5.4.1 Motion of a plucked string as a modal sum

A string of tension $T$ and mass per unit length $m$ has length $L$, and is fixed at the two ends $x=0$ and $x=L$. A transverse force $F$ is applied to the string at position $x=a$, as sketched in Fig. 1. Static equilibrium of the forces, assuming that the displacement is small, requires

$$F \approx T \dfrac{X}{a}+ T \dfrac{X}{L-a} \tag{1}$$

where $X$ is the displacement at the pluck point. This equation is based on small-angle approximations for the two angles between the string segments and the x-axis: note that Fig. 1 uses an exaggerated vertical scale so that the angles may not look small. We can deduce from eq. (1) that

$$X=\dfrac{Fa(L-a)}{TL}. \tag{2}$$

Figure 1. The string is pulled into a triangular shape by a force $F$ before release.

We already know that the mode shapes on the string are sinusoidal, and that the natural frequencies are integer multiples of a fundamental frequency $\Omega = \pi c/L$, where $c=\sqrt{T/m}$ is the wave speed on the string. The motion following the release can be expressed as a linear combination of modal contributions:

$$w(x,t)=\sum_n{c_n u_n(x) e^{i \omega_n t}}=\sum_n{c_n \sin \dfrac{n \pi x}{L} e^{in \Omega t}} \tag{3}$$

where $c_n$ is the ‘amount’ of the $n$th mode involved in the vibration. Since $c_n$ may be complex, because it contains phase information as well as amplitude information, it is convenient to write it explicitly as

$$c_n=a_n +i b_n \tag{4}$$

so that

$$w(x,t)=\sum_n{\left[ a_n \cos n \Omega t -b_n \sin n \Omega t \right] \sin \dfrac{n \pi x}{L}} \tag{5}$$

after taking the real part.

The constants $a_n$ and $b_n$ are determined from the initial conditions. At $t=0$ the string is at rest in the shape sketched in Fig. 1. So

$$w(x,0) = \left\{ \begin{array}{ll} \dfrac{xX}{a} \mathrm{~~~for~~~} 0\le x\le a \\ \dfrac{X(L-x)}{L-a} \mathrm{~~~for~~~} a\le x\le L \end{array} \right. \tag{6}$$


$$\dfrac{\partial w(x,0)}{\partial t}=0 . \tag{7}$$

The second condition (7) requires

$$\sum_n{n \Omega b_n \sin\dfrac{n \pi x}{L}}=0 \tag{8}$$

for all values of $x$. This is only possible if $b_n=0$ for all values of $n$.

Now the condition (6) requires

$$ \sum_n{a_n \sin \dfrac{n \pi x}{L}} = \left\{ \begin{array}{ll} \dfrac{xX}{a} \mathrm{~~~for~~~} 0\le x\le a \\ \dfrac{X(L-x)}{L-a} \mathrm{~~~for~~~} a\le x\le L \end{array} \right. \tag{9}$$

In other words, we must express the initial deflected shape as a Fourier sine series with coefficients $a_n$. This can be done using the standard formula for Fourier coefficients:

$$a_n= \dfrac{2}{L} \left\lbrace \int_0^a{ \left[ \dfrac{xX}{a} \right] \sin \dfrac{n \pi x}{L} dx } + \int_a^L{\left[ \dfrac{X(L-x)}{L-a} \right] \sin \dfrac{n \pi x}{L} dx } \right\rbrace . \tag{10}$$

Integration by parts gives

$$a_n=\dfrac{2X}{La} \left\lbrace \left[ -x \dfrac{\cos (n \pi x/L)}{n \pi/L} \right] _0^a + \dfrac{L}{n \pi} \int_0^a{\cos \dfrac{n \pi x}{L} dx}\right\rbrace $$

$$+\dfrac{2X}{L(L-a)} \left\lbrace \left[ -(L-x) \dfrac{\cos (n \pi x/L)}{n \pi/L} \right]_a^L – \dfrac{L}{n \pi} \int_a^aL{\cos \dfrac{n \pi x}{L} dx}\right\rbrace $$

$$ = \dfrac{2}{n \pi}\left\lbrace \dfrac{X}{a}\left( -a \cos \dfrac{n \pi a}{L}\right) +\dfrac{LX}{n \pi a}\sin \dfrac{n \pi a}{L} \right. $$

$$\mathrm{~~~~~~~~~~~}\left. +\dfrac{X}{L-a}\left[ (L-a) \cos \dfrac{n \pi a}{L} +\dfrac{L}{n \pi}\sin \dfrac{n \pi a}{L}\right] \right\rbrace $$

$$= \dfrac{2XL}{n^2 \pi^2} \left( \dfrac{1}{a}+\dfrac{1}{L-a}\right) \sin \dfrac{n \pi a}{L}$$

so finally

$$a_n= \dfrac{2XL^2}{n^2 \pi^2 a(L-a)} \sin \dfrac{n \pi a}{L}= \dfrac{2LF}{n^2 \pi^2 T} \sin \dfrac{n \pi a}{L} \tag{11}$$

using eq. (2). The motion of the string is thus described by

$$w(x,t)= \sum_n{\dfrac{2LF}{n^2 \pi^2 T} \sin \dfrac{n \pi x}{L} \sin \dfrac{n \pi a}{L} \cos n \Omega t } \tag{12}$$

from eq. (5)

There is an alternative approach to deriving this result, making use of a general formula for the step response of a linear system. For any system with mode shapes $u_n(x)$ and corresponding natural frequencies $\omega_n$, the response to this kind of downward step input is given by the general formula

$$w(x,t)= F \sum_n{\frac{u_n(x) u_n(a) \cos(\omega_n t)}{\omega_n^2}}. \tag{13}$$

Applying this to the particular modes and natural frequencies of the ideal string, the motion following the pluck is given by

$$w(x,t)= \frac{2F}{mL} \sum_n{\frac{1}{ n^2 \Omega^2} \sin \frac{n \pi x}{L} \sin \frac{n \pi a}{L} \cos n \Omega t} \tag{14}$$

The factor $2/mL$ arises because the mode shapes have to be normalised according to eq. (10) of section 2.2.5. Applied to the string, this requires

$$\int_0^L{m u_n^2(x) dx}=1 \tag{15}$$

so that

$$u_n(x) = \sqrt{\frac{2}{mL}} \sin \frac{n \pi x}{L}. \tag{16}$$

Substituting the value $\Omega= \pi c/L$, eq. (14) gives exactly the same result as eq. (12).