As you might already know, frequency of musical notes is arranged in a such a way that if, for example, an A note has frequency of $x$, another A note which is placed one octave higher would produce frequency of $2x$.

So here's my childhood toy glockenspiel:

toy glockenspiel

This is where my question raised. If wave speed inside the bars were the same, as in strings of a guitar, then we would expect two notes that are one octave apart, like the two C's, would have a 2:1 length ratio. But measurement shows that they are designed in approximately 7:5 ratio or maybe $\sqrt2$.

Now I'm wondering what exactly causes this speed variation? Thickness or width doesn't vary meaningfully so it must be about the length but how would length of a bar affect wave speed inside it is what I'm asking here. And why this doesn't happen for strings.

I'll measure and report length, width and thickness of the bars if necessary.

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    $\begingroup$ Incidentally that's more like a glockenspiel than a vibraphone, although many people seem not to care much about naming for toy instruments. $\endgroup$ – David Z Jul 24 '16 at 13:10
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    $\begingroup$ In particular, all such instruments are part of the xylophone family, including xylophones proper, marimbas, etc., and you can't go wrong calling it a xylophone. Vibraphones by definition include a motorized component that modulates the sound of sustained notes. $\endgroup$ – user10851 Jul 24 '16 at 17:20
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    $\begingroup$ What's really bothering me is that there's a B flat in there. Did they do that on purpose? Do kids like Mixolydian mode better or something? $\endgroup$ – knzhou Jul 24 '16 at 20:47
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    $\begingroup$ @knzhou, don't worry about the Bb. In the left it has a pane for 2 extra bars so I believe there must have been a B too. $\endgroup$ – Moctava Farzán Jul 24 '16 at 22:23
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    $\begingroup$ @knzhou I'd say it was designed for playing tunes in F major, with some notes available below the keynote as well as above. More kids' glockenspiels should be so thoughtfully made. $\endgroup$ – nekomatic Jul 25 '16 at 8:48

The answer to this question has significant overlap with my answer on piano tuning. There, I discuss how a thick wire has an extra restoring force, in addition to its tension, from its resistance to bending. This modifies the usual wave equation to $$v^2 \frac{\partial^2 y}{\partial x^2} - A \frac{\partial^4 y}{\partial x^4} = \frac{\partial^2 y}{\partial t^2}.$$ This case is the other way around: the tension is negligible, so we only have the 'extra' term. The wave equation becomes $$-A \frac{\partial^4 y}{\partial x^4} = \frac{\partial^2 y}{\partial t^2}.$$ Plugging in an ansatz of $\cos(kx-\omega t)$ gives the dispersion relation $$Ak^4 = \omega^2.$$ That is, $\omega \propto k^2$. Since $k$ is inversely proportional to length, $$\omega \propto 1/L^2$$ as desired. A bar $\sqrt{2}$ times shorter makes a tone twice as high.

As you saw, the wave speed must change for the results to make sense. The phase velocity of a wave is $v_p = \omega / k$, and this is constant only for the simplest dispersion relation, the ideal wave equation $\omega = vk$. In this case, we have $\omega \propto k^2$, which implies $v_p \propto k$. Waves with shorter wavelength, like the ones on the smaller bars, travel faster.

But this doesn't mean anything about the smaller bars is different. The phase velocity changes because wave propagation is fundamentally different on bars than strings; it exhibits dispersion.

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    $\begingroup$ It may be of interest to note, as well, that the mass of the bars is carefully controlled; they're not usually just rectangular blocks. They have mass removed from the underside. E.g., see inverse.com/article/…. Wooden bars have a "scoop" underneath, as well: lafavre.us/tuning-marimba.htm . $\endgroup$ – Joshua Taylor Jul 25 '16 at 12:58
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    $\begingroup$ Indeed. My answer only addresses the fundamental modes of each of the bars, and I believe it holds regardless of the shape of the bars. But the nontrivial dispersion relation $\omega \propto k^2$ means that the overtones on an individual bar don't line up as they would on a stringed instrument. For example, the second harmonic (called the 'third transverse mode' in your article) of a uniform bar would naively be 9 times higher than the fundamental, which is somewhat unpleasant. $\endgroup$ – knzhou Jul 25 '16 at 20:02
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    $\begingroup$ So the purpose of removing mass is to further change the dispersion relation, so that higher harmonics end up in nicer places. But then the calculations get much harder. I'm thinking one might be able to get a crude approximation by letting $A$ vary with $x$ and applying WKB. $\endgroup$ – knzhou Jul 25 '16 at 20:02

As knzhou identifies, the key difference between vibrations of a free beam and a string is that the restoring force is now provided by bending moments (proportional to $\frac{d^4y}{dx^4}$) rather than linear tension (proportional to $\frac{d^2y}{dx^2}$).

The consequence, as this source shows, is that for free beams like the bars of the glockenspiel, angular frequency $\omega=2\pi f$ and wavenumber $k=\frac{2\pi}{\lambda}$ are related by

$$\omega^2=\frac{YI}{\rho A} k^4$$

where $I=\frac{1}{12}bh^3$ is 2nd moment of cross-sectional area about a horizontal axis through the centre, and $A=hb$ is the cross-sectional area.

Because the bars are unconstrained at any point, the wavelength does not correspond exactly to a multiple of the bar length. However, in each mode the same relation applies for different bar lengths. In the fundamental mode the bars vibrate with nodes at approx. $0.224L$ from each end (source), so that $\lambda=1.104L$. This leads (I think) to


which is the same form as the formula quoted here, but the leading factor does not quite agree.

Frequency is inversely proportional to square of length, so halving the frequency requires increasing the length by a factor of $\sqrt2=1.414$ as you found.

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  • $\begingroup$ What a cool and counterintuitive (to me) fact: thanks! $\endgroup$ – tfb Jul 24 '16 at 15:57

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