Fundamental frequency of a pipe half-filled with water

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Homework Statement



A very tall pipe is partially filled as shown. (A vertical pipe is filled with water about halfway, and .85m of air is in the pipe to the open end. The bottom is closed).
|**| ^
|**| |
|**| .85m
|__| _|_
|__|
|__|
|__|
|__|
|__|

^^ My attempt at a drawing, where |**| is air and |__| is water
1)Draw the fundamental wavelength of the closed pipe.

Homework Equations


The Attempt at a Solution



This is the first part of an 8 part question for a post-lab. I don't recall any lecture or lab problem like this, so I really don't know where to start.
How does the water change the wavelength? I know for an empty pipe it would be a node at the closed end and an antinode at the open end.
Does the water in the pipe change the fundamental frequency? I would assume so, but how?
 
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The water just changes the length of air able to vibrate.
The fundamental frequency of this pipe would be the same as one without water that was .85m in length.
In other words, it's just the length of air that determines the fundamental frequency.
It's a useful arrangement in labs, because you can have a variable length of vibrating air while only owning one pipe! Just add or subtract water. :)
 
OK, thank you, that makes sense.
I calculated the wavelength by 4L, where L is .85m to get 3.4m.

Part two was to calculate the fundamental frequency.
Easy enough, f=340 m/s (given) / 3.4 m = 100Hz.

In part 3, a tuning fork is of 500 Hz is brought near the open end of the pipe. Which harmonic of the pipe is resonant with the fork? I simply noticed 5(100)=500, making it the fifth harmonic.

Part four asks to draw the standing wave created by the fork in the tube. I like to think of a harmonic as a football, so I would add four footballs to the first (fundamental) frequency, giving me 4.5 footballs in the pipe, Is this correct? I also need to label pressure nodes and antinodes, which are 90 degrees offset, which is easy enough after drawing the wave.

Edit: I just noticed the football might not make sense. I mean that if you were to draw a fundamental frequency of a one end closed pipe, it would make the shape of a half football. A full wave would be two footballs. Hopefully that is clear!
 
It's clear.
The fundamental has a quarter wavelength in the tube, giving the wavelength as 4L.
Yes. The 5th harmonic is 5 times the frequency of the fundamental.
All correct
So, you will get 5 (quarter wavelengths) in the tube now.
By my reckoning, and using your football analogy, that should give 2 and a half footballs.
Fundamental |<
5th harmonic |<><><

One football <>
 
OK, so for part five I need to calculate the new wavelength by the tuning fork: =(4/5) L = .68m

Part six requests me to draw what the wave will look like when the water is lowered to reach the next resonance. I added another football! |<><><><

Part 7: Which harmonic of the pipe is now being driven by the tuning fork? I tend to mess these up. I know one end closed, so n = 1,3,5..., and there are 3.5 footballs (assuming part six is correct), meaning third harmonic, and if I only count the odd ones, it's the second harmonic, right? Like I said, I am not to good at this part.

Part 8: i need to figure out how far the water moved down for the part 6 resonance. I know the wavelength is .68m (assuming part five was correct), so can I simply (.68)(5/7)=.49m?
 
Part 5 is correct.
For part 6 the pattern is correct |<><><>< for the next resonance.
What happens here is that the water lowers down a distance equal to one "football" (half wavelength).
Here the frequency hasn't changed. It's still 500Hz, and the wavelength is still 2 footballs. (0.68m).
What has changed is that the tube is longer, so its fundamental frequency has changed.
The pattern we now see is the 7th harmonic of that wave. We know this because there are 7 quarter wavelengths in the tube. 3.5 "footballs" is 7 quarters because a "football" is 2 quarters.
For part 8. The distance the water has moved is one "football". One half wavelength of the part six value.