# 3.6: Finite Length Strings

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We now return to the physical example of wave propagation in a string. We found that the general solution can be represented as a sum over product solutions. We will restrict our discussion to the special case that the initial velocity is zero and the original profile is given by $$u(x, 0)=f(x)$$. The solution is then $u(x, t)=\sum_{n=1}^{\infty} A_{n} \sin \frac{n \pi x}{L} \cos \frac{n \pi c t}{L}\label{eq:1}$ satisfying $f(x)=\sum_{n=1}^{\infty} A_{n} \sin \frac{n \pi x}{L} .\label{eq:2}$ We have seen that the Fourier sine series coefficients are given by $A_{n}=\frac{2}{L} \int_{0}^{L} f(x) \sin \frac{n \pi x}{L} d x .\label{eq:3}$

We can rewrite this solution in a more compact form. First, we define the wave numbers, $k_{n}=\frac{n \pi}{L}, \quad n=1,2, \ldots,\nonumber$ and the angular frequencies, $\omega_{n}=c k_{n}=\frac{n \pi c}{L} .\nonumber$ Then, the product solutions take the form $\sin k_{n} x \cos \omega_{n} t\nonumber$ Using trigonometric identities, these products can be written as $\sin k_{n} x \cos \omega_{n} t=\frac{1}{2}\left[\sin \left(k_{n} x+\omega_{n} t\right)+\sin \left(k_{n} x-\omega_{n} t\right)\right] .\nonumber$ Inserting this expression in the solution, we have $u(x, t)=\frac{1}{2} \sum_{n=1}^{\infty} A_{n}\left[\sin \left(k_{n} x+\omega_{n} t\right)+\sin \left(k_{n} x-\omega_{n} t\right)\right] .\label{eq:4}$

Since $$\omega_{n}=c k_{n}$$, we can put this into a more suggestive form: $u(x, t)=\frac{1}{2}\left[\sum_{n=1}^{\infty} A_{n} \sin k_{n}(x+c t)+\sum_{n=1}^{\infty} A_{n} \sin k_{n}(x-c t)\right] .\label{eq:5}$

We see that each sum is simply the sine series for $$f(x)$$ but evaluated at either $$x+c t$$ or $$x-c t$$. Thus, the solution takes the form $u(x, t)=\frac{1}{2}[f(x+c t)+f(x-c t)] .\label{eq:6}$ If $$t=0$$, then we have $$u(x, 0)=\frac{1}{2}[f(x)+f(x)]=f(x)$$. So, the solution satisfies the initial condition. At $$t=1$$, the sum has a term $$f(x-c)$$.

Recall from your mathematics classes that this is simply a shifted version of $$f(x)$$. Namely, it is shifted to the right. For general times, the function is shifted by $$c t$$ to the right. For larger values of $$t$$, this shift is further to the right. The function (wave) shifts to the right with velocity $$c$$. Similarly, $$f(x+c t)$$ is a wave traveling to the left with velocity $$-c$$.

Thus, the waves on the string consist of waves traveling to the right and to the left. However, the story does not stop here. We have a problem when needing to shift $$f(x)$$ across the boundaries. The original problem only defines $$f(x)$$ on $$[0, L]$$. If we are not careful, we would think that the function leaves the interval leaving nothing left inside. However, we have to recall that our sine series representation for $$f(x)$$ has a period of $$2 L$$. So, before we apply this shifting, we need to account for its periodicity. In fact, being a sine series, we really have the odd periodic extension of $$f(x)$$ being shifted. The details of such analysis would take us too far from our current goal. However, we can illustrate this with a few figures.

We begin by plucking a string of length $$L$$. This can be represented by the function $f(x)=\left\{\begin{array}{cc} \frac{x}{a} & 0 \leq x \leq a \\ \frac{L-x}{L-a} & a \leq x \leq L \end{array}\right.\label{eq:7}$ where the string is pulled up one unit at $$x=a$$. This is shown in Figure $$\PageIndex{1}$$.

Next, we create an odd function by extending the function to a period of $$2 L$$. This is shown in Figure $$\PageIndex{2}$$.

Finally, we construct the periodic extension of this to the entire line. In Figure $$\PageIndex{3}$$ we show in the lower part of the figure copies of the periodic extension, one moving to the right and the other moving to the left. (Actually, the copies are $$\frac{1}{2} f(x \pm c t)$$.) The top plot is the sum of these solutions. The physical string lies in the interval $$[0,1]$$. Of course, this is better seen when the solution is animated.

The time evolution for this plucked string is shown for several times in Figure $$\PageIndex{4}$$. This results in a wave that appears to reflect from the ends as time increases.

The relation between the angular frequency and the wave number, $$\omega=$$ $$c k$$, is called a dispersion relation. In this case $$\omega$$ depends on $$k$$ linearly. If one knows the dispersion relation, then one can find the wave speed as $$c=\frac{\omega}{k}$$. In this case, all of the harmonics travel at the same speed. In cases where they do not, we have nonlinear dispersion, which we will discuss later.

This page titled 3.6: Finite Length Strings is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Russell Herman via source content that was edited to the style and standards of the LibreTexts platform.