A closed organ pipe has a length of 2.40 m?

Click For Summary
A closed organ pipe with a length of 2.40 m produces a frequency of 35.7 Hz based on the speed of sound at 343 m/s. When a second pipe is played, a beat frequency of 1.40 Hz indicates that the second pipe is too long by 0.10 m. The discussion emphasizes the need for relevant equations and methods to solve the problems presented. Participants encourage sharing any attempted solutions to facilitate assistance. Clear guidance on how to approach the problem is sought, particularly for part b.
TheCuriosity
Messages
3
Reaction score
0
A closed organ pipe has a length of 2.40 m.

a.) What is the frequency of the note played by the pipe? Use
343 m/s as the speed of sound.

b.) When a second pipe is played at the same time, a 1.40 Hz beat note is heard.
By how much is the second pipe too long?

The a.) problem's solution is 35.7 Hz. and b.) problem's solution is 0.10 m

The actual problem here is that I don't know how to properly solve b.)
 
Physics news on Phys.org
Please show some work.
- What are the relevant equations for this problem?
- What have you done to try to solve the problem?
 
Curiosity,there's a template provided.Use that and you won't get into any trouble
 
I'm lost... I'm new here.I just don't know here to post my question lol. I wasn't trying to solve this problem. Can you guys help me and tell me how to solve both of a.) and b.)and tell me where can I post this question if this was the wrong section.Thanks in advanced :)
 
I just read Mentor's message. I'm sorry for messing up xDI understand that you guys won't do my homework, but could you PLEASE help me and explain how to solve b.) ?
 
TheCuriosity said:
I understand that you guys won't do my homework, but could you PLEASE help me and explain how to solve b.) ?
This is the correct section.So as the Mentor said,
- What are the relevant equations for this problem?
- What have you done to try to solve the problem?

Only then will we be able to help you.Sorry :frown:
If that was a question given to you,you must be knowing some relevant equations or some method to solve it.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 31 ·
2
Replies
31
Views
3K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K