Harmonics of a Noon Hour Whistle in a Closed Air Column

Click For Summary

Homework Help Overview

The problem involves determining the fundamental frequency and the next harmonic of a noon hour whistle, modeled as a closed air column with a length of 0.750 m. The speed of sound in air is given as 342 m/s.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between the length of the air column and the wavelength, with some suggesting to divide the length by 4 to find the fundamental frequency. Others question how to find the next harmonic and whether to multiply or divide frequencies.

Discussion Status

There is active engagement with various interpretations of how to calculate the fundamental frequency and the next harmonic. Some participants have provided guidance on using the length of the tube to determine the wavelength and frequency, while others are seeking clarification on the steps involved.

Contextual Notes

Participants are navigating through the constraints of the problem, including the specific setup of a closed air column and the definitions of harmonics in this context.

dance_sg
Messages
113
Reaction score
0

Homework Statement


A small town’s noon hour whistle is an air column 0.750 m long and closed at one end. Find the fundamental frequency and the next harmonic produced by this whistle when the speed of sound in air is 342 m/s.


Homework Equations


f= v/λ,


The Attempt at a Solution


would i first find the frequency, then divide that by 3 to get the 3rd harmonic? then take the same number i divided by 3, but divde it by two to find the second harmonic?
 
Physics news on Phys.org


In the closed pipe, length of the pipe for the fundamental frequency is equal to λ/4.
The next harmonics is three times the fundamental frequency.
 


so divide 0.750m by 4 to find the fundamental frequency? then how would i find the necxt harmoinic?
 


dance_sg said:
so divide 0.750m by 4 to find the fundamental frequency? then how would i find the necxt harmoinic?
No.
λ = 4* length of the tube.
 


so I am multiplying the wavelength by 4...
 


dance_sg said:
so I am multiplying the wavelength by 4...
No. In the problem the length of the tube is given, not the wavelength.
If you see the stranding wave in the tube closed at one end, you can see that length of then tube is equal to λ/4.
 


so how do i solve for the wavelength then?
 


dance_sg said:
so how do i solve for the wavelength then?
Wavelength = 4*length of the tube.
 


0.750 x 4 then..
 
  • #10


dance_sg said:
0.750 x 4 then..
Use the relevant formula which you have mentioned, and find the frequency.
 
  • #11


ok. so that would be the frequency. in the question tho it says find the next harmonic. how do i do that?
 
  • #12


dance_sg said:
ok. so that would be the frequency. in the question tho it says find the next harmonic. how do i do that?
Next frequency is equal to three times the fundamental frequency.
 
  • #13


ok. so multiply what i got for the frequency by 3.
 
  • #14


dance_sg said:
ok. so multiply what i got for the frequency by 3.
Yes.
 
  • #15


thank you so much
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
9
Views
18K