What Water Levels Produce Resonance Frequencies with a Tuning Fork?

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SUMMARY

The discussion focuses on determining the water levels that produce resonance frequencies with a tuning fork at 300 Hz placed atop a metal cylinder. The natural frequencies in an open-closed system are calculated using the formula f_n = nv/4L, where n represents the harmonic number. The first resonance frequency occurs at a water level of L_1 = 0.286 m. The subsequent resonance frequency can be calculated using the same formula, but the exact method for determining the second level is questioned.

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  • Understanding of resonance frequencies in open-closed systems
  • Familiarity with the formula f_n = nv/4L
  • Basic knowledge of tuning forks and sound waves
  • Concept of beat frequency in acoustics
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  • Explore the relationship between water levels and sound resonance
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dekoi
There's a metal cylinder from which water is flowing out at a rate of 10^-6 m^3/s. A tuning fork of 300 Hz is placed on top of the cylinder. At what water levels will the first two resonance frequencies be heard?

It doesn't seem like a difficult problem, and it really isn't. I just can't seem to see it.

I don't think the flow rate has any signifiance in calculation; it's only useful to know that the water level is decreasing.

It's an open-closed system, so the natural frequencies are [tex]f_n = \frac{nv}{4L}[/tex] where n = 1, 3, 5, 7...

So in the first case, i think the first frequency heard would be 300 Hz. Hence, one can solve for [tex]L_1[/tex] to attain 0.286m. But how about for the next? Will it just be 3 times L_1 ? Or how do you solve for it otherwise?Thanks
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Question 2: Is it logical to ask whether the beat frequency is e.g. +4 or -4? Afterall, [tex]f_{beat} = |f_2 - f_1|[/tex] so because of the absolute value,it is always +4.
 
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