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2.6), and taking the limit by22
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0 (problem 2.7), the displacement of the string is given
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A sinwt y(x, t) = wv SIn (L w/ v ) .
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([x + xo]w/v) ([x + xo]w/v) -
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< x < xo < L/2 < xo < x < L/2.
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Looking at (2.83), we can see that unlike the infinite string, there are no traveling waves, only a standing wave. 23 This happens because the ends of the string are fixed and reflect back the waves. Moreover, there is now a sin(wL/v) in the denominator of the second term in (2.83). When this sine vanishes, the displacement actually becomes infinite. This happens for wL/v = mr, where n is an integer. These frequencies Wn = (1r / L )nv are the resonance frequencies of our one-dimensional resonator. We have seen that the confinement makes the resonances appear. Being resonant to specific discrete frequencies is a very important characteristic of resonators. It is so important that it gave them their name. The second important feature is the specific field patterns that form inside the resonator when it vibrates at one of its resonance frequencies (i.e., the modes). Unfortunately, we cannot see these patterns using (2.83), as this solution diverges when the driving force vibrates at one of the resonator resonances. In practice, the displacement can become very large but never really diverges because of damping. Real resonators are never perfect and always let the fields leak a bit from their interior (or even absorb them a little), which leads to the damping of the vibrations. We could introduce damping, as is often done in textbook treatments of the harmonic oscillator, for example. However, then the field patterns that we would see would not correspond to a mode of a perfect resonator, which clearly show the ideas we are discussing, but to a more complicated sort of mode. We discuss damping in resonators later (you will get a flavor of it in 6 and a more detailed discussion will be given in 9). To see the patterns that develop in our one-dimensional string resonator, we will simply remove the driving force and assume that we have somehow set the string in oscillation so that it oscillates as a perfect sine at a point x = Xo: for example,
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As there is no driving force, the displacement satisfies (2.74). Then the general solution24 must be of the form
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22The previous solution can be recovered by taking the limit L ..... 00 and then f ..... 0 (Problem 2.8). 23In fact, this standing wave can be written as the sum of two traveling waves: one that travels to the right and another that travels to the left. 24The solution obtained by imposing only the boundary conditions y( -L/2, t) = y(L/2, t) = O.
At x = Xo (2.85) must coincide with (2.84):
~ { an sin ( n Ivt) + bn cos ( n Ivt) } sin ( n I [xo + ~ ]) =
A sin wt.
(2.86) Now, as this is a solution for all times, we can multiply (2.86) by (1/21r) exp( -iw't) and integrate over t from -00 to 00. We find that
~ansin(nI [xo+~]) {6(nI v - w') -6(nIv+w')}
+i~ bnsin (nI
= A {6 (w
+~]) {6 (nIv -Wi) + 6(nIv +Wl)}
- Wi) - 6 (w + Wi)} .
As an, bn , and A are all real, it follows that (2.88) and
~ an sin (nI
= A {6 (w -
[XO + ~]) {6 ( n IV - w') - 6 ( nIv + Wi) }
Wi) - 6 (w
+ Wi)}.
Now if w =f n( 1r/ L)v for all n = 1,2, ... there is an e such that the interval (w - e, w+ e) does not contain any ofthe points n(1r / L )v. Integrating over Wi on this interval, we find that A and all the an must vanish for (2.86) to hold. In other words, our simple one-dimensional resonator can only support free persistent oscillations of frequenciesw n = n(1r/ L)v. Having bothw =f Wn and A =f 0 in (2.86) is inconsistent. When w = Wn for a given fi, the integration over Wi mentioned above yields (2.90) and an