# 7.4.1 Nonlinearity and longitudinal string motion

Transverse vibration of a string inevitably causes an increase in length of any small element of the string, leading to a change in local tension and to excitation of longitudinal vibration. In this section we will derive a simple approximate equation governing the process, then use it to explore the consequences for the frequency content of a plucked note. As in previous derivations of governing equations, we will look at the balance of forces on a small element of the string, sketched in Fig. 1. The transverse displacement is $w(x,t)$, and we now also allow longitudinal displacement $\xi(x,t)$. Because of the longitudinal stretch, the tension may also vary:

$$T(x,t) = T_0 +EA\epsilon(x,t) \tag{1}$$

where $E$ is the Young’s modulus of the string, $A$ is the cross-sectional area, $\epsilon$ is the longitudinal strain and $T_0$ is the original string tension. Figure 1. Sketch of an element of the string. Longitudinal displacement is indicated in blue, tension is indicated in red.

This element of string originally had length $\delta x$. Denoting the stretched length by $\delta s$, Pythagoras’ theorem gives

$$\delta s^2 \approx (\delta x + \delta \xi)^2 + \delta w^2 \tag{2}$$

where $\delta \xi=\xi(x + \delta x) – \xi(x)$ and $\delta w=w(x + \delta x) – w(x)$, so that

$$\delta s \approx \delta x \left[ \left(1 + \dfrac{\partial \xi}{\partial x}\right)^2 + \left(\dfrac{\partial w}{\partial x}\right)^2 \right]^{1/2} . \tag{3}$$

We are most interested in the case of very small longitudinal motion $\xi$ driven by fairly small transverse displacement $w$. With those approximations, we can expand using the binomial theorem to see that the leading-order contribution to string stretching is given by

$$\delta s \approx \delta x \left[ 1 + \dfrac{\partial \xi}{\partial x} + \dfrac{1}{2}\left(\dfrac{\partial w}{\partial x}\right)^2 \right] . \tag{4}$$

It follows that the strain is

$$\epsilon \approx \dfrac{\delta s – \delta x}{\delta x} \approx \dfrac{\partial \xi}{\partial x} + \dfrac{1}{2}\left(\dfrac{\partial w}{\partial x}\right)^2 \tag{5}$$

and the tension at position $x$ is

$$T \approx T_0 + EA\left[\dfrac{\partial \xi}{\partial x} + \dfrac{1}{2}\left(\dfrac{\partial w}{\partial x}\right)^2 \right] . \tag{6}$$

Now we can apply Newton’s law to the net horizontal motion of the small element of string:

$$m \delta x \dfrac{\partial^2 \xi}{\partial t^2} \approx T(x+\delta x) – T(x) \tag{7}$$

where $m$ is the mass per unit length of the string, and the angles $\theta_1$ and $\theta_2$ are assumed to be small. It follows that

$$m \dfrac{\partial^2 \xi}{\partial t^2} \approx \dfrac{\partial T}{\partial x} \approx EA\left[ \dfrac{\partial^2 \xi}{\partial x^2} + \dfrac{1}{2} \dfrac{\partial}{\partial x} \left(\dfrac{\partial w}{\partial x}\right)^2 \right] \tag{8}$$

so that finally the governing equation we want is

$$m \dfrac{\partial^2 \xi}{\partial t^2} – EA \dfrac{\partial^2 \xi}{\partial x^2} \approx \dfrac{EA}{2} \dfrac{\partial}{\partial x} \left(\dfrac{\partial w}{\partial x}\right)^2 . \tag{9}$$

This is the usual linear wave equation for longitudinal motion, with a source term on the right-hand side determined by a quadratic function of the transverse motion of the string. We can immediately see what the frequency content of longitudinal motion must consist of. In terms of the usual jargon for solving differential equations, it could involve a mixture of a “complementary function” and a “particular integral”. The complementary function consists of solutions of the wave equation without the forcing term: in other words, a linear combination of the longitudinal resonances of the string.

The particular integral is the forced part of the motion, arising directly from the right-hand side forcing term. The quadratic nature of that term tells us what possible frequencies will be present. The slope $\partial w/\partial x$ is a linear function of the transverse motion, so it must consist of a linear combination of the transverse resonance frequencies of the string. This linear combination is then squared, but other than that only linear operations are involved. So to find the possible frequencies, it is sufficient to look at a squared combination of the form

$$\left[\sum_j{a_j \sin \omega_j t} \right]^2 \tag{10}$$

where $\omega_j$ is the $j$th transverse resonance frequency. We can ignore the effects of damping for this qualitative discussion. This squared sum will involve terms like

$$\sin^2 \omega_j t = (1 – \cos 2 \omega_j t)/2 \tag{11}$$

and

$$\sin \omega_j t \sin \omega_k t = [\cos (\omega_j -\omega_k)t – \cos (\omega_j +\omega_k)t]/2 . \tag{12}$$

So the particular integral must consist of a linear combination of sinusoidal terms with frequencies which are either frequency-doubled transverse frequencies ($2 \omega_j$) or else sums and differences of pairs of transverse frequencies ($\omega_j \pm \omega_k$).

For an ideal flexible string the sum in Eq. (10) would involve harmonically-related frequencies, so it would look like a Fourier series. All possible frequency doublings or sum and difference frequencies would also be harmonics of the fundamental frequency: they would automatically give results that exactly matched other frequencies in the series. However, for the real string the effect of bending stiffness produces non-harmonic frequencies. It is useful to look at explicit results for the resulting combination frequencies. We showed in section 5.4.3 that

$$\omega_j \approx j \omega_1 (1+\alpha j^2) \tag{13}$$

where $\alpha=\frac{EI\pi^2}{2T_0 L^2}$ in terms of the string length $L$ and the bending stiffness $EI$. So, for example, the frequency-doubled version of $\omega_3$ will be $6 \omega_1 (1+9 \alpha)$. Unless $\alpha = 0$, this is a different frequency from $\omega_6 = 6 \omega_1 (1+36 \alpha)$. Similarly, the sum $\omega_3 + \omega_4$ will be $3 \omega_1 (1+9 \alpha) + 4 \omega_1(1+16 \alpha) = \omega_1 (7 + 91 \alpha)$, which is a different frequency from $\omega_7 = 7 \omega_1 (1+49 \alpha)$. The “phantom partials” generated by the square law inside Eq. (9) will not match any of the original frequencies $\omega_j$. This accounts for the complicated patterns of peaks revealed in Figs. 5 and 6 of section 7.4.

Finally, our interest is not directly in the longitudinal string motion as such, but in the force exerted on the soundboard, which then produces sound radiation and contributes to the sound of the plucked note. For that, we need to look back at Eq. (6). Apart from the steady tension $T_0$, there are two terms. One involves $\partial \xi/\partial x$, and will consist of the frequency components we have just described. The final term involves the squared slope of the transverse string shape, at the termination point on the soundboard. This is another quadratic combination of the transverse motion, so it will involve the same set of possible frequency components, but with a different pattern of amplitudes.

This distinction turns out to be important for the harp, and also for the piano. The term involving $\partial \xi/\partial x$ will always be present, but the final term involving the transverse motion will depend on a detail we have ignored up to now when discussing stretched strings. The stiff-string equation involves a fourth derivative in $x$, so that it requires two boundary conditions at each end of the string, rather than just one as was the case for the ideal string.

The extra condition is familiar from thinking about bending beams (see section 3.2.1). For a string, anchored at the ends, there are two standard choices: a pinned boundary has no bending moment at the end, so that it allows a non-zero slope $\partial w/\partial x$ at the end, and so the final term in Eq. (6) can contribute. On the other hand, a clamped boundary imposes a condition of zero rotation, so that this slope is zero and the final term in Eq. (6) is also zero. The termination of a real string will probably be somewhere in between these two textbook cases, depending on small details of how the instrument maker has adjusted the bridge/string interaction at the termination.

The two cases produce different patterns of phantom partials. Comparison with measurements on both the piano and the harp gives a strong suggestion that the real boundary condition is closer to clamped than pinned. This has an audible consequence for the sound of both instruments. Some sound examples for the harp are included in Section 7.4.