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To explain, I'm doing an experiment that tries to prove the relationship of frequency being proportional to the square root of tension by measuring the string extension of a guitar strings I've measured. I don't understand how the spring constant is just neglected as it is in this argument:

\begin{align} f &\propto \sqrt{T} \\ F&=-xk \qquad\text{(Hooke's Law)}\\ & \downarrow \\ f &\propto \sqrt{kx} \\ &\downarrow \\ \underline{\Delta} f &\propto \underline{\Delta} \sqrt{x} \end{align}

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    $\begingroup$ Do you know what the symbol $\propto$ means? $\endgroup$
    – knzhou
    Jan 25, 2020 at 21:20
  • $\begingroup$ proportionality $\endgroup$
    – Milan Pin
    Jan 25, 2020 at 21:22

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The spring constant hasn't been neglected. If f is proportional to $\sqrt{kx}$ then they differ only by a constant multiple. Since the spring constant is a constant, f is also a constant multiple of $\sqrt{x}$, so f is proportional to $\sqrt{x}$.

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The forces on elements of the string that are responsible for the oscillations of the string are not due to changes in tension. Provided that amplitude << $\lambda$, the tension is almost constant, as the fractional change in the string's length due to the wave is very small. So Hooke's law isn't involved (except in determining the constant tension). The transverse forces responsible for the oscillations are due to string not being straight, owing to the wave profile!

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  • $\begingroup$ tension is not almost constant $\endgroup$
    – Milan Pin
    Jan 25, 2020 at 21:50
  • $\begingroup$ @ Milan Pin In showing that disturbances propagate as transverse waves at a speed of $v=\sqrt{\frac{T}{\mu}}$, we neglect any changes in the tension due to the wave. See any first year university text that deals with the topic. $\endgroup$ Jan 25, 2020 at 22:15

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