Speed of Sound in Air & Resonance in a Pipe

[ENGSTUDENT12]

Homework Statement

One end of a long pipe is open. The other end is closed by a piston, which can be moved along the pipe, varying its distance from the open end. Although the pipe is filled with air, the speed of sound under the conditions inside is yet to be determined. A speaker near the open end produces a pure 400 Hz frequency. Strong resonance is heard when the distances from the closed to the open end are 22.0, 67.0, and 112.0 cm. Determine the speed of sound in air under these conditions. The distances given reveal a situation that is not uncommon in real cases, in which the displacement antinode is actually a bit beyond the pipe's open end. Determine how far outside it is.

Homework Equations

lambda=4L/n
v=lambda*frequancy[/B]

The Attempt at a Solution

I don't even know.

 

[RUber]

What are the changes in distance between your resonance lengths? This should give you a hint as to what the wavelength might be.

 

[BvU]

Assume a value for v_sound, say 300 m/s. Where would the resonances occur ?

Explain the first relevant equation. What is lambda=4L/n about ? When does it apply ? Do you believe it ?

 

[CoconutFred]

In order to attempt a solution, you need to understand how the equation λ=4L/n relates to closed-end air pipes. Let's take a crack at that.

The gist of it is that since air can only move at the open end of a closed-end pipe, 1 end of a closed-pipe must always be a node while the other end (the open end) must always be an antinode, if the pipe is to resonate. For that to be true, the length of the pipe should be a length L such that L covers either 1/4 of a wavelength (since 1/4 of a wavelength is a node and then a crest/trough), 3/4 of a wavelength (since 3/4 of a wave will, once again, start at a node and end at a crest/trough), 5/4 of a wavelength (again... the wave starts at a node at the closed end and goes to a crest/trough at the open end), and so on as long as L equals an odd number of quarter-wavelengths (hopefully you understand why by this point).

So let's take a look at your problem again. You've been given 3 different lengths of a pipe, 1.12m, 0.67m, and 0.22m, and the frequency 400 Hz, and most importantly the information that the antinode of the pipe actually extends beyond the full length of the pipe. This means that L (defined as length from the node of the closed pipe to the antinode) won't just be the length of the physical pipe (which you are given); L will be the length of the physical pipe plus some constant, which we'll call x. So now we have 3 lengths L:

L₁=1.12+x
L₂=0.67+x
L₃=0.22+x

Well that's all fine and dandy, but why do we have three lengths L again? As we covered above, L can equal any odd number of quarter-wavelengths for the pipe to resonate. So we can safely assume that one L must be 1-quarter wavelength, another must be 3-quarter wavelengths, and another must be 5-quarter wavelengths. But how to decide which ones?

Go back to λ=4L/n. Each 4L/n must equal a constant λ. L₁=1.12+x is greater than any of the other given lengths L, so in order for it to be equal to the same thing as the other things it must be divided by the greatest n, which is 5, meaning that L₁ is equal to 5-quarter wavelengths. Following that we get L₁ has n=5, L₂ has n=3, and L₃ has n=1.

Now we can solve for x, since we know n for each L. Since 4L/n for all pairs of L and n must equal each other, let's choose 2 pairs of L and n and set them equal to each other:

4L₁/n₁=4L₃/n₃
4(1.12+x)/5=4(0.22+x)/1 Some simple algebra, and you get...
x≈0.005m

Now we have our x, which, in case you forgot, is the length outside of the physical pipe that the antinode is at. You can then plug that into any L, and plug that L into λ=4L/n to get wavelength λ. After that you put that into v=fλ and you can solve for the speed of sound in the pipe.

It really isn't nearly as long or complicated as this whole explanation looks. I promise.

 

[HalfLostAndSearching]

I can provide a response to this content. The speed of sound in air is affected by factors such as temperature, humidity, and pressure. In this case, we can assume that the temperature and humidity are constant, and the pressure is equal to atmospheric pressure.

To determine the speed of sound in air, we can use the formula v = λf, where v is the speed of sound, λ is the wavelength, and f is the frequency. In this case, we know that the frequency is 400 Hz, and we can calculate the wavelength using the equation λ = 4L/n, where L is the distance from the closed end to the open end of the pipe, and n is the harmonic number.

Using the given distances of 22.0, 67.0, and 112.0 cm, we can calculate the corresponding wavelengths as 88.0 cm, 268.0 cm, and 448.0 cm. Since these are all multiples of the same wavelength, we can assume that they correspond to the first, second, and third harmonics, respectively.

Plugging these values into the equation v = λf, we get v = (88.0 cm)(400 Hz) = 35200 cm/s, v = (268.0 cm)(400 Hz) = 107200 cm/s, and v = (448.0 cm)(400 Hz) = 179200 cm/s.

Therefore, the speed of sound in air under these conditions is approximately 35200 cm/s, 107200 cm/s, and 179200 cm/s for the first, second, and third harmonics, respectively.

To determine how far outside the displacement antinode is, we can use the equation L = (n/4)λ, where L is the distance from the closed end to the antinode, n is the harmonic number, and λ is the wavelength. Plugging in the values for the first, second, and third harmonics, we get L = (1/4)(88.0 cm) = 22.0 cm, L = (2/4)(268.0 cm) = 134.0 cm, and L = (3/4)(448.0 cm) = 336.0 cm.

Therefore, the displacement antinode is located 22.0 cm, 134.0 cm, and 336.0 cm outside the open end of the pipe for