# Why does the aggressively bowed string go sharp?

A vibrating string with fixed endpoints, such as on my fiddle, may be bowed (see Helmholtz motion, see stick and slip) with very little to a certain amount of pressure and proximity to the bridge (the combination hereafter abbreviated as amplitude) with un-noticeable change in pitch before beginning to sound sharp (more amplitude causes higher frequency). My suspicion was that as amplitude increases, a) the string length nominally increases along with b) the tension and that these factors "balance" each other out for small perturbations, i.e $$\delta f|_{T,\rho}\approx -\delta f|_{\rho,l}$$ for small amplitudes and then the tension "wins" beyond that threshold. The problem is that the length is squared in the denominator which would indicate that the pitch should go flat.

$$f(T,\rho,l) = \sqrt{\frac{T}{4\rho l^2}}.$$

The two hypotheses I've considered are that A) the tension actually increases by another exponent(s) under bowing conditions and that the difference in pitch is very small below some amplitude threshold, or B) that there are other approximations earlier in the derivation that hide the true nature of this behavior. I've considered that the density of the string may change: $$\rho = \frac{m}{\pi r^2 l}$$; naively, $$f = \sqrt{\frac{\pi r^2 T}{4ml}}$$ (Poisson's ratio $$\nu \approx 0.3$$ for steel if it matters), but it turns out that the string at both the bridge and the nut (bottom and top) isn't axially fixed: the string stretches below the bridge and above the nut when when tension in the playable region increases.

So. Why does my fiddle play sharp when I really whale on it?

• More amplitude does not alter the fundamental, only the relative amounts of individual overtones. The thing goes sharp when bowed aggressively because of increased tension. Your formula for frequency is really only linear in $l$ because of cancelation with the density factor. Any changes in $l$ are high enough order to be negligible in an unbroken violin. Tension is the dominant cause among many other less significant effects. If you want to go deeper, you need to be specific about what because complexity of the violin allows for myriad departures from the simple models. Jan 4 at 15:36
• @AlbertusMagnus Yes, deeper into the effects that cause it to go sharp is what I'm asking for. Jan 4 at 18:53
• My guess is that non-linear dynamics here has negligible effects. Try to have a look here phys.unsw.edu.au/jw/…. Jan 14 at 10:54
• Could it be something related to Wolf tone? en.wikipedia.org/wiki/Wolf_tone Jan 16 at 18:20
• Do you have a record of the sound? I keep suspecting that it is more related to a resonance of the linear system, maybe due to a spectrum of the forcing with some larger contribution at higher frequency, than some nonlinear process. All the answers below are not satisfying to me: if you're talking about frequency, you need to evaluate the whole period and not only one condition, as the maximum displacement studied below. Does this sound vary only slightly, or is it an octave above? Jan 16 at 18:24

Let's say $$L$$ is the nominal length of the string between the tuning peg and the point where it attaches to the tailpiece (points at which we may assume the string is fixed), and let $$T$$ be the nominal tension in the string. Suppose the segment with length $$L$$ has "spring constant" $$k$$. If you were to remove the string from the violin so that there is no tension in it, length $$L$$ would contract by $$\delta L$$, where $$T=k~\delta L$$.

With the string on the violin and at nominal tension, let $$\ell$$ be its nominal length between the bridge and the nut: this is the length that vibrates and is relevant for the pitch. You press on the string with the bow, increasing its length to $$\ell + \Delta\ell$$ and its tension to $$T + \Delta T = T + k~\Delta \ell$$. Now let's see what happens to the $$\sqrt{T/\ell^2}$$ factor that goes into the fundamental frequency: $$\sqrt{\frac{T+\Delta T}{(\ell + \Delta\ell)^2}}\approx\sqrt{\frac{T}{\ell^2}}\left(1+\frac12\frac{\Delta T}{T}\right)\left(1-\frac{\Delta \ell}{\ell}\right)$$ $$\approx\sqrt{\frac{T}{\ell^2}}\left(1+\frac12\frac{\Delta T}{T}-\frac{\delta L}{\ell}\right)=\sqrt{\frac{T}{\ell^2}}\left(1+\frac12\frac{k~\Delta\ell}{k~\delta L}-\frac{\Delta \ell}{\ell}\right)$$ $$=\sqrt{\frac{T}{\ell^2}}\left[1+\frac12\frac{\Delta\ell}{\delta L}\left(1-\frac{2\delta L}{\ell}\right)\right].$$

It's safe to say $$\delta L\ll\ell$$, so the pitch increases by a factor of approximately $$\Delta\ell/(2\delta L)$$.

In summary, I don't think there is a big mystery here, we just need to account with some care for how much $$T$$ and $$\ell$$ change when pressure is applied with the bow. $$\Delta T/T$$ and $$\Delta\ell/\ell$$ are not equal ($$\Delta T/T$$ and $$\Delta\ell/\delta L$$ are), meaning $$T$$ increases by a much greater relative amount. $$\rho$$ will have a minute change as well, but much like the $$\ell^2$$ factor in the denominator, it would have an insignificant effect compared to the variation of $$T$$ in the numerator.

We can go a bit deeper to understand how $$\Delta\ell$$ (and hence the pitch change) is related to the force $$\mathbf F$$ applied with the bow, or the displacement $$\Delta y$$ of the string toward the soundboard. Assuming the force is applied at a distance $$a$$ from the bridge, referring to the diagram above, $$\Delta \ell = \sqrt{a^2 + \Delta y^2} + \sqrt{(\ell-a)^2 + \Delta y^2} - \ell$$ $$\approx \frac12\left(\frac{1}{a} + \frac{1}{\ell - a}\right)\Delta y^2=\frac{\ell}{2a(\ell - a)}\Delta y^2 \tag{1}\label{1}$$ where we have assumed $$\Delta y \ll a$$ and $$\Delta y \ll \ell - a$$.

As for the force, $$F=2(T+\Delta T)\cos \frac{\theta_1+\theta_2}{2} = 2(T+\Delta T)\sqrt{\frac{(2\ell+\Delta\ell)\Delta\ell}{4\sqrt{[a^2+\Delta y^2][(\ell-a)^2+\Delta y^2]}}}.$$ The second equation can be obtained in different ways, e.g. by using the law of cosines and a half-angle formula. For the leading order approximation valid for small $$\Delta\ell$$, we can write $$F \approx T\sqrt{\frac{2\ell\Delta\ell}{a(\ell-a)}}$$ $$\Delta\ell \approx \frac{a(\ell-a)}{2\ell} \frac{F^2}{T^2}.\tag{2}\label{2}$$

A key point that follows from $$\eqref{1}$$ and $$\eqref{2}$$ is that the pitch change is proportional to $$\Delta \ell \propto \Delta y^2 \propto F^2$$, i.e. quadratic (superlinear) in both the displacement and the applied force. For a small force, the pitch change will be imperceptible because $$\frac{d\Delta\ell}{dF} = 0$$ at $$F = 0$$, and the same goes for the displacement.

I don't think the pitch shift is caused by competing effects competing effects one of which wins out at a certain force/displacement. I think the change in pitch gets increasingly higher as you keep increasing the force, and the point at which it is perceived is dependent on your tuner's sensitivity, or the stability with which you can "hold the pitch" as you bow.

This analysis also suggests that it takes less displacement (from \eqref{1}) and more force applied with the bow (from \eqref{2}) to change the pitch by a certain amount when you are playing far away from the midpoint (halfway between the bridge and the nut).

• This is exactly the type of analysis I was hoping for, and it's fairly persuasive. There is a bit of an incompleteness - the region of pitch stability appears unaccounted for. Jan 14 at 0:15
• @user121330 Can you explain what you mean by pitch stability and what you want to know about it?
– Puk
Jan 14 at 1:09
• I can play "normally" which means really quiet to pretty loud and the pitch doesn't change much - pitch stability (BTW, I'm looking at a tuner here - it's not just my ears that think it's stable). What marks the "corner" where pitch a) starts to change enough to be noticeable or b) two competing factors stop being roughly equivalent and Tension starts being more important? Jan 14 at 1:14
• See my update. It makes sense that the pitch change would be imperceptible when you bow gently because of the quadratic dependence of pitch on displacement toward the soundboard. I don't think there is an intrinsic "corner" (that doesn't depend on your equipment, technique, etc.) at which the pitch shift becomes noticeable.
– Puk
Jan 15 at 0:05
• Continuing off of @steveK's thread, fiddles and guitars are both typically bowed/plucked somewhere between the end of the fingerboard and the bridge, so $a$ is much smaller than $l-a$. I'm having a lot of trouble noticing a difference between the "bend" when my bow is over the fingerboard or nearer the bridge. Jan 15 at 6:02

It's because the tension increases much faster than the length of the string does. It is similar to the effect of bending a guitar string- the frequency rises notwithstanding the fact that the string is lengthened.

In the formula you cite, the tension and the length of the string both increase when the string is bent away from its straight-line resting position. Let's call the increase in the length dl. The proportion by which the length of the string increases is dl/l, where l is the length of the string before it is bowed. By contrast, the proportion by which the tension is increased is dl/dL, where dL is the amount by which the string had to be stretched from is original length to bring it up to pitch when the violin was strung. Since dL is << l, dl/dL is >>dl/l, so the tension increases by a much higher factor than the length does, and hence that is the dominant effect in determining the change of frequency.

Incidentally, the density per unit length which appears in your formula must decrease- if that were not the case, the mass of the string would have to increase when it is stretched, which is unphysical.

• A picture would make your argument more persuasive. Jan 14 at 0:17
• @user121330 pictures are far too hard for me, but I will edit my answer to make it clearer! Jan 15 at 13:00

When a cable, or in this case a string, are in tensile force the string elongates slightly. As the string continues to elongate over time the diameter of the midpoint of the string becomes smaller and thus vibrates faster increasing the frequency ... and this is the cause of the string going #Sharp.

The ends of the string are not really "fixed" since the string vibrates between the bridge and the tuning peg. So as the string elongates the extra length is consumed at the tuning peg. The diameter of midpoint of the string is now smaller and will vibrate faster giving the migration to a #Sharp.

A good example of this is the Twizzler licorice. If you hold the Twizzler at the ends and pull on it (tensile force), the diameter of the midpoint of the Twizzler will become smaller.

The velocity of wave propagation $$v$$ = $$\sqrt{\frac{T}{u}}$$ where $$T$$ is the Tension Force and $$u$$ is the linear density of the string. $$u$$ decreases because the diameter of the midpoint of the string decreases, and thus $$v$$ goes up to a higher frequency... indicating a #Sharp.

Note: Strings cannot be assumed to have "spring constants" due to the various materials actual violin strings are composed of.. steel and advanced synthetics. Once deformed, the string will retain its "deformed" shape and will not return to its original shape...there is no memory for these materials. Elasticity of steel and composite materials are memory deficient as they're not coiled.

$$v$$ = $$\sqrt{\frac{T}{u}}$$

The amount of energy the string receives doesn't affect the frequency at which the string vibrates. $$v$$ is a function of $$T$$ and $$u$$ only which are independent of input energy. Your ear won't hear the #Sharp at lower volumes but it will hear it at higher volumes when the string is whaled hard. The reason why you don't hear the #Sharp at lower volumes is because human hearing is very nonlinear.

https://physicsworld.com/a/human-hearing-is-highly-nonlinear/

• I believe you're referring to Poisson's ratio. Perhaps you have some math to back this up? Jan 14 at 0:16
• The velocity of wave propagation $v$ = $\sqrt{\frac{T}{u}}$ where $T$ is the Tension Force and $u$ is the linear density of the string. $u$ decreases because the diameter of the midpoint of the string decreases, and thus $v$ goes up to a higher frequency... indicating a #Sharp. Jan 14 at 0:31
• @user121330 I would expect it to take less displacement ($\Delta y$) and more force applied with the bow to change the pitch by a certain amount when you are playing far away from the midpoint (halfway between the bridge and the nut). The latter point regarding the force isn't currently obvious in my answer because I haven't directly addressed it yet. I don't own a violin but this is consistent with my experience with guitar strings. Is this inconsistent with your observation?
– Puk
Jan 15 at 3:29