How many wavelengths are in this pipe?

[dedicate]

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I thought this was a somewhat straightforward problem. I counted the wavelengths on the 2 waves, but apparently that is wrong.

What is the correct way to see this problem?

 

[dualityim]

ah standing waves . . .

I count two wavelengths, open at both ends . . . I don't see how it could be four.

 

[dedicate]

ah standing waves . . .

I count two wavelengths, open at both ends . . . I don't see how it could be four.

My interpretation was 2 waves and each one repeats twice.

 

[0Complications]

ah standing waves . . .

I count two wavelengths, open at both ends . . . I don't see how it could be four.

I was thinking this as well, but physics is the bane of my existence so my opinion doesn't count for much here.

 

[dedicate]

I was thinking this as well, but physics is the bane of my existence so my opinion doesn't count for much here.

Just remember, physics doesn't suck.

 

[Marjan Islam]

Well, I see 4 wavelenghts: 2 upright, and 2 inverted. Also, I see both wavelenghts have antinodes at the sides so both ends of the pipe are open! Right?

 

[0Complications]

Just remember, physics doesn't suck.

I try to tell myself this daily...

 

[dedicate]

Well, I see 4 wavelenghts: 2 upright, and 2 inverted. Also, I see both wavelenghts have antinodes at the sides so both ends of the pipe are open! Right?

That's how I saw it too, but that is wrong! The answer is b

 

[dualityim]

My interpretation was 2 waves and each one repeats twice.

In pipes, there are always two individual waves that run anti-parallel to each other. In conjunction, these two are referred to as a single standing wave. With this in mind, I'm sure you can see why there are only two wavelengths

 

[dedicate]

In pipes, there are always two individual waves that run anti-parallel to each other. In conjunction, these two are referred to as a single standing wave. With this in mind, I'm sure you can see why there are only two wavelengths

Hmm, I can see why the answer is the answer now, but I still don't know why exactly a standing wave is two waves referred to as 1?

 

[LeftyUM]

In pipes, there are always two individual waves that run anti-parallel to each other. In conjunction, these two are referred to as a single standing wave. With this in mind, I'm sure you can see why there are only two wavelengths

he got it 👍

To get the answer of 2 wavelengths, you have to see how many antinodes there are. From one antinode to the next = 1/2 wavelength...

 

[Vihsadas]

Hmm, I can see why the answer is the answer now, but I still don't know why exactly a standing wave is two waves referred to as 1?

A good way to visualize "intuitively" why this is, is to think about what happens when you are holding a string tightly in both your hands, and someone plucks it. What does the resultant wave look like? If two people plucked the string at the same time at exactly 1/4 distance from each end one plucking up and one plucking down, then what would the resultant string look like? (See how when you pluck the string the standing wave you generate is actually both waveforms?)

 

[HurricaneKatt]

2 wavelengths, open at both ends. If it was closed at either end there would be a node there rather than an anitnode.

 

[ak00]

Easy question:

1) Quick glance at the answer choices: you have to determine whether its a closed or open end.
2) Determine number of waves

For 1) Ask yourself: "At the start of the pipe, are there "nodes" or are there "antinodes"?

A: I see a antinode. hmm..this means: it is Open: Since open pipes, always have antinodes in the beginning of the pipe. (Closed: start with nodes)

Now that you have determined its a Open Pipe: count the number of nodes. (=4)
The rule for open pipes: # of nodes/2= number of waves in the pipe.

Number of nodes 👎= 4
4/2= 2 waves.

This rule doesn't work for antinodes/ or closed pipes. This relationship only holds true for nodes and its connection to number of waves for open pipes.

Hope this helps!

More reading on this topic: http://www.physicsclassroom.com/class/sound/U11L5c.cfm