# Tuning fork clanging mode: boundary conditions?

The equation of motion of a tuning fork tine is given by: $$EI\frac{\partial^4z}{\partial x^4}+m\frac{\partial^2z}{\partial t^2}=0$$ or in shorthand: $$EIz_{xxxx}+mz_{tt}=0$$ Boundary conditions: $$z(0,t)=0$$ $$z_x(0,t)=0$$ $$z_{xx}(L,t)=0$$ $$z_{xxx}(L,t)=0$$ Initial condition: $$z_t(x,0)=0$$ Using the Ansatz: $$z(x,t)=X(x)T(t)$$ ... the PDE can be separated to: $$\frac{X''''}{X}=-\frac{m}{EI}\frac{T''}{T}=k^4$$ So: $$X''''-k^4X=0$$ ... which has four roots: $+k$, $-k$, $+ki$ and $-ki$ and the general solution: $$X=A\cosh kx+B\sinh kx+C\cos kx+D\sin kx$$ Using the boundary conditions we get: $$B=\frac{\sinh kL-\sin kL}{\cosh kL+\cos kL}A=\gamma A$$ $$C=-A$$ $$D=-B$$ $$\cos kL=-\frac{1}{\cosh kL}$$ The latter with first root: $$kL=0.597\pi$$ Solving the DE for $T(t)$ then gives the angular velocity and fundamental frequency (not shown).

For $X(x)$ we then have: $$X(x)=A(\cosh kx+\gamma \sinh kx-\cos kx-\gamma \sin kx)$$ A dimensionless plot for that first root gave me:

But there are several modes of oscillation available to a tuning fork. Two important ones are:

Left: fundamental mode (derived above), right: clanging mode. The latter is said to be obtained by hitting the tine "hard" and has a frequency roughly 6.26 times higher than the fundamental.

I suspect that this hitting "hard" changes the boundary condition(s), leading to a new set of coefficients for $X(x)$. I suspect some boundary condition may have to be applied to $x=L/2$ rather than $x=L$ but cannot see how hitting the tine hard can have that effect.

Can anyone see it?

• I don't think your drawing on the left shows the fundamental mode. – M. Enns Jul 3 '16 at 16:39
• @Gert's drawing does match the fundamental mode as defined in T. D. Rossing, D. A. Russell, and D. E. Brown, ‘‘On the acoustics of tuning forks,’’ Am. J. Phys. 60 , 620–626 ~ 1992 here Looking over the paper it seems the clang mode is a higher order in plane normal mode of treating the fork as a cantilevered beam, i.e same boundary conditions as the in plane fundamental, but a higher order mode. – paisanco Jul 3 '16 at 16:45
• A better picture has been inserted. And great animations can be found here: acs.psu.edu/drussell/Demos/TuningFork/fork-modes.html – Gert Jul 3 '16 at 16:49
• @paisanco: not sure what you mean by higher order. A harmonic? The first harmonic is for $kL=1.492\pi$ and gives the same shape for $X(x)$ as the fundamental mode. But the frequency would be $(1.492/0.597)^2\approx 6.3$ times higher, so that would fit. – Gert Jul 3 '16 at 17:04
• Yes that's what I mean, fundamental the first normal mode, next higher order mode (first harmonic ) the "clang" mode. Are you sure the X(x) is the same shape though? I'd expect the wavelength of the clang mode to be shorter. – paisanco Jul 3 '16 at 17:09

Based on T. D. Rossing, D. A. Russell, and D. E. Brown, ‘‘On the acoustics of tuning forks,’’ Am. J. Phys. 60 , 620–626 ~ 1992, I'd conclude that the fundamental and "clang" modes are both in plane modes (X plane in the questions' notation ) and have the same boundary conditions.

So I don't think the "clang" mode has a different boundary condition from the fundamental mode. They are simply different normal modes of vibration , the "fundamental" mode being the first normal mode, and the "clang" mode being the second normal mode (first harmonic is really a misnomer here as the frequencies aren't harmonics). Both of these modes have boundary conditions of treating the fork as a cantilevered beam (ie. one end, the stem end, fixed).

I suspect the comment about striking the fork hard to excite the "clang" mode really refers to exciting the motion of the "clang" mode by hitting the whole fork at a hard surface. This would excite motion at the antinodes of the "clang" mode along the fork , since it would have shorter wavelength than the fundamental mode. The fundamental mode would best be excited by tapping the open end of the fork.

It's similar in idea to exciting natural harmonics on a guitar- you fix the string at the 12th fret with your finger and pluck along the string, exciting it at at the antinode.

The treatment by @Gert in the question is using the Euler-Bernoulli (E-B) beam theory. There is a decent Wikipedia article on this in which the in plane transverse normal modes of vibration of a cantilevered bar (free at one end, clamped at the other end in this case $x=0$) are computed and plotted.

Here is the plot of the shape of the bar displacement due to vibration, for the first few normal modes (normalized to the length of the bar being 1.0) from the Wikipedia article:

Mode 1 is the fundamental mode and has no node or antinode along the bar.

Mode 2 is the next normal mode (the "clang" mode) and shows a node (zero displacement) at just below 0.8 times the length of the bar, and an antinode (local maximum of displacement) at a shorter distance along the bar.

The higher order normal modes also show nodes and antinodes along the bar.

The statement that the clang mode displacement does not show optima or roots for $x<L$ isn't consistent with the second drawing of the tuning fork in the question showing the clang mode having a maximum and minimum of displacement along the length of the fork. Each term in the E-B equation's solution should indeed have a different constant.

Of course both analyzing a tuning fork in terms of a analytical solution using Euler-Bernoulli theory, and in analogy to a cantilevered bar, are approximations. A more accurate treatment would use Timoshenko beam theory and finite element analysis to find a solution.

• $X(x)=A(\cosh kx+\gamma \sinh kx-\cos kx-\gamma \sin kx)$ shows no optima, it's not wavy at all. But the first harmonic's frequency does fit the 6.26 ratio to the fundamental frequency. – Gert Jul 3 '16 at 17:49
• X_m(x) should depend on wavenumber of the m-th normal mode. k should be a wavenumber depending on the mode. There should be units of frequency in the initial equation's second term. It's not simple harmonic (sinusoidal) wave motion, but it's definitely "wavy" -the drawings of the modes from the papers show nodes and antinodes. – paisanco Jul 3 '16 at 18:32
• $X(x)$ does depend on order: both $\gamma$ and $k$ depend on it. But for $kL=1.494\pi$ (first harmonic) the function does not become wavy, it continues to grow for $x$, without maxima or minima. – Gert Jul 3 '16 at 21:34
• As regards the Am. J. Phys. 60 paper, I've had it for a while (the second graphic in my question was lifted from it) and while interesting it doesn't contain any ab initio derivations at all. – Gert Jul 3 '16 at 21:49
• For the second normal mode $kL = 1.49\pi$ yes, it will grow without bound for $x>L$. But there should be a zero of $X(x)$ for a position on the fork $x<L$ (a node) and somewhere below the position of the node, the displacement $X(x)$ should reach a local maximum (antinode). – paisanco Jul 4 '16 at 1:14