Xylophone Octave Radical!
Strings
First a bit of background. Why did I expect halving the length to result in a note one octave higher (i.e. double the frequency)? The answer is because I'd learned that this was the case with stringed instruments. Assume that the tension is a constant $T$, and let $z(y,t)$ be the vertical displacement of the string at position $y$ and time $t$. Then the upward force on a small element of size $\delta y$ is approximately$$
T \left(\left. \frac{\partial z}{\partial y}\right|_{y+\delta y} - \left.\frac{\partial z}{\partial y}\right|_y\right) = T\delta y \frac{\partial ^2 z}{\partial y^2}
$$
Newton's 2nd law $F=ma$ then becomes
$$
T\delta y \frac{\partial ^2 z}{\partial y^2} = \rho \delta y \frac{\partial ^2 z}{\partial t^2}
$$
or
$$
\frac{\partial ^2 z}{\partial y^2} - \frac{1}{c^2}\frac{\partial ^2 z}{\partial t^2}=0
$$
where $c$ is a velocity that, importantly, is independent of the string length $l$. The general solution of the above equation is $z=f(y-ct) + g(y+ct)$ for arbitrary functions $f$ and $g$. But when we put in the condition that $z = 0$ at $y = 0$ this becomes $z = f(y-ct) - f(-y - ct)$. And when we put in the condition that $z=0$ at $y=l$ we get the result that $f$ is periodic with period $2l$. So, if the longest wavelength is $2l$ then the lowest frequency generated by the vibrating string is $\frac{c}{2l}$. The lowest frequency is the dominant one, and this explains why you need to halve the length of the string to double the frequency, i.e. go up one octave.
Xylophones
So why doesn't this work with xylophones?The calculation above assumed a model which doesn't really apply to xylophones. All models are lies, usually convenient ones. But to work they can't lie completely - they're only allowed to ignore facts which aren't significant. For example, we assumed that the tension $T$ was constant over all time and along the whole length of the string. This isn't completely true, but if the string is tightened enough the variation will be unimportant and ignoring it won't change the result much.
A xylophone bar is not under tension and so we need to come up with a different model! Let's suppose that the bar bends in just one plane, and assume that although the radius of curvature $r$ varies with time it does not vary with position along the bar. This is reasonable since if $r$ were greater in one place than in another then the tension/compression would vary along the length of the bar at the same point in time, and it might be that such spatial difference equalize really quickly.
Next we're going to suppose when the bar bends the inner surface remains at the same length $l$ but the outer surface stretches to $l+x$ and that the only force acting on the bar is tension due to the extension $x$. Now we could instead assume that the restoring forces are split between tensile forces on the outer surface and compressive forces on the inner one (in some fixed ratio) but my instinct is that it will just complicate things and not change the result.
Hooke's Law says the tension $F$ is related to the extension $x$ by a simple constant of proportionality
$$
F = kx
$$
However, note that $k$ is only constant for a given spring (or xylophone bar). $k$ is in fact inversely proportional to the bar length $l$ since if you have a longer bar made of the same material, and under the same tension, you get a proportionally greater extension. So let's rewrite it as
$$
F = \frac{Kx}{l}
$$
Additionally, $x$ can be rewritten as $hls$ where $h$ is the thickness of the bar and $s = r^{-1}$. Substituting gives
$$
F = Khs
$$
This allows us to calculate the potential energy $V$ in terms of $s$
$$
\begin{align}
V &= \int F(x) dx \\
&= hl \int F(s) ds \\
&= lKh^2\frac{s^2}{2}
\end{align}
$$
All we need now is to be able to work out the kinetic energy $T$ in terms of $s$ and we'll be able to write the Lagrangian $L = T - V$ in terms of $s$. This will allow us to write the Euler Lagrange equation of motion - a differential equation describing how $s$ evolves in time.
To work out $T$ let's assume that all the mass is located in at the very centre of the bar. This obviously isn't true, but the velocity of any other part of the bar should be a fixed multiple of the velocity of the centre, so this assumption should cause the result to be out by just a fixed scale factor. Next, impose a coordinate system $(y,z)$ as shown in the diagram above. The velocity we are looking for is $\dot{z}\left(\frac{l}{2}\right)$. Finally note that a little bit of geometry can tell us that
$$
s \approx \frac{\partial^2 z}{\partial y^2}
$$
This allow us to write $z$ in terms of $y$ as follows
$$
\begin{align}
z(y) &= \int_0^y \frac{\partial z(\mu,t)}{\partial \mu} d\mu + C \\
&= \int_0^y \int_0^{\mu} \frac{\partial ^2 z(\nu,t)}{\partial \nu^2}d\nu + D d\mu + C \\
&= \int_0^y \int_0^{\mu} \frac{\partial ^2 z(\nu,t)}{\partial \nu^2}d\nu d\mu +Dy + C \\
&= \int_0^y \int_0^{\mu} s d\nu d\mu +Dy + C \\
&= \frac{y^2}{2}s +Dy + C \\
\implies z\left(\frac{l}{2}\right) &= \frac{l^2}{8}s + \frac{Dl}{2} + C \\
\implies \dot{z}\left(\frac{l}{2}\right) &= \frac{l^2 }{8}\dot{s}
\end{align}
$$
And this gives us the kinetic energy
$$
\begin{align}
T &= \frac{1}{2} m v^2 \\
&= \frac{1}{128}ml^4\dot{s}^2 \\
&= \frac{1}{128}\rho h w l^5\dot{s}^2 \\
\end{align}
$$
Where $w$ is the bar width and $\rho$ is the bar density. Now we can finally write the Lagrangian
$$
\begin{align}
L &= T - V \\
&= \frac{1}{128}\rho h w l^5\dot{s}^2 - lKh^2\frac{s^2}{2}
\end{align}
$$
And the E-L equation of motion
$$
\begin{align}
&\frac{d}{dt}\frac{\partial L}{\partial \dot{s}} = \frac{\partial L}{\partial s} \\
\implies & \frac{1}{64}\rho h w l^5\ddot{s} = - lKh^2s \\
\implies & \ddot{s} + \omega^2 s = 0 \\
\end{align}
$$
where
$$
\omega = \sqrt{\frac{64Kh}{\rho w l^4}}
$$
So this model produces the same result as reality: If you divide $l$ by $\sqrt{2}$ you get double the frequency! This doesn't prove the model is correct, but it does prove that we haven't proven that the model isn't correct. Which is probably the best you can hope for.
Summary
Why does halving the length double the frequency in a guitar string but quadruple it in a xylophone key? Because in both the acceleration is proportional to the curvature and the tension, but in the xylophone key the tension isn't constant - it's proportional to the curvature.
Comments
Post a Comment