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Velocity of sound in air?

  • Thread starter pivoxa15
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1. Homework Statement
A glass tube has one end dipping into a large container of water and its height above the water is slowly increased until it resonates to a tuning fork of frequency 512Hz. It is then raised further until it resonates to a tuning fork of frequency 280Hz. The distance the tube has been raised in passing from one resonant length to the other is 0.139m. What is the velocity of sound in air at the temperature of the experiment?


2. Homework Equations
v=wavelength * frequency
wavelength=4L/(2n-1) for open-closed tube, L is length of tube, n is number of segments or number of antinodes.



3. The Attempt at a Solution

The resonation is done in a tube with one end open and one end closed. So the frequency must be odd integer multiples of each other. However 512/280=64/35 suggesting that 512 is an even number in fact 64 times the fundamental frequency. So is that a mistake in the problem?
 
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Answers and Replies

andrevdh
Homework Helper
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When the tube is raised it will resonate at the point when

[tex]\lambda = 4L[/tex]

We therefore find that a standing wave forms when the air column is a quarter wavelength long. The next length that will form a standing wave is three quarters of a wavelength.
The air column length for the second (lower) frequency will be given by a similar formula since its wavelength is longer than that of the higher frequency. Another way to describe the relationship between the various quantitites in this case is therefore given by:

[tex]L = \frac{v}{4f}[/tex]
 
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AlephZero
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There is nothing special about the frequencies 512 and 280 and the ratio between them. The two frequencies could be anything.

If the length of the resonating tube was x meters for the resonance at 512 Hz, it was x + 0.139 meters for the resonance at 280 Hz. Use the relevant equations you stated twice, to eliminate x.
 
2,234
1
When the tube is raised it will resonate at the point when

[tex]\lambda = 4L[/tex]

We therefore find that a standing wave forms when the air column is a quarter wavelength long. The next length that will form a standing wave is three quarters of a wavelength.
The air column length for the second (lower) frequency will be given by a similar formula since its wavelength is longer than that of the higher frequency. Another way to describe the relationship between the various quantitites in this case is therefore given by:

[tex]L = \frac{v}{4f}[/tex]
How do you know it's the first and second harmonics?

I am trying to find the harmonic number by looking at the frequency but one of them gives an even number which shouldn't happen for a closed and open pipe. Or is there something I am missing?
 
2,234
1
There is nothing special about the frequencies 512 and 280 and the ratio between them. The two frequencies could be anything.

If the length of the resonating tube was x meters for the resonance at 512 Hz, it was x + 0.139 meters for the resonance at 280 Hz. Use the relevant equations you stated twice, to eliminate x.
But what is n? The quesiton didn't state which harmonic they were resonating at. If I assume they were resonating at 1 and 2 harmonics than I get the speed of sound in air to be 61.4m/s which is not the suggested answer at the back of the book.
 
andrevdh
Homework Helper
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Since the length of the air column is increased from zero the standing wave will form when it can accomodate a quarter of a wavelength.
 
2,234
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Since the length of the air column is increased from zero the standing wave will form when it can accomodate a quarter of a wavelength.
This wasn't explicitly stated in the problem.

But if we assume this than if you follow the calculations you should get the speed of air to be around 62m/s. Do you also get this? But the answers at the back of the book suggested 344m/s
 
andrevdh
Homework Helper
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Well I assure you that is what they intended to convey (which they did a bad job of if that was the original wording):

[tex]f_1 = 512 Hz[/tex]

[tex]f_2 = 280 Hz[/tex]

[tex]\Delta L = 0.139 m[/tex]


[tex]v = 4\ \Delta L \left(\frac{1}{f_2} - \frac{1}{f_1}\right)^{-1}[/tex]
 
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2,234
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I see, they change the length of the tube so the frequency dosen't change, i.e it is always the first fundalmental harmonic. Hence the two frequencies are not related to each other. They should have stated this in the problem. When I did it, I intepreted that in the first case it was the 1st harmonic frequency and the second it was the 2nd harmonic even though the length of the tubes were different.
 
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andrevdh
Homework Helper
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The wavelenght of the second tuning fork is longer than that of the first. That way resonance will also occur when a quarter of its wavelength fits into the air column as it is lengthened beyond the first resonance.
 
2,234
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How did you get this from the (ambigous) way the problem was stated?
 
andrevdh
Homework Helper
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The first resonance is for a tuning fork of frequency 512Hz. The second resonance is for a tuning fork of frequency 280Hz, which generates a sound wave with a longer frequency than the first sound wave.

Remember that the wavelength of a particular tuning fork is fixed. So if the air column is slowly lengthened a quarter of a wavelength will fit for the first (and second) resonance lengths.
 

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