Waves in a closed organ pipe (homework check)

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Homework Help Overview

The problem involves a closed organ pipe of length 1.00m filled with gas and an open pipe of length 1.30m filled with air, both resonating at their fundamental modes. The task includes drawing wave diagrams and finding the ratio of sound velocities in the two mediums.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the relationships between frequency and velocity in closed and open pipes, referencing relevant equations. There is an exploration of algebraic rearrangements to solve for velocity, with some questioning the accuracy of these steps.

Discussion Status

Some participants have pointed out potential algebraic mistakes in the original poster's calculations. Suggestions for clearer variable naming conventions have been made to avoid confusion in future work. The discussion is ongoing, with participants actively engaging in clarifying the problem setup.

Contextual Notes

There is an emphasis on ensuring clarity in the equations used, particularly regarding the different lengths and velocities for the open and closed pipes. The original poster is encouraged to consider variable distinctions to enhance understanding.

Sullo
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Homework Statement


A closed organ pipe of length 1.00m is filled with a gas and is found to give the same note as an open pipe of length 1.30m filled with air, when both are resonating at their fundamental mode of vibration

a) draw diagrams to show the nature of the waves in each pipe and use them to assist in your working
b) find the ratio of the velocity of sound in the gas compared to the velocity of sound in air.

Homework Equations


b)
f(open) = v/2L
f(closed) = v/4L

The Attempt at a Solution


a) The diagrams I've drawn look like this:

open:
MXcmJ5h
https://imgur.com/a/MXcmJ5h closed: https://imgur.com/a/cK79JjL

upload_2018-5-20_11-22-22.png


b)
f(open) = v/2L
v = 1/2Lf
-> 1/2(1.30)f
=1/2.60f

f(closed) = v/4L
v = 1/4Lf
v = 1/4(1.00)f
v = 1/4.00f

ratio = vclosed/vopen

= (1/4.00f) / (1/2.60f)
= 2.60f/4.00f
= 2.60/4.00 (f cancels, same pitch)
= 0.650
 

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Sullo said:
b)
f(open) = v/2L
v = 1/2Lf
-> 1/2(1.30)f
=1/2.60f

f(closed) = v/4L
v = 1/4Lf
v = 1/4(1.00)f
v = 1/4.00f
Check your algebra for both: you've made the same mistake rearranging the expressions to solve for v in each case.
 
gneill said:
Check your algebra for both: you've made the same mistake rearranging the expressions to solve for v in each case.
Thank you for pointing this out.

So it should be v = 2lf v = 4lf ?
 
Sullo said:
Thank you for pointing this out.

So it should be v = 2lf v = 4lf ?
Yes.

I might also suggest that you alter the variable names to distinguish the two cases right from the start. This can help prevent mistakenly combining or cancelling variables that really are different when you work the algebra (Not that you've done this here, I'm just saying for future work where the equations might be more complicated). It will also make your work more clear and easy to follow for others.

It's clear that there are two different lengths and velocities involved in the problem, one for the open pipe and one for the closed pipe. Why not write them as ##L_o## and ##v_o##, and ##L_c## and ##v_c## for example?
 

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