Waves in air in a tube that is closed

AI Thread Summary
In a closed pipe with an adjustable plunger, sound waves resonate at a frequency of 256 Hz, with the shortest length for a loud note calculated to be 0.32 meters. As the pipe length increases to 1.5 meters, the discussion focuses on identifying the number of loud notes produced, with calculations indicating that only two resonant frequencies fit within this length. The importance of understanding harmonics and the relationship between wavelength and pipe length is emphasized, particularly noting that each resonance requires an antinode at the open end. The final consensus confirms that two distinct loud notes can be heard as the plunger is withdrawn. This analysis highlights the principles of standing waves in closed pipes and their harmonic relationships.
the-pal
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1. The air in a closed pipe with an adjustable plunger in is made to vibrate at a frequency 256 Hz over its open end. As the length of the pipe is increased, loud notes are heard as the standing wave in the pipe resonates with the tuning fork.

(a) What is the shortest length that will cause a loud note?
(b) If the pipe is 1.5 m long, how many loud notes will you hear as the plunger is withdrawn?

2. Relevant formulae are

For closed pipe, first harmonic λ = 4L , f¹ = v/4L f³ = 3v/4L = 3 * f¹

3. Answers
(a) Rearranging the formula to get:
L = v/4f¹
L = 330/4*256
L = 330/1024
L = 0.32m or 32.2cm

(b) i thinking that i need to look for the 3rd and 5th harmonics to see how the length I'd affected but i calculate this to be smaller and therefore not really affected by a longer tube! Help?
 
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the-pal said:
(b) i thinking that i need to look for the 3rd and 5th harmonics to see how the length I'd affected but i calculate this to be smaller and therefore not really affected by a longer tube! Help?

Hello the-pal. Welcome to PF!

Can you show your work for the "3rd and 5th harmonics". You should be getting longer lengths. Note that the frequency of the tuning fork is fixed, so the frequency at each resonance will be the same. What does that mean for the wavelength of the sound for each resonance? For these types of problems, it can be very helpful to draw a picture which shows the resonant standing waves inside the columns of different length. By noting the number of nodes and antinodes, you can easily figure out the length of the resonating column of air.
 
Thanks,

So is it that as I've drawn that only one more full wave will fit in the tube therefore the answer is 2? The third wave would overlap the end and they're not produce the standing wave? Right?

So actually it is not about the other harmonics?
 
Remember that you must have an antinode at the open end and a node at the water surface. So, if you think about it, how far should the water be lowered to go from one resonance to the next?
 
Yes. I know that is what my diagram shows but I didn't seem to upload from the physics forum app?

The wave will look like this ><> right?
 
the-pal said:
The wave will look like this ><> right?

Right:

>

><>

><><>

Cool :cool:
 
So my answer of 2 is right as ><><> would be 1.61 so will not fit in the 1.5m pipe.

Boom.

Thanks
 
Looks good!
 
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