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I was studying vibrating strings and in my teacher's notes I found that, generically, if I change the tension on the string by $\Delta T$ then, the speed percentage change can be written as: $\frac{\Delta v}{v}=\frac{1}{2} \frac{\Delta T}{T}$, knowing that $v=\sqrt{\frac{T}{\mu}}$, where $\mu$ is the linear mass density. I've been trying to prove this relation for quite a while, but still didn't get the result I want. How can I prove it?

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  • $\begingroup$ which relation do you want to prove? $\endgroup$ – TotallyRhombus May 4 '16 at 2:00
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This is a typical related rates calculation. The square root formula implies that $v^2=\frac{T}{\mu}$. Differentiating it we have $2v\,dv=\frac{dT}{\mu}$. Dividing the second formula by the first gives $2\,\frac{dv}{v}=\frac{dT}{T}$, which is equivalent to your increment formula. Of course, I replaced finite increments with differentials, so strictly speaking this holds only when $\Delta v, \Delta T$ are "infinitesimal". In practice, $\Delta v\ll v$ and $\Delta T\ll T$ suffice.

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You can't; it isn't true as you've written it. You can trivially show this by taking some numbers (any numbers at all will do) for $T, \Delta T, v, \mu$ and plugging them in.

What's going on is something else. $\Delta v/v$ can be written as a power series in $\Delta T/T$. What your professor put in the notes is the first-order term only. Thus, it is an approximation that is good when $\Delta T/T$ is small.

If you get stuck proving this, refresh yourself on the binomial theorem and try applying that to the square root function.

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  • $\begingroup$ Got it! After applying taylor series! Thank you! :) $\endgroup$ – RicardoP May 4 '16 at 10:35

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