A person uses a 512 Hz tuning fork to find the first two resonant lengths of a closed air column in a very warm classroom. The first length measured was 17.1 cm

Question

A person uses a 512 Hz tuning fork to find the first two resonant lengths of a closed air column in a very warm classroom. The first length measured was 17.1 cm and the second resonant length was 51.3 cm.

(b) Calculate the wavelength of the tuning fork at that temperature.

(c) Use the lab observations and analysis to find the speed of sound and the temperature of the room.

Answer 1

I think there is a way to solve this if you interpret "very warm" to mean that the speed of sound is very high (per the equation that Steven gave), and hence has a very long wavelength for the fixed frequency of the tuning fork, the latter being a fixed constant of 512 Hz. Then assume the wavelength λ is so big that λ/4 fills the entire length of the tube for the first harmonic. (At resonance, a tube that is open at one end and closed at the other has an odd multiple of quarter-wavelengths end-to-end.)

Under these assumptions, the first resonance would give

λ/4 = 17.1 cm = 0.171 m ⇒ λ = 0.684 m

Then the speed of sound is

a = λf = (0.684 m)(512 Hz) = 350 m/s

The next resonance is at 51.3 cm. This time,

3λ/4 = 51.3 cm = 0.513 m, which also gives

λ = 0.684 m

This result is compatible with the 17.1 cm resonance. The speed of sound will be the same, since λf is unchanged.

Finally, using the equation Steven gave,

T = (v / 20.05)² K = (350 / 20.05)² K = 305 K

This is about 32ºC, a very warm temperature.

Answer 2

Let me flesh out my comment, and describe what I mean, with a little work.  The fundamental fact to start from is that, for one temperature, a tuning fork of 512 Hz will ideally produce one wavelength of sound.  This is because:

 

λ = v/f

 

where

 

λ = wavelength

v = speed of sound

f = frequency

 

The speed of sound is fundamentally determined by the properties of the medium the sound is traveling in... properties including temperature.  But once a temperature is chosen (assuming no other transient conditions) the speed of sound is set (constant).  Thus, if we plug in one frequency, we can solve for one wavelength of sound corresponding to that frequency.

 

For a pipe closed at one end, the harmonic wavelengths for sound in the pipe are given by the expression:

 

λ = 4L/n

 

where

 

L = length of pipe

n = harmonic index number  (n=1, 3, 5... and so on)

 

Often, once the length of the pipe is set, we change n to determine the allowed resonant wavelengths within that pipe.  But, in this case (as with a slide whistle), we change L instead.

 

That is sometimes done.  The problem for me is that the statement (form above):

 

"The first length measured was 17.1 cm and the second resonant length was 51.3 cm."

 

does not tell me which value of n we are dealing with in either case.  Often, in such situations, we assume n = 1, since the fundamental harmonic is usually the one most strongly triggered in any resonant situation.  However, if we assume n = 1 in both cases, this means:

 

λ_17.1 = 4(0.171 m)/1 = 0.684 m

λ_51.3 = 4(0.513 m)/1 = 2.052 m

 

This would suggest that the same tuning fork is creating two different wavelengths of sound, which is not possible.

 

The only way this works is if the tuning fork is exciting two DIFFERENT harmonics at these different lengths, so that n has a different value in each case.

 

I notice that the second length is three times the first.  If the tuning fork were exciting the SECOND (n=3) harmonic of the tube when it is at the longer length, this might be consistent.  Then we would have:

 

λ_51.3 = 4(0.513 m)/3 = 0.684 m, consistent with the wavelength value calculated for the other length.

 

(now that I look at it, maybe that is what that phrasing in the question above meant; but it was not clear to me right away)

 

So I would say that λ = 0.684 m is the wavelength that is asked for.

 

Then use the λ = v/f expression to solve for v, the speed of sound in air indicated by these numbers.

 

Once you have v, you can relate it to temperature (assuming dry air, which I think you are meant to) by:

 

v = 20.05(m/s)*√T_K

 

where the second term is the square root of the temperature (in units of kelvin).  If you put in the value for v you computed, you can solve for the temperature.

 

If you want to look at any of this in more detail, just let me know.