Fundamental Frequency of an organ

[Foxhound101]

Homework Statement

An open organ pipe (i.e., a pipe open at both ends) of length L₀ has a fundamental frequency f₀.

Part A
If the organ pipe is cut in half, what is the new fundamental frequency?

4f₀
2f₀
f₀
f₀
f₀

Part B
Part C

This part will be visible after you complete previous item(s).

Homework Equations

f=v/2L

The Attempt at a Solution

I am really confused by the standing waves and fundamental frequencies. The book does not do a good job explaining how this all works.

Anyways...for this individual problem I was thinking it might be 2f₀.

if L is half as long, then the frequency is twice as big?

 

[Kurdt]

Sounds good to me.

 

[Foxhound101]

Thank you Kurdt. 2f₀ was the correct answer.

Part B has revealed itself.

Part B
After being cut in half in Part A, the organ pipe is closed off at one end. What is the new fundamental frequency?

Homework Equations

f=v/2L

The Attempt at a Solution

The fundamental frequency of an open-closed tube is half that of an open-open or a closed-closed tube of the same length.

So...that means that the answer is f₀/2?

 

[Kurdt]

Well be careful because remember the pipe was halved as well.

 

[Foxhound101]

Hm...so...

Cutting it in half made the frequency 2f₀

Then making it open-closed...

2f₀/2 = f₀?

 

[Kurdt]

Yes. That seems fine.

 

[Foxhound101]

Part C
The air from the pipe in Part B (i.e., the original pipe after being cut in half and closed off at one end) is replaced with helium. (The speed of sound in helium is about three times faster than in air.). What is the approximate new fundamental frequency?

3f₀
2f₀
f₀
f₀/2
f₀/3


I'm thinking the frequency gets bigger...so...3f₀?

This is the last part of this question.

 

[Kurdt]

Yes that seems Ok too.