nu than the nu provided in the PF symbols' menu which looks exactly like vee. I guess this confusion between vee and nu is the reason why "f" is preferred to denote frequency.