General Time Evolution of Uniform String

In the preceding section, we found the normal modes of a uniform string of length $l$, both ends of which are attached to immovable walls. These modes are spatially-periodic solutions of the wave equation, (4.30), that oscillate at unique frequencies and satisfy the spatial boundary conditions (4.34) and (4.35). Because the wave equation is linear [i.e., if $y(x,t)$ is a solution then so is $\alpha\,y(x,t)$, where $\alpha$ is an arbitrary constant], it follows that its most general solution is a linear combination of all of the normal modes. In other words,

$$y(x,t) = \sum_{n'=1,\infty}y_{n'}(x,t) = \sum_{n'=1,\infty} A_{n'... ...'\,\pi\,\frac{x}{l}\right)\,\cos\left(n'\,\pi\,\frac{v\,t}{l}-\phi_{n'}\right),$$ (4.44)

where use has been made of Equations (4.37) and (4.38). The preceding expression is a solution of Equation (4.30), and also automatically satisfies the boundary conditions (4.34) and (4.35). As we have already mentioned, the constants $A_n$ and $\phi_n$ are determined by the initial conditions. Let us see how this comes about in more detail.

Suppose that the initial displacement of the string at $t=0$ is

$$y_0(x) \equiv y(x,0).$$ (4.45)

Likewise, let the initial velocity of the string be

$$v_0(x) \equiv \frac{\partial y(x,0)}{\partial t}.$$ (4.46)

For consistency with the boundary conditions, we must have $y_0(0)=y_0(l)=v_0(0)=v_0(l)=0$. It follows from Equation (4.44) that

$$y_0(x) =\sum_{n'=1,\infty} A_{n'}\,\cos\phi_{n'}\,\sin\left(n'\,\pi\,\frac{x}{l}\right),$$ (4.47)

$$v_0(x) = \frac{v}{l}\sum_{n'=1,\infty} n'\,\pi\,A_{n'}\,\sin\phi_{n'}\,\sin\left(n'\,\pi\,\frac{x}{l}\right).$$ (4.48)

It is readily demonstrated that

$$\frac{2}{l}\int_0^l\sin\left(n\,\pi\,\frac{x}{l}\right)\,\sin\left(n'\,\pi\,\frac{x}{l}\right)dx={} \frac{2}{\pi}\int_0^\pi \sin\left(n\,\theta\right)\,\sin\left(n'\,\theta\right)d\theta$$

$$={} \frac{1}{\pi}\int_0^\pi \cos\left[(n-n')\,\theta\right]d\theta$$

$$-\frac{1}{\pi}\int_0^\pi\cos\left[(n+n')\,\theta\right] d\theta$$

$$={} \frac{1}{\pi}\left[\frac{\sin\left[(n-n')\,\theta\right]}{n-n'}- \frac{\sin\left[(n+n')\,\theta\right]}{n+n'}\right]_0^{\pi}$$

$$={} \frac{\sin\left[(n-n')\,\pi\right]}{(n-n')\,\pi}- \frac{\sin\left[(n+n')\,\pi\right]}{(n+n')\,\pi},$$ (4.49)

where $n$ and $n'$ are (possibly different) positive integers, $\theta=\pi\,x/l$, and use has been made of the trigonometric identity $2\,\sin a\,\sin b \equiv \cos(a-b)-\cos(a+b)$. (See Appendix B.) Furthermore, if $k$ is a non-zero integer then

$$\frac{\sin(k\,\pi)}{k\,\pi} =0.$$ (4.50)

On the other hand, $k=0$ is a special case, because both the numerator and the denominator in the preceding expression become zero simultaneously. However, application of l`Hopital's rule yields

$$\lim_{x\rightarrow 0} \frac{\sin x}{x} = \lim_{x\rightarrow 0} \frac{d(\sin x)/dx}{dx/dx} = \lim_{x\rightarrow 0} \frac{\cos x}{1} = 1.$$ (4.51)

It follows that

$$\frac{\sin(k\,\pi)}{k\,\pi} = \left\{ \begin{array}{ccc} 1 &\mbox{\hspace{0.5cm}}&k=0\\ [0.5ex] 0&&k\neq 0 \end{array}\right.,$$ (4.52)

where $k$ is any integer. This result can be combined with Equation (4.49), recalling that $n$ and $n'$ are both positive integers, to give

$$\frac{2}{l}\int_0^l\sin\left(n\,\pi\,\frac{x}{l}\right)\,\sin\left(n'\,\pi\,\frac{x}{l}\right)dx = \delta_{n,n'}.$$ (4.53)

Here, the quantity

$$\delta_{n,n'} = \left\{ \begin{array}{ccc} 1 &\mbox{\hspace{0.5cm}}&n=n'\\ [0.5ex] 0&&n\neq n' \end{array}\right.,$$ (4.54)

where $n$ and $n'$ are integers, is known as the Kronecker delta function.

Let us multiply Equation (4.47) by $(2/l)\,\sin (n\,\pi\,x/l)$, and integrate over $x$ from 0 to $l$. We obtain

$$\frac{2}{l}\int_0^l y_0(x)\,\sin\left(n\,\pi\,\frac{x}{l}\right)dx =\sum_{n'=1,\infty} \!\!\!A_{n'}\,\cos\phi_{n'}\,\frac{2}{l}\int_0^\pi\sin\left(n'\,\pi\,\frac{x}{l}\right)\,\sin\left(n\,\pi\,\frac{x}{l}\right)dx$$

$$=\sum_{n'=1,\infty} A_{n'}\,\cos\phi_{n'}\,\delta_{n,n'} = A_n\cos\phi_n,$$ (4.55)

where use has been made of Equations (4.53) and (4.54). Similarly, Equation (4.48) yields

$$\frac{2}{v}\int_0^l v_0(x)\,\sin\left(n\,\pi\,\frac{x}{l}\right)dx = n\,\pi \,A_n\,\sin\phi_n.$$ (4.56)

Thus, defining the integrals

$$C_n = \frac{2}{l}\int_0^l y_0(x)\,\sin\left(n\,\pi\,\frac{x}{l}\right)dx,$$ (4.57)

$$S_n = \frac{2}{n\,\pi\,v}\int_0^l v_0(x)\,\sin\left(n\,\pi\,\frac{x}{l}\right)dx,$$ (4.58)

for $n=1,\infty$, we obtain

$$C_n = A_n\,\cos\phi_n,$$ (4.59)

$$S_n = A_n\,\sin\phi_n,$$ (4.60)

and, hence,

$$A_n = (C_n^{\,2} + S_n^{\,2})^{1/2},$$ (4.61)

$$\phi_n = \tan^{-1}(S_n/C_n).$$ (4.62)

Thus, the constants $A_n$ and $\phi_n$, appearing in the general expression (4.44) for the time evolution of a uniform string with fixed ends, are ultimately determined by integrals over the string's initial displacement and velocity of the form (4.57) and (4.58).

As an example, suppose that the string is initially at rest, so that

$$v_0(x) = 0,$$ (4.63)

but has the initial displacement

$$y_0(x) = 2\,A\left\{ \begin{array}{lcc} x/l&\mbox{\hspace{0.5... ... x < l/2\\ [0.5ex] 1-x/l &&l/2\leq x\leq l \end{array}\right. .$$ (4.64)

This triangular pattern is zero at both ends of the string, rising linearly to the peak value $A$, halfway along the string, and is designed to mimic the initial displacement of a guitar string that is plucked at its mid-point. See Figure 4.10. A comparison of Equations (4.58) and (4.63) reveals that, in this particular example, all of the $S_n$ coefficients are zero. Hence, from Equations (4.61) and (4.62), $A_n = C_n$ and $\phi_n=0$ for all $n$. Thus, making use of Equations (4.44), (4.57), and (4.64), the time evolution of the string is governed by

$$y(x,t) = \sum_{n=1,\infty}A_n\,\sin\left(n\,\pi\,\frac{x}{l}\right)\,\cos\left(n\,2\pi\,\frac{t}{\tau}\right),$$ (4.65)

where $\tau= 2\,l/v$ is the oscillation period of the $n=1$ normal mode, and

$$A_n = \frac{2}{l}\int_0^{l/2}2\,A\,\frac{x}{l}\,\sin\left(n\,\pi\... ...l/2}^{l}2\,A\left(1-\frac{x}{l}\right)\sin\left(n\,\pi\,\frac{x}{l}\right)\,dx.$$ (4.66)

The preceding expression transforms to

$$A_n = A\left(\frac{2}{\pi}\right)^2\left\{ \int_0^{\pi/2}\theta\,... ...eta)\,d\theta + \int_{\pi/2}^\pi(\pi-\theta)\,\sin(n\,\theta)\,d\theta\right\},$$ (4.67)

where $\theta=\pi\,x/l$. Integration by parts (Riley 1974) yields

$$A_n = 2\,A\,\frac{\sin(n\,\pi/2)}{(n\,\pi/2)^{\,2}}.$$ (4.68)

It follows that $A_n=0$ whenever $n$ is even. We conclude that the triangular initial displacement pattern (4.64) only excites normal modes with odd mode numbers.

When a stringed instrument, such as a guitar, is played, a characteristic pattern of normal mode oscillations is excited on the plucked string. These oscillations excite sound waves of the same frequency, which propagate through the air and are audible to a listener. The normal mode (of appreciable amplitude) with the lowest oscillation frequency is called the fundamental harmonic, and determines the pitch of the musical note that is heard by the listener. For instance, a fundamental harmonic that oscillates at $261.6$ Hz corresponds to “middle C”. Those normal modes (of appreciable amplitude) with higher oscillation frequencies than the fundamental harmonic are called overtone harmonics, because their frequencies are integer multiples of the fundamental frequency. In general, the amplitudes of the overtone harmonics are much smaller than the amplitude of the fundamental. Nevertheless, when a stringed instrument is played, the particular mix of overtone harmonics that accompanies the fundamental determines the timbre of the musical note heard by the listener. For instance, when middle C is played on a piano and a harpsichord, the same frequency fundamental harmonic is excited in both cases. However, the mix of excited overtone harmonics is quite different. This accounts for the fact that middle C played on a piano can be easily distinguished from middle C played on a harpsichord.

Figure 4.9: Relative amplitudes of the overtone harmonics of a uniform guitar string plucked at its mid-point.

Figure 4.9 shows the ratio $A_n/A_1$ for the first ten overtone harmonics of a uniform guitar string plucked at its midpoint; that is, the ratio $A_n/A_1$ for odd-$n$ modes, with $n>1$, calculated from Equation (4.68). It can be seen that the amplitudes of the overtone harmonics are all small compared to the amplitude of the fundamental. Moreover, the amplitudes decrease rapidly in magnitude with increasing mode number, $n$.

Figure 4.10: Reconstruction of the initial displacement of a uniform guitar string plucked at its mid-point. The long-dashed line shows a reconstruction made with only the largest-amplitude normal mode, the short-dashed line shows a reconstruction made with the four largest-amplitude normal modes, and the solid line shows a reconstruction made with the sixteen largest-amplitude normal modes.

In principle, we must include all of the normal modes in the sum on the right-hand side of Equation (4.65). In practice, given that the amplitudes of the normal modes decrease rapidly in magnitude as $n$ increases, we can truncate the sum, by neglecting high-$n$ normal modes, without introducing significant error into our calculation. Figure 4.10 illustrates the effect of such a truncation. In fact, this figure shows the reconstruction of $y_0(x)$, obtained by setting $t=0$ in Equation (4.65), made with various different numbers of normal modes. It can be seen that sixteen normal modes is sufficient to very accurately reconstruct the triangular initial displacement pattern. Indeed, a reconstruction made with only four normal modes is surprisingly accurate. On the other hand, a reconstruction made with only one normal mode is fairly inaccurate.

Figure 4.11: Time evolution of a uniform guitar string plucked at its mid-point. This evolution is reconstructed from Equation (4.65) using the sixteen largest-amplitude normal modes of the string. The upper solid, upper short-dashed, upper long-dashed, upper dot-short-dashed, dot-long-dashed, lower dot-short-dashed, lower long-dashed, lower short-dashed, and lower solid curves correspond to $t/\tau = 0$, $1/16$, $1/8$, $3/16$, $1/4$, $5/16$, $3/8$, $7/16$, and $1/2$, respectively.

Figure 4.11 shows the time evolution of a uniform guitar string plucked at its mid-point. It can be seen that the string oscillates in a rather peculiar fashion. The initial kink in the string at $x=l/2$ splits into two equal kinks that propagate in opposite directions along the string, at the velocity $v$. The string remains straight and parallel to the $x$-axis between the kinks, and straight and inclined to the $x$-axis between each kink and the closest wall. When the two kinks reach the wall, the string is instantaneously found in its undisturbed position. The kinks then reflect off the two walls, with a phase change of $\pi$ radians. When the two kinks meet again at $x=l/2$ the string is instantaneously found in a state that is an inverted form of its initial state. The kinks subsequently pass through one another, reflect off the walls, with another phase change of $\pi$ radians, and meet for a second time at $x=l/2$. At this instant, the string is again found in its initial position. The pattern of motion then repeats itself ad infinitum. The period of the oscillation is the time required for a kink to propagate two string lengths, which is $\tau= 2\,l/v$. This is also the oscillation period of the $n=1$ normal mode.