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24 CHAPTER 4. TRANSVERSE WAVES ON A STRING One fixed end, one free end Now consider the case where one end is fixed and the other is free. Let’s take the fixed end to be the one at x = 0. The case where the fixed end is located at x = L gives the same general result; it’s just the mirror image of the result we will obtain here. The two boundary conditions are ψ(0, t) = 0 and ∂ψ/∂x|x=L = 0 for all t. Eq. (56) again gives the most general form of a wave satisfying the first of these conditions. So we just need to demand that the second one is also true. From Eq. (56), the slope ∂ψ/∂x is proportional to cos kx. The only way for this to be zero at x = L is for kL to equal (n + 1/2)π for some integer n. So (n + 1/2)π kn = . (65) L n starts at zero here. Unlike in the two-fixed-ends case, the n = 0 value now gives a nontrivial wave. The wavelength is λn = 2π/kn = 2L/(n + 1/2). These wavelengths are most easily seen in pictures, and snapshots of the first few modes are shown below in the second set of <strong>waves</strong> in Fig. 24. The easiest way to describe the wavelengths in words is to note that the number of oscillations that fit on the string is L/λn = n/2 + 1/4. So for the lowest mode (the n = 0 one), a quarter of a wavelength fits on the string. The higher modes are then obtained by successively adding on half an oscillation (which ensures that x = L is always located at an antinode, with slope zero). The frequency νn can be found via Eq. (62), and we can summarize the results: λn = 2L n + 1/2 and νn = (n + 1/2)v 2L Similar to Eq. (64), the most general motion of a string with one fixed end and one free end is a linear combination of the solutions in Eq. (56): ψ(x, t) = ∞ Bn sin(ωnt + φn) sin knx where kn = n=0 (n + 1/2)π L (66) , and ωn = vkn. (67) If we instead had the left end as the free one, then Eq. (60) would be the relevant equation, and the sin kx here would instead be a cos kx. As far as the dependence on time goes, it doesn’t matter whether it’s a sine or cosine function of t, because a redefinition of the t = 0 point yields a phase shift that can turn sines into cosines, and vice versa. We aren’t free to redefine the x = 0 point, because we have a physical wall there. Two free ends Now consider the case with two free ends. The two boundary conditions are ∂ψ/∂x|x=0 = 0 and ∂ψ/∂x|x=L = 0 for all t. Eq. (60) gives the most general form of a wave satisfying the first of these conditions. So we just need to demand that the second one is also true. From Eq. (60), the slope ∂ψ/∂x is proportional to sin kx. The only way for this to be zero at x = L is for kL to equal nπ for some integer n. So kn = nπ L , (68) which is the same as in the case of two fixed ends. The wavelength is λn = 2π/kn = 2L/n. So the possible wavelengths are all integral divisors of 2L, again the same as in the case of two fixed ends. Snapshots of the first few modes are shown below in the third set of <strong>waves</strong> in Fig. 24.
4.5. STANDING WAVES 25 The n values technically start at n = 0. In this case ψ has no dependence on x, so we simply have a flat line (not necessarily at ψ = 0). The line just sits there at rest, because the frequency is again given by Eq. (62) and is therefore zero (the kn values, and hence ωn values, are the same as in the two-fixed-ends case, so ω0 = 0). This case isn’t as trivial as the n = 0 case for two fixed ends; the resulting constant value of ψ might be necessary to satisfy the initial conditions of the string. This constant value is analogous to the a0 term in the Fourier-series expression in Eq. (3.1). As with two fixed ends, an integral number of half oscillations must fit into L, but they now start and end at antinodes instead of nodes. The frequency νn is the same as in the two-fixed-ends case, so as in that case we have: λn = 2L n and νn = nv 2L Similar to Eqs. (64) and (67), the most general motion of a string with two free ends is a linear combination of the solutions in Eq. (60): ψ(x, t) = ∞ n=0 (69) Bn cos(ωnt + φn) cos knx where kn = nπ L , and ωn = vkn. (70) Fig. 24 summarizes the above results. 2L/λ 4 7/2 3 5/2 2 3/2 1 1/2 0 Figure 24 Power in a standing wave both ends fixed one fixed, one free both ends free fixed end free end We saw above in Section 4.4 that not only do traveling <strong>waves</strong> contain energy, they also contain an energy flow along the string. That is, they transmit power. A given point on the string does work (which may be positive or negative, depending on the direction of the wave’s velocity) on the part of the string to its right. And it does the opposite amount of work on the string to its left. A reasonable question to ask now is: Is there energy flow in standing <strong>waves</strong>? There is certainly an energy density, because in general the string both moves and stretches. But is there any energy transfer along the string? Intuitively, a standing wave is the superposition of two oppositely-moving traveling <strong>waves</strong> with equal amplitudes. These traveling <strong>waves</strong> have equal and opposite energy flow, on
Page 1 and 2: Chapter 4 Transverse</stron
Page 3 and 4: 4.1. THE WAVE EQUATION 3 In Fig. 2,
Page 5 and 6: 4.1. THE WAVE EQUATION 5 The physic
Page 7 and 8: 4.2. REFLECTION AND TRANSMISSION 7
Page 9 and 10: 4.2. REFLECTION AND TRANSMISSION 9
Page 11: 4.2. REFLECTION AND TRANSMISSION 11
Page 14 and 15: 14 CHAPTER 4. TRANSVERSE WAVES ON A
Page 17 and 18: 4.3. IMPEDANCE 17 your mind, you mi
Page 19 and 20: 4.4. ENERGY 19 work, and this work
Page 21 and 22: 4.5. STANDING WAVES 21 following ch
Page 23: 4.5. STANDING WAVES 23 4.5.2 Finite
Page 27 and 28: 4.6. ATTENUATION 27 therefore can d

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