Resonance and rules for mathematical equations

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Discussion Overview

The discussion revolves around the calculation of the wavelength of the lowest note that can resonate in an air column of 42 cm in length, which is closed at both ends. Participants explore the mathematical relationships involved in deriving the wavelength from the length of the air column.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • Scott presents a homework problem involving the wavelength of sound in a closed air column and seeks clarification on the derivation of the formula.
  • Another participant explains the algebraic steps to rearrange the equation from l = nλ/2 to λ = 2l/n, detailing each manipulation of the equation.
  • Scott expresses gratitude for the assistance provided in understanding the mathematical steps.
  • A participant suggests online resources for algebra to help Scott with foundational concepts related to the problem.

Areas of Agreement / Disagreement

Participants generally agree on the mathematical steps involved in the derivation, but the discussion remains focused on the explanation rather than any competing views or unresolved issues.

Contextual Notes

The discussion does not address any assumptions or limitations in the mathematical derivation, nor does it explore the physical principles underlying resonance in air columns.

Who May Find This Useful

This discussion may be useful for individuals seeking help with algebraic manipulation in physics problems, particularly those related to sound and resonance in closed systems.

Spookie71
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Homework Statement



What is the wavelength of the lowest note that can resonate within an air column 42 cm in length and closed at both ends.

Homework Equations



Given: l = 42 cm
n = 1

Required: [tex]\lambda[/tex]

Analysis: l = [tex]\frac{n\lambda}{2}[/tex]

therefore [tex]\lambda[/tex] = [tex]\frac{2l}{n}[/tex]

I don't know how this came to be, could someone please explain.

I've just gone back to high school after 15 years and would appreciate if you could also forward me on to some good links for rules for mathematical equations such as these.

Thanks
Scott
 
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The steps to go from:

[tex]l = \frac{n\lambda}{2}[/tex] to [tex]\lambda = \frac{2l}{n}[/tex] ?

start at:

[tex]l = \frac{n\lambda}{2}[/tex]

first multiply both sides by 2.

that gives:

[tex]2l = 2\times \frac{n\lambda}{2}[/tex]

on the right side, the 2 in the numerator cancels with the 2 in the denominator, so

[tex]2l = n\lambda[/tex]

Then divide both sides by n.

[tex]\frac{2l}{n} = \frac{n\lambda}{n}[/tex]

on the right side, the n in the numerator cancels with the n in the denominaotr. so,

[tex]\frac{2l}{n} = \lambda[/tex]

Then just switch sides.

[tex]\lambda = \frac{2l}{n}[/tex]
 
Thanks Learningphysics for helping

Scott
 

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