# How does Quincke's tube work?

How exactly does Quincke's tube work? I've been trying to understand this from various sources but I still don't get it. I only have a basic knowledge of interference and path length. Would someone please give me an intuitive explanation? For instance, does the wave from the source split and go into the two sides? When the movable tube is not extended what happens to the waves? What happens when it is extended? What is the difference? Why doesn't the same wave come out through R in both instances? The wave in the movable tube will have a greater number of wavelengths, but how will that affect the resultant wave? Does it change the number of maxima? I was doing a question that said that when the movable tube is displaced by 5 cm or net length increases by 10 cm, and during displacement maximum intensity is detected 10 times. The solution said that the path difference is equal to 10 times the wavelength which is 10 cm and therefore wavelength is 1 cm. But why is this so? My concept is not clear at all. Would someone please explain how this works?

• This is a apparatus in which standing waves can be produced for a tube which is closed at both ends. Have a look at this video and the next one in the series. khanacademy.org/science/ap-physics-1/… By sliding one tube relative to the other you are changing the length of the vibration air column and hence the frequencies of maximum loudness. – Farcher May 27 '18 at 8:30

does the wave from the source split and go into the two sides?

Yes.

When the movable tube is not extended what happens to the waves?

They travel along the air in the tube and meet again at "R".

What happens when it is extended?

The same thing.

What is the difference?

The difference is the difference in path length in both cases.

Why doesn't the same wave come out through R in both instances?

It does.

The wave in the movable tube will have a greater number of wavelengths, but how will that affect the resultant wave?

At "R" we have interference of the waves that travelled through each tube. If the difference in length of both tubes is an integer multiple of the wavelength, we have constructive interference, if we have an integer multiple plus half of the wavelength, we have destructive interference (idealized that would mean silence at "R").

Now assume that the apparatus is configured such that we have constructive interference, i.e. maximum loudness. We can't tell how long each path along the tubes is, but we do know that they differ by an integer multiple of the wavelength $\lambda$:

$$\Delta s_1 = k \cdot \lambda,~~ k = 0, 1, 2, \dots$$

Now imagine that the tube is extended by a certain amount $x$. The path on the right hand side is now $2x$ longer. Further assume that $x$ is exactly the distance needed to go to the next maximum at "R". Now the right path is one whole wavelength longer than before and we have

$$\Delta s_2 = (k+1) \cdot \lambda$$

And because $2x = \Delta s_2 - \Delta s_1 = \lambda$, this apparatus can be used to measure the wavelength of sound of a fixed frequency.

The example with 10 wavelength is just an extension of this principle.

• I have a doubt, what do you mean by distance required to go to next maximum? For instance, if there are 4 crests and troughs in the fixed tube, even if the movable tube has 6 or 8 or 10 crests and troughs won't the number of maxima be still 4? How does extension of movable tube affect this? Please help me. – Hema May 28 '18 at 3:29
• You have to be careful when you are talking about "the number of maxima". Within the tube, there are no maxima or minima, just a propagating wave. At "R" you have a maximum or a minimum or anything in between depending on the right tube. If you move the right tube by a certain amount, you'll experience $x$ maxima while moving it. – Jasper May 28 '18 at 14:39