Normal Modes of Uniform String

Consider a uniformly-beaded string in the limit in which the number of beads, $N$, becomes increasingly large, while the spacing, $a$, and the individual mass, $m$, of the beads become increasingly small. Let the limit be taken in such a manner that the length, $l=(N+1)\,a$, and the average mass per unit length, $\rho=m/a$, of the string both remain constant. As $N$ increases, and becomes very large, such a string will more and more closely approximate a uniform string of length $l$ and mass per unit length $\rho$. What can we guess about the normal modes of a uniform string on the basis of the analysis contained in the preceding section? First of all, we would guess that a uniform string is an infinite degree of freedom system, with an infinite number of unique normal modes of oscillation. This follows because a uniform string is the $N\rightarrow\infty$ limit of a beaded string, and a beaded string possesses $N$ unique normal modes. Next, we would guess that the normal modes of a uniform string exhibit smooth sinusoidal spatial variation in the $x$-direction, and that the angular frequency of the modes is directly proportional to their wavenumber. These last two conclusions follow because all of the normal modes of a beaded string are characterized by $k\,a\ll 1$, in the limit in which the spacing between the beads becomes zero. Let us see whether these guesses are correct.

Consider the transverse oscillations of a uniform string of length $l$ and mass per unit length $\rho$ that is stretched between two immovable walls. It is again convenient to define a Cartesian coordinate system in which $x$ measures distance along the string from the left wall, and $y$ measures the transverse displacement of the string. Thus, the instantaneous state of the system at time $t$ is determined by the function $y(x,t)$ for $0\leq x\leq l$. This function consists of an infinite number of different $y$ values, corresponding to the infinite number of different $x$ values in the range 0 to $l$. Moreover, all of these $y$ values are free to vary independently of one another. It follows that we are indeed dealing with a dynamical system possessing an infinite number of degrees of freedom.

Let us try to reuse some of the analysis of the previous section. We can reinterpret $y_i(t)$ as $y(x,t)$, $y_{i-1}(t)$ as $y(x-\delta x,t)$, and $y_{i+1}(t)$ as $y(x+\delta x,t)$, assuming that $x_i=x$ and $a=\delta x$. Moreover, $\ddot{y}_i(t)$ becomes $\partial^{\,2} y(x,t)/\partial t^{\,2}$; namely, a second derivative of $y(x,t)$ with respect to $t$ at constant $x$. Finally, $\omega_0^{\,2}= T/(m\,a)$, where $T$ is the tension in the string, can be rewritten as $T/[\rho\,(\delta x)^{\,2}]$, because $\rho=m/\delta x$. Incidentally, we are again assuming that the transverse displacement of the string remains sufficiently small that the tension is approximately constant in $x$. Thus, the equation of motion of the beaded string, (4.8), transforms into

$$\frac{\partial^{\,2} y(x,t)}{\partial t^{\,2}} = \frac{T}{\rho}\left[\frac{y(x-\delta x,t)-2\,y(x,t)+y(x+\delta x,t)}{(\delta x)^{\,2}}\right].$$ (4.26)

However, Taylor expanding $y(x+\delta x,t)$ in $x$ at constant $t$ (see Appendix B), we obtain

$$y(x+\delta x,t) = y(x,t) + \frac{\partial y(x,t)}{\partial x}\,\d... ...,t)}{\partial x^{\,2}}\,(\delta x)^{\,2} + {\cal O}\left(\delta x^{\,3}\right).$$ (4.27)

Likewise,

$$y(x-\delta x,t) = y(x,t) - \frac{\partial y(x,t)}{\partial x}\,\d... ...,t)}{\partial x^{\,2}}\,(\delta x)^{\,2} + {\cal O}\left(\delta x^{\,3}\right).$$ (4.28)

It follows that

$$\left[\frac{y(x-\delta x,t)-2\,y(x,t)+y(x+\delta x,t)}{(\delta x)... ...}}\right] =\frac{\partial^{\,2} y(x,t)}{\partial x^{\,2}} + {\cal O}(\delta x).$$ (4.29)

Thus, in the limit that $\delta x\rightarrow 0$, Equation (4.26) reduces to

$$\frac{\partial^{\,2} y}{\partial t^{\,2}} = v^{\,2}\,\frac{\partial^{\,2} y}{\partial x^{\,2}},$$ (4.30)

where

$$v = \sqrt{\frac{T}{\rho}}$$ (4.31)

is a quantity having the dimensions of velocity. Equation (4.30), which is the transverse equation of motion of the string, is an example of a very famous partial differential equation known as the wave equation. The quantity $v$ turns out to be the propagation velocity of transverse waves along the string. (See Section 6.2.)

By analogy with Equation (4.12), let us search for a solution of the wave equation of the form

$$y(x,t) = A\,\sin(k\,x)\,\cos(\omega\,t-\phi),$$ (4.32)

where $A>0$, $k>0$, $\omega>0$, and $\phi$ are constants. We can interpret such a solution as a standing wave of wavenumber $k$, wavelength $\lambda=2\pi/k$, angular frequency $\omega$, peak amplitude $A$, and phase angle $\phi$. Substitution of the preceding expression into Equation (4.30) yields the dispersion relation [cf., Equation (4.16)]

$$\omega = k\,v.$$ (4.33)

The standing wave solution (4.32) is subject to the physical constraint that the two ends of the string, which are both attached to immovable walls, remain stationary. This leads directly to the spatial boundary conditions

$$y(0, t) =0,$$ (4.34)

$$y(l,t) =0.$$ (4.35)

It can be seen that the solution (4.32) automatically satisfies the first boundary condition. However, the second boundary condition is only satisfied when $\sin(k\,l)=0$, which immediately implies that

$$k = n\,\frac{\pi}{l},$$ (4.36)

where the mode number, $n$, is an integer. We, thus, conclude that the possible normal modes of a taut string, of length $l$ and fixed ends, have wavenumbers which are quantized such that they are integer multiples of $\pi/l$. Moreover, this quantization is a direct consequence of the imposition of the physical boundary conditions at the two ends of the string.

It follows, from the previous analysis, that the $n$th normal mode of the string is associated with the pattern of motion

$$y_n(x,t) = A_n\,\sin\left(n\,\pi\,\frac{x}{l}\right)\,\cos(\omega_n\,t-\phi_n),$$ (4.37)

where

$$\omega_n = n\,\frac{\pi\,v}{l}.$$ (4.38)

Here, $A_n$ and $\phi_n$ are constants that are determined by the initial conditions. (See Section 4.4.) How many unique normal modes are there? The choice $n=0$ yields $y_0(x,t)=0$ for all $x$ and $t$, so this is not a real normal mode. Moreover,

$$\omega_{-n} =-\omega_n,$$ (4.39)

$$y_{-n}(x,t) = y_n(x,t),$$ (4.40)

provided that $A_{-n}=-A_n$ and $\phi_{-n}=-\phi_n$. We conclude that modes with negative mode numbers give rise to the same patterns of motion as modes with corresponding positive mode numbers. However, modes with different positive mode numbers correspond to unique patterns of motion that oscillate at unique frequencies. It follows that the string possesses an infinite number of normal modes, corresponding to the mode numbers $n=1,2,3,$ et cetera. Recall that we are dealing with an infinite-degree-of-freedom system, which we would expect to possess an infinite number of unique normal modes. The fact that we have actually obtained an infinite number of such modes suggests that we have found all of the normal modes.

Figure: 4.6 First eight normal modes of a uniform string. The modes are all shown at the instances in time when they attain their maximum amplitudes; namely, at $\omega_n\,t-\phi_n=0$.

Figure 4.6 illustrates the spatial variation of the first eight normal modes of a uniform string with fixed ends. It can be seen that the modes are all smoothly-varying sine waves. The low-mode-number (i.e., long-wavelength) modes are actually quite similar in form to the corresponding normal modes of a uniformly-beaded string. (See Figure 4.3.) However, the high-mode-number modes are substantially different. We conclude that the normal modes of a beaded string are similar to those of a uniform string, with the same length and mass per unit length, provided that the wavelength of the mode is much larger than the spacing between the beads.

Figure 4.7: Time evolution of the $n=4$ normal mode of a uniform string. The mode is shown at $\omega_4\,t-\phi_4=0$, $\pi /8$, $\pi /4$, $3\pi /8$, $\pi /2$, $5\pi /8$, $3\pi /2$, $7\pi /8$ and $\pi$.

Figure 4.7 illustrates the temporal variation of the $n=4$ normal mode of a uniform string. All points on the string attain their maximal transverse displacements, and pass through zero displacement, simultaneously. Moreover, the $n=4$ mode possesses five nodes (i.e., points where the string remains stationary). Two of these are located at the ends of the string, and three in the middle. In fact, according to Equation (4.37), the nodes correspond to points where $\sin [n\,(x/l)\,\pi]=0$. Hence, the nodes are located at

$$x_{n,j} = \left(\frac{j}{n}\right)l,$$ (4.41)

where $j$ is an integer lying in the range 0 to $n$. Here, $n$ indexes the normal mode, and $j$ the node. Thus, the $j=0$ node lies at the left end of the string, the $j=1$ node is the next node to the right, et cetera. It is apparent, from the preceding formula, that the $n$th normal mode has $n+1$ nodes that are uniformly spaced a distance $l/n$ apart.

Figure 4.8: Normal frequencies of the first eight normal modes of a uniform string.

Finally, Figure 4.8 shows the first eight normal frequencies of a uniform string with fixed ends, plotted as a function of the mode number. It can be seen that the angular frequency of oscillation increases linearly with the mode number. Recall that the low-mode-number (i.e., long-wavelength) normal modes of a beaded string also exhibit a linear relationship between normal frequency and mode number of the form [see Equation (4.25)]

$$\omega_n = \frac{n\,\pi}{N+1}\,\omega_0= \frac{n\,\pi}{N+1}\left(\frac{T}{m\,a}\right)^{1/2}.$$ (4.42)

However, $m=\rho\,a$ and $l=(N+1)\,a$, so we obtain

$$\omega_n = \frac{n\,\pi}{l}\left(\frac{T}{\rho}\right)^{1/2} =n\, \frac{\pi\,v}{l},$$ (4.43)

which is identical to Equation (4.38). We, thus, conclude that the normal frequencies of a uniformly-beaded string are similar to those of a uniform string, with the same length and mass per unit length, as long as the wavelength of the associated normal mode is much larger than the spacing between the beads.