Third resonance length frequency(which formula do i use?)

AI Thread Summary
To find the frequency for the third resonant length in an open-ended column, the universal wave equation v=fλ can be used, where f is frequency and λ is wavelength. The calculated wavelengths for the first and second modes are based on the lengths of the column, with the first mode having a wavelength of 1.08m and the second mode equal to the length of the column. The confusion arises from incorrectly adding lengths for resonant frequencies; the third resonant length must be derived from the correct wavelength associated with its mode. For open and closed-ended columns, the conditions for nodes and anti-nodes differ, affecting the wavelengths and frequencies that can exist. Understanding these principles is crucial for accurately calculating resonant frequencies in different types of columns.
supernova1203
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for 15 a) (question in attachment)

using given 0.54m i calculated L1

also after rearranging L1=λ/2 i got wavelength which is 1.08m

L2 = λ

and since the distance between all the resonant lengths is always the same, i just added 0.54 to length 2 to figure out length 3

which is L3=1.62m
my question is i can't seem to be able to find a formula for open ended column resonant frequencies for resonant length 3should i just use the universal wave equation? v=fλ

and re arrange and use f=v/λ

or is there a different equation for finding frequency of 3rd resonant length in open air columns?


how do i find the frequency for third resonant length?
 

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On a side note, I am curious are there different formulas for resonant lengths and frequencies for open and closed end colums or are some of them the same and others different?

For instance are they the same for lengths but different for frequencies?
 
anyone?
 
I am confused as to your adding lengths, this may just by my understanding though. If I were to do this question then I would draw a diagram, importantly the open ends mean both ends must have nodes. So the first mode as you mentioned will have no nodes inside the pipe, therefore the wavelength will equal twice the length L=0.54 of the pipe.

The second will have one node in the centre, so will have a wavelength exactly equal to L, the third will have 2 nodes, evenly distributed. So you need to calculate the wavelength of this mode. And then to find the frequency use the wave equation as you suggested.

I think you have confused yourself, you write that L3=1.62m. Draw a sine wave with wavelength 1.62m inside a 0.52m tube and you will see what this does not make sense (high frequency means shorter wavelength).

On your side note a closed end mean you change the condition that the ends are nodes, in fact you can show a closed end must be an anti-node, so for the lowest order mode the wavelength is 4*L, but then the only modes which can exist are those for which the end is an anti-node so modes such as the third order mode will not work. You can test this by getting a role of wrapping paper and blowing in an end then blocking and unblocking the other end.
 
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