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Show that if the tension $F$ in a string is changed by a small amount $\mathrm dF$, the fractional change in frequency of a standing wave, $\frac{\mathrm df}{f}$ is given by:

$$\frac{\mathrm df}{f}~=~0.5\frac{\mathrm dF}{F}.$$

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1 Answer 1

up vote 4 down vote accepted

Let me use a bit more standard notation.

If you pull the string from equilibrium then to the first order there will be a force $F$ pulling it back that is proportional to tension $T$ and the distance $u$ that the string is moved from the equilibrium $$F = -A T u$$ where $A$ is some constant not interesting to the further discussion.

For a piece of string with mass $m$ we get to the first order the equation for harmonic oscillator

$$ 0 = m \ddot{u} - F = m \ddot{u} + A T u $$ which can be solved by $u \sim \exp(i\omega t)$ giving us the relation $$\omega \sim \sqrt{T}$$ and differentiating both sides gives you the relation you wanted.

Now, this was actually just a quick and dirty derivation. For a full treatment one needs to account for the infinite number of degrees of freedom of the string. However, this derivation only rests on the fact that the pull-back force is proportional to the tension and that resulting equations are in the form of harmonic oscillator. First point translates to the full string directly while the second is correct because the string is described by a wave equation which can indeed be decomposed into simple harmonic oscillators.

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