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Normal Modes of a Beaded String

Consider a mechanical system consisting of a taut string that is stretched between two immovable walls. Suppose that $ N$ identical beads of mass $ m$ are attached to the string in such a manner that they cannot slide along it. Let the beads be equally spaced a distance $ a$ apart, and let the distance between the first and the last beads and the neighboring walls also be $ a$ . (See Figure 19.) Consider transverse oscillations of the string: that is, oscillations in which the string moves up and down (i.e., in the $ y$ -direction). It is assumed that the inertia of the string is negligible with respect to that of the beads. It follows that the sections of the string between neighboring beads, and between the outermost beads and the walls, are straight. (Otherwise, there would be a net tension force acting on the sections, and they would consequently suffer an infinite acceleration.) In fact, we expect the instantaneous configuration of the string to be a set of continuous straight-line segments of varying inclinations, as shown in the figure. Finally, assuming that the transverse displacement of the string is relatively small, it is reasonable to suppose that each section of the string possesses the same tension, $ T$ . (See later.)

Figure 19: A beaded string.
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It is convenient to introduce a Cartesian coordinate system such that $ x$ measure distance along the string from the left wall, and $ y$ measures the transverse displacement of the string from its equilibrium position. (See Figure 19.) Thus, when the string is in its equilibrium position it runs along the $ x$ -axis. We can define

$\displaystyle x_i = i\,a,$ (225)

where $ i=1,2,\cdots, N$ . Here, $ x_1$ is the $ x$ -coordinate of the closest bead to the left wall, $ x_2$ the $ x$ -coordinate of the second closest bead, et cetera. The $ x$ -coordinates of the beads are assumed to remain constant during their transverse oscillations. We can also define $ x_0=0$ and $ x_{N+1}= (N+1)\,a$ as the $ x$ -coordinates of the left and right ends of the string, respectively. Let the transverse displacement of the $ i$ th bead be $ y_i(t)$ , for $ i=1,N$ . Since each displacement can vary independently, we are dealing with an $ N$ degree of freedom system. We would, therefore, expect such a system to possess $ N$ unique normal modes of oscillation.

Figure 20: A short section of a beaded string.

Consider the section of the string lying between the $ i-1$ th and $ i+1$ th beads, as shown in Figure 20. Here, $ x_{i-1}=x_i-a$ , $ x_i$ , and $ x_{i+1}=x_i+a$ are the distances of the $ i-1$ th, $ i$ th, and $ i+1$ th beads, respectively, from the left wall, whereas $ y_{i-1}$ , $ y_i$ , and $ y_{i+1}$ are the corresponding transverse displacements of these beads. The two sections of the string that are attached to the $ i$ th bead subtend angles $ \theta_i$ and $ \theta_{i+1}$ with the $ x$ -axis, as illustrated in the figure. Standard trigonometry reveals that

$\displaystyle \tan\theta_i = \frac{y_i-y_{i-1}}{x_i-x_{i-1}} = \frac{y_i-y_{i-1}}{a},$ (226)


$\displaystyle \tan\theta_{i+1} = \frac{y_{i+1}-y_{i}}{a}.$ (227)

However, if the transverse displacement of the string is relatively small--that is, if $ \vert y_i\vert\ll a$ for all $ i$ --which we shall assume to be the case, then $ \theta_i$ and $ \theta_{i+1}$ are both small angles. Thus, we can use the small angle approximation $ \tan\theta\simeq \theta$ . (See Appendix B.) It follows that

$\displaystyle \theta_i$ $\displaystyle \simeq \frac{y_i-y_{i-1}}{a},$ (228)
$\displaystyle \theta_{i+1}$ $\displaystyle \simeq \frac{y_{i+1}-y_{i}}{a}.$ (229)

Let us find the transverse equation of motion of the $ i$ th bead. This bead is subject to two forces: namely, the tensions in the sections of the string to the left and to the right of it. (Incidentally, we are neglecting any gravitational forces acting on the beads, compared to the tension forces.) These tensions are of magnitude $ T$ , and are directed parallel to the associated string sections, as shown in Figure 20. Thus, the transverse (i.e., $ y$ -directed) components of these two tensions are $ -T \sin\theta_i$ and $ T \sin\theta_{i+1}$ , respectively. Hence, the transverse equation of motion of the $ i$ th bead becomes

$\displaystyle m \ddot{y}_i = -T \sin\theta_i + T \sin\theta_{i+1}.$ (230)

However, because $ \theta_i$ and $ \theta_{i+1}$ are both small angles, we can employ the small angle approximation $ \sin\theta\simeq \theta$ . It follows that

$\displaystyle \ddot{y}_i\simeq \frac{T}{m}\left(\theta_{i+1}-\theta_i\right).$ (231)

Finally, making use of Equations (228) and (229), we obtain

$\displaystyle \ddot{y}_i = \omega_0^{\,2}\left(y_{i-1}-2\,y_i+y_{i+1}\right),$ (232)

where $ \omega_0 = \sqrt{T/ m a}$ . Because there is nothing special about the $ i$ th bead, we deduce that the preceding equation of motion applies to all $ N$ beads: that is, it is valid for $ i=1,N$ . Of course, the first ($ i=1$ ) and last ($ i=N$ ) beads are special cases, because there is no bead corresponding to $ i=0$ or $ i=N+1$ . In fact, $ i=0$ and $ i=N+1$ correspond to the left and right ends of the string, respectively. However, Equation (232) still applies to the first and last beads, as long as we set $ y_0=0$ and $ y_{N+1}=0$ . What we are effectively demanding is that the two ends of the string, which are attached to the left and right walls, must both have zero transverse displacement.

Incidentally, we can prove that the tensions in the two sections of the string shown in Figure 20 must be equal by considering the longitudinal equation of motion of the $ i$ th bead. This equation takes the form

$\displaystyle m\,\ddot{x}_i=-T_i\,\cos\theta_i +T_{i+1}\,\cos\theta_{i+1},$ (233)

where $ T_i$ and $ T_{i+1}$ are the, supposedly different, tensions in the sections of the string to the immediate left and right of the $ i$ th bead, respectively. We are assuming that the motion of the beads is purely transverse: that is, all of the $ x_i$ are constant in time. Thus, it follows from the preceding equation that

$\displaystyle T_i\,\cos\theta_i = T_{i+1}\,\cos\theta_{i+1}.$ (234)

However, if the transverse displacement of the string is such that all of the $ \theta_i$ are small then we can make use of the small angle approximation $ \cos\theta_i\simeq 1$ . (See Appendix B.) Hence, we obtain

$\displaystyle T_i\simeq T_{i+1}.$ (235)

A straightforward extension of this argument reveals that the tension is the same in all sections of the string.

Let us search for a normal mode solution to Equation (232) that takes the form

$\displaystyle y_i(t) = A \sin(k x_i) \cos(\omega t-\phi),$ (236)

where $ A>0$ , $ k>0$ , $ \omega>0$ , and $ \phi$ are constants. This particular type of solution is such that all of the beads execute transverse simple harmonic oscillations in phase with one another. (See Figure 22.) Moreover, the oscillations have an amplitude $ A \sin(k x_i)$ that varies sinusoidally along the length of the string (i.e., in the $ x$ -direction). The pattern of oscillations is thus periodic in space. The spatial repetition period, which is generally termed the wavelength, is $ \lambda=2\pi/k$ . [This follows from Equation (236) because $ \sin\theta$ is a periodic function with period $ 2\pi$ : i.e., $ \sin(\theta+2\pi)\equiv \sin\theta$ .] The constant $ k$ , which determines the wavelength, is usually referred to as the wavenumber. Thus, a small wavenumber corresponds to a long wavelength, and vice versa. The type of solution specified in Equation (236) is generally known as a standing wave. It is a wave because it is periodic in both space and time. (An oscillation is periodic in time only.) It is a standing wave, rather than a traveling wave, because the points of maximum and minimum amplitude oscillation are stationary (in $ x$ ). (See Figure 22.)

Substituting Equation (236) into Equation (232), we obtain

$\displaystyle -\omega^2\,A\,\sin(k\,x_i)\,\cos(\omega\,t-\phi)$ $\displaystyle =\omega_0^{\,2}\,A\left[\sin(k\,x_{i-1})-2\,\sin(k\,x_i)\right.$    
  $\displaystyle  \left.+\sin(k x_{i+1})\right] \cos(\omega t-\phi),$ (237)

which yields

$\displaystyle -\omega^2\,\sin(k\,x_i)=\omega_0^{\,2}\left(\sin[k\,(x_i-a)] -2\,\sin(k\,x_i) +\sin[k\,(x_i+a)]\right).$ (238)

However, because $ \sin(a+b)\equiv \sin a \cos b + \cos a \sin b$ (see Appendix B), we get

$\displaystyle -\omega^2\,\sin(k\,x_i)=\omega_0^{\,2}\left[\cos(k\,a)-2+\cos(k\,a)\right]\sin(k\,x_i),$ (239)

which gives

$\displaystyle \omega = 2 \omega_0 \sin(k a/2),$ (240)

where use has been made of the trigonometric identity $ 1-\cos\theta \equiv 2 \sin^2(\theta/2)$ . (See Appendix B.) An expression, such as Equation (240), that determines the angular frequency, $ \omega $ , of a wave in terms of its wavenumber, $ k$ , is generally termed a dispersion relation.

The solution (236) is only physical provided $ y_0=y_{N+1}=0$ . In other words, provided the two ends of the string remain stationary. The first constraint is automatically satisfied, because $ x_0=0$ [see Equation (225)]. The second constraint implies that

$\displaystyle \sin(k x_{N+1}) = \sin[(N+1) k a] = 0.$ (241)

This condition can only be satisfied if

$\displaystyle k = \frac{n}{N+1}\,\frac{\pi}{a},$ (242)

where the integer $ n$ is known as a mode number. A small mode number translates to a small wavenumber, and, hence, to a long wavelength, and vice versa. We conclude that the possible wavenumbers, $ k$ , of the normal modes of the system are quantized such that they are integer multiples of $ \pi/[(N+1)\,a]$ . Thus, the $ n$ th normal mode is associated with the characteristic pattern of bead displacements

$\displaystyle y_{n,i}(t)=A_n\,\sin\left(\frac{n\,i}{N+1}\,\pi\right)\,\cos(\omega_n\,t-\phi_n),$ (243)


$\displaystyle \omega_n = 2\,\omega_0\,\sin\left(\frac{n}{N+1}\,\frac{\pi}{2}\right).$ (244)

Here, the integer $ i=1,N$ indexes the beads, whereas the mode number $ n$ indexes the normal modes. Furthermore, $ A_n$ and $ \phi_n$ are arbitrary constants determined by the initial conditions. Of course, $ A_n$ is the peak amplitude of the $ n$ th normal mode, whereas $ \phi_n$ is the associated phase angle.

How many unique normal modes does the system possess? At first sight, it might seem that there are an infinite number of normal modes, corresponding to the infinite number of possible values that the integer $ n$ can take. However, this is not the case. For instance, if $ n=0$ or $ n=N+1$ then all of the $ y_{n,i}$ are zero. Clearly, these cases are not real normal modes. Moreover, it can be demonstrated that

$\displaystyle \omega_{-n}$ $\displaystyle =-\omega_n,$ (245)
$\displaystyle y_{-n,i}(t)$ $\displaystyle = y_{n,i}(t),$ (246)

provided $ A_{-n}=-A_n$ and $ \phi_{-n}=-\phi_n$ , as well as

$\displaystyle \omega_{N+1+n}$ $\displaystyle =\omega_{N+1-n},$ (247)
$\displaystyle y_{N+1+n,i}(t)$ $\displaystyle =y_{N+1-n,i}(t),$ (248)

provided $ A_{N+1+n}=-A_{N+1-n}$ and $ \phi_{N+1+n}=\phi_{N+1-n}$ . We, thus, conclude that only those normal modes that have $ n$ in the range $ 1$ to $ N$ correspond to unique modes. Modes with $ n$ values lying outside this range are either null modes, or modes that are identical to other modes with $ n$ values lying within the prescribed range. It follows that there are $ N$ unique normal modes of the form (243). Hence, given that we are dealing with an $ N$ degree of freedom system, which we would expect to only possess $ N$ unique normal modes, we can be sure that we have found all the normal modes.

Figure 21: Normal modes of a beaded string with eight equally spaced beads.

Figure 21 illustrates the spatial variation of the normal modes of a beaded string possessing eight beads. That is, an $ N=8$ system. The modes are all shown at the instances in time when they attain their maximum amplitudes: in other words, at $ \omega_n\,t-\phi_n=0$ . It can be seen that the low-$ n$ --that is, long wavelength--modes cause the string to oscillate in a fairly smoothly varying (in $ x$ ) sine wave pattern. On the other hand, the high-$ n$ --that is, short wavelength--modes cause the string to oscillate in a rapidly varying zig-zag pattern that bears little resemblance to a sine wave. The crucial distinction between the two different types of mode is that the wavelength of the oscillation (in the $ x$ -direction) is much larger than the bead spacing in the former case, while it is similar to the bead spacing in the latter. For instance, $ \lambda=18\,a$ for the $ n=1$ mode, $ \lambda = 9\,a$ for the $ n=2$ mode, but $ \lambda = 2.25 a$ for the $ n=8$ mode.

Figure 22: Time evolution of the $ n=2$ normal mode of a beaded string with eight equally spaced beads.
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Figure 23 displays the temporal variation of the $ n=2$ normal mode of an $ N=8$ beaded string. The mode is shown at $ \omega_2\,t-\phi_2=0$ , $ \pi/8$ , $ \pi/4$ , $ 3\pi/8$ , $ \pi/2$ , $ 5\pi/8$ , $ 3\pi/2$ , $ 7\pi/8$ and $ \pi$ . It can be seen that the beads oscillate in phase with one another. In other words, they all attain their maximal transverse displacements, and pass through zero displacement, simultaneously. Moreover, the mid-way point of the string always remains stationary. Such a point is known as a node. The $ n=1$ normal mode has two nodes (counting the stationary points at each end of the string as nodes), the $ n=2$ mode has three nodes, the $ n=3$ mode four nodes, et cetera. In fact, the existence of nodes is one of the distinguishing features of a standing wave.

Figure 23: Normal frequencies of a beaded string with eight equally spaced beads.
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Figure 23 shows the normal frequencies of an $ N=8$ beaded string plotted as a function of the normalized wavenumber. Recall that, for an $ N=8$ system, the relationship between the normalized wavenumber, $ k\,a$ , and the mode number, $ n$ , is $ k\,a = (n/9)\,\pi$ . It can be seen that the angular frequency increases as the wavenumber increases, which implies that shorter wavelength modes have higher oscillation frequencies. The dependence of the angular frequency on the normalized wavenumber, $ k\,a$ , is approximately linear when $ k\,a\ll 1$ . Indeed, it can be seen from Equation (244) that if $ k\,a\ll 1$ then the small angle approximation $ \sin\theta\simeq \theta$ yields a linear dispersion relation of the form

$\displaystyle \omega_n \simeq (k\,a)\,\omega_0 = \left(\frac{n}{N+1}\right)\pi\,\omega_0.$ (249)

We, thus, conclude that those normal modes of a uniformly beaded string whose wavelengths greatly exceed the bead spacing (i.e., modes with $ k\,a\ll 1$ ) have approximately linear dispersion relations in which their angular frequencies are directly proportional to their mode numbers. However, it is evident from the figure that this linear relationship breaks down as $ k\,a\rightarrow 1$ , and the mode wavelength consequently becomes similar to the bead spacing.

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Next: Normal Modes of a Up: Transverse Standing Waves Previous: Transverse Standing Waves
Richard Fitzpatrick 2013-04-08