What is the Frequency of the Third Harmonic in a Heated, Stretched Yarn?

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The discussion focuses on determining the frequency of the third harmonic in a heated, stretched yarn. The problem involves a yarn with specific properties and a metal bar whose coefficient of expansion changes with temperature. Participants analyze the relationship between the tension in the yarn, its change in length due to temperature, and the resulting frequency formula. A key point of confusion arises regarding the correct application of harmonic frequencies, with one participant initially mixing up harmonics and overtones. Ultimately, the correct frequency formula is established as a function of the elastic constant, temperature change, and the yarn's mass.
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Homework Statement


A yarn of material that cannot dilate, length L, mass m and elastic constant K is trapped and stretched with negligible tension between the two supports A and B attached to the ends of the metal bar, CD, whose coefficient of expansion varies linearly from to , increasingly with temperature in the range of interest of the question. Determine the frequency of the third harmonic that is established in the rope when heated ΔT.


The Attempt at a Solution



\alpha _{eq} = \dfrac{\alpha 1 + \alpha 2}{2}

Since the metal bar expands, separation between A and B increases. This creates a tension in the string. The change in length is given by LαΔT.
F = KLαΔT
Frequency of third harmonic = 4v/2L
where v=\sqrt{\dfrac{FL}{m}}

If I substitute the value of F, the answer comes out to be wrong.
 
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utkarshakash said:
...linearly from to , increasingly ...

Something seems to be missing here.

The Attempt at a Solution



\alpha _{eq} = \dfrac{\alpha 1 + \alpha 2}{2}
[STRIKE]I am not sure but I think this is incorrect. The question says that coefficient of linear expansion varies linearly with temperature so I think you should find it as a function of temperature and then obtain the change in length through integration.[/STRIKE]

EDIT: Sorry, that is correct, integration yields the same result. So the only possible error is in your formula for frequency of third harmonic.
 
Last edited:
Pranav-Arora said:
Something seems to be missing here.


[STRIKE]I am not sure but I think this is incorrect. The question says that coefficient of linear expansion varies linearly with temperature so I think you should find it as a function of temperature and then obtain the change in length through integration.[/STRIKE]

EDIT: Sorry, that is correct, integration yields the same result. So the only possible error is in your formula for frequency of third harmonic.

Ah! That was a silly mistake. I confused "harmonics" with "overtones". Thanks for pointing out.
 
Here's the correct answer

\dfrac{3}{2} \sqrt{\dfrac{K Δ T (\alpha_1 + \alpha_2)}{2m}}
 
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