Standing Waves in a tube closed at both ends

[physicslover2012]

Homework Statement

A sound tube, which measures 50 cm, is closed AT BOTH ENDS and has a reed that produces vibrations in the middle. The air is at 0°C and the speed of sound is c= 340 m/s.
a) What is the fundamental frequency of the tub?
b) How much should the air be cooled in order to obtain the next harmonic?

Relevant Equations

λ = v / f

During our classes, we haven't discussed the situation of a tube closed at both ends. But, assuming the position of the nodes and antinodes, I think it's a case similar to the one where the tube is open at both ends, so I think that f = v/λ = nv/(2L). Using the numeric data, my frequency would be 340 Hz. But I am not sure and I need some other opinions. As for the second question, I found a formula online that shows the dependence between the speed of sound and temperature, but using the formula I get an absurdly big number. If you can, please help me understand this particular case!

 

[haruspex]

similar to the one where the tube is open at both ends

Yes, just swapping nodes with antinodes. But you have not taken into account that the sound source is in the middle. I can see why that might change things.. not sure.

using the formula I get an absurdly big number

Please post your working and result.

 

[physicslover2012]

Please post your working and result.

The fundamental frequency I calculated is f=c/(2*l)= 340 Hz. I calculated the next harmonic, which was f1=2*c/(2*l)=680 Hz. Then, I wrote again the formula, using this time another speed: f1=c'/(2*l) From here, I deducted that c'=680 m/s, and I found this formula online: c'=c+0,6*t. From here I deducted the temperature, t=566,66 s. But, to me, it seems a number way too big.
Yes, just swapping nodes with antinodes. But you have not taken into account that the sound source is in the middle. I can see why that might change things.

Yes, just swapping nodes with antinodes. But you have not taken into account that the sound source is in the middle. I can see why that might change things.. not sure.

I assumed that it was a similar case because we learned that where the reed is situated, there is an antinode.

 

[haruspex]

From here I deducted the temperature, t=566,66 s.

You have slightly misused the formula. The base speed in that (the c on the right) is at 0C, so is 331m/s. But your answer is about right.

we learned that where the reed is situated, there is an antinode.

That's what bothered me. If the tube is open at both ends, for a wavelength of 2L, where are the nodes and antinodes?

 

[physicslover2012]

You have slightly misused the formula. The base speed in that (the c on the right) is at 0C, so is 331m/s. But your answer is about right.

In the problem, it was assumed that at 0C, the speed of sound is 340 m/s. I also found out that the speed of sound at 0C is 331, so I think it is a mistake on the teachers' part.

That's what bothered me. If the tube is open at both ends, for a wavelength of 2L, where are the nodes and antinodes?

If the tube is open at both ends, we have the antinodes at the ends of the tube. I watched an online lesson where the subject of a tube closed at both ends was discussed, and, despite the fact that practically it isn't possible to exist (since the sound must exit the tube somewhere), the teacher used this graphic for the waves:
Technically, if the sound starts in the middle, I thought that there would be antinodes that go both ways, thus making a full form.

 

[haruspex]

In the problem, it was assumed that at 0C

Sorry, I missed that.

despite the fact that practically it isn't possible to exist (since the sound must exit the tube somewhere)

You maybe misunderstand what nodes and antinodes are. A node is where the displacement amplitude of the oscillating molecules is zero (and correspondingly where the pressure amplitude is max). At an antinode, displacement amplitude is max and pressure variation is zero.
At a closed end the molecules have nowhere to move, so that's a node; at an open end they can move freely but are at constant pressure. (In practice, the pressure variation extends a bit beyond the end, leading to a slightly longer effective tube length.)

I thought that there would be antinodes that go both ways,

In the diagrams, an antinode is not a shape; it is the point where the curved lines are furthest apart. They don't "go" anywhere.

If the tube is open at both ends, we have the antinodes at the ends of the tube.

Yes, but in the question you also have an antinode in the middle. So what does the diagram look like?