The frequencies of the normal modes form an arithmetic series ν 1, 2ν 1, 3ν 1,. The fundamental frequency ν 1 is thus the difference between the frequencies of two adjacent modes, i.e. ν 1 = ν = ν n+1 ν n. (3.26) Example: a 130 cm long string of mass 5.0 g oscillates in its n = 3 mode with a frequency of 175 Hz and a maximum amplitude of 5.0 mm. Calculate both the wavelength of the standing wave and tension in the string. Solution: so the string, fixed at both ends, form standing waves. Recall that the wavelength of the third harmonic (n = 3) is given by λ n = 2L n λ 3 = 2L 3 = 2.60 3 If the speed of the waves in the string is v = ν 3 λ 3, then = 0.867 m. v = ν 3 λ 3 = (175) (0.867) = 151.7 m s 1. Of course, the speed of the waves is also given by v = T s /µ, and so the tension, T s, is T s = µv 2 = m L v2 = ( ) 0.005 (151.7) 2 = 88.5 N. 1.30 3.6 Examples of Standing Waves Transverse standing waves in a string are only one example. Let us consider other common cases of transverse and longitudinal standing waves: 3.6.1 Electromagnetic Waves 102
Standing electromagnetic waves can be established between two parallel mirrors that reflect light back and forth. The mirrors are boundaries, analogous to the clamped/fixed points acting as boundaries at the end of the string. Two such facing mirrors (figure 56 47 ) form a laser cavity. Figure 56: Standing light waves in a laser cavity. The boundary conditions are that the light must have a node at the surface of each mirror. To allow light to escape from the cavity, and form a laser beam, on of the mirrors is only partially reflective note that this doesn t affect the boundary condition. Under such circumstances, we may apply equations (3.21) (3.25): The primary difference between the laser cavity and string is that a typical laser cavity has a length L 30 cm. Visible light has a wavelength λ 600 nm. 47 Knight, Figure 21.13, page 654 103
So we expect a standing light wave in a laser cavity to have a mode number n (i.e. number of antinodes) of m = 2L λ (2.0) (0.30) (600 10 9 ) = 106, (3.27) i.e. a standing light wave inside a laser cavity has approximately one million antinodes. 3.6.2 Sound Waves A long, narrow column of air (e.g. in a tube, pipe) can support longitudinal standing sound waves. As sound travels through the tube, the air oscillates parallel to the tube. If the tube is closed at one end, the air at the closed end cannot oscillate. So the closed end of a column of air must be a node. Figures 57 and 58 48 show the n = 2 standing wave inside a column of air that is closed at both ends a so-called closed closed tube. So standing sound waves are analogous to those standing waves discussed previously for the string. Closed-closed tubes are of very limited interest unless you are inside the column itself. Columns of air that emit sound are open at one or both ends. example, in musical instruments, For Open both ends flute. Open one end trumpet. Recall that at a discontinuity, some of the energy of the wave is partially transmitted and partially reflected. 48 Knight, Figures 21.14(a),(b), page 656 104
Figure 57: Physical representation of a standing longitudinal wave. Figure 58: Graphical representation of a standing longitudinal wave. 105
When sound wave traveling through the tube reaches the open end: Some of the wave s energy is transmitted out of the tube (e.g. in musical instrument, sound that we hear). Some of the wave s energy is reflected back into the tube. These reflections allow standing sound waves to exist in so-called open open and open closed tubes. At closed end, air has no room to vibrate i.e. node exists. At open end, the air does have room to vibrate so at the open end of an air column, there must be an antinode. Note: in reality, we find that the antinode is just outside the open end, which leads to something called an end correction, but we will ignore this, and assume the antinode is at the opening. Let us consider the standing waves for the three categories of closed closed, open open and open closed tubes in turn: Closed closed tube: figure 59. 49 Open open tube: figure 60. 50 Open closed tube: figure 61. 51 Note: in both closed closed and open open tubes, there are n half-wavelength segments between the ends. So the frequencies and wavelengths in both types of tube are the same as those of the string tied at both ends. For closed closed and open open tubes, λ n = 2L n, 49 Knight, Figure 21.15(a), page 657 50 Knight, Figure 21.15(b), page 657 51 Knight, Figure 21.15(c), page 657 ν n = nν 2L = nν 1, where n = 1.2.3.. (3.28) 106
Figure 59: The first three standing sound wave modes for a closed closed tube. 107
Figure 60: The first three standing sound wave modes for an open open tube. 108
Figure 61: The first three standing sound wave modes for an open closed tube. 109
The open-open tube case is different: The fundamental frequency ν 1 is half that for the closed closed and open open tubes of the same length. For open open tubes, λ n = 4L n, ν n = nν 4L = nν 1, where n = 1, 3, 5,, (3.29) i.e. n takes on only odd values. Note: figures 59, 60 and 61 are not pictures of the wave in the same way as they are for a string wave: The standing waves in this case are longitudinal. The figures show the displacement x parallel to the axis, against position x. The tube itself is shown merely to indicate the location of the open and closed ends. The diameter of the tube is not related to the amplitude of the displacement. Example: consider a sound wave in the 80 cm long tube shown in figure 62. 52 The tube is filled with an unknown gas. What is the speed of sound in the gas? Solution: referring to the figure, we see that the tube in which the sound waves will form standing waves is an open open tube, The gas molecules at the ends of the tube exhibit maximum displacement, i.e. forming antinodes. There is another antinode in the middle of the tube. Therefore, this is the 52 Knight, Figure ex21.15, page 677 110
Figure 62: Standing sound wave propagating through a tube. n = 2 mode, and the wavelength of the standing wave is equal to the length of the tube, i.e. λ = 0.80 m. Since ν = 500 Hz, the speed of the sound waves is v = νλ = (500) (0.80) = 400 m s 1. 3.7 Interference of Waves A basic characteristic of waves is their ability to combine into a single wave whose displacement is given by the principle of superposition. This combination/superposition is often called interference. From the past few sections, we recall that standing waves is the interference pattern produced when two waves of equal frequency travel in opposite directions. What happens when two sinusoidal waves, traveling in the same direction interfere? 3.7.1 One-dimensional Interference 111
Let us initially consider the case of two traveling waves of Equal amplitude a. Equal frequency ν = ω/2π. Equal speeds and so equal λ and hence wave number k = 2π/λ. Same direction along the +x-axis. Suppose that their displacements u 1 and u 2 are given by and u 1 (x 1, t) = a sin (kx 1 ωt + φ 10 ), (3.30) u 2 (x 2, t) = a sin (kx 2 ωt + φ 20 ), (3.31) where φ 10 and φ 20 are the phase constants of the waves. Note: φ 10 and φ 20 are characteristics of the sources, not the medium. Figures 63, 64 and 65 53 show snapshot graphs at t = 0 for waves emitted by three sources with phase constants φ 0 = 0 rad, φ 0 = π/2 rad and φ 0 = π rad. It can be seen that the phase constant tells us what the source is doing at t = 0. For example, a loudspeaker at its centre position, and moving backward at t = 0 has φ 0 = 0 rad. Identical sources have the same phase constants. Let us consider the superposition of u 1 and u 2 graphically (figures 66 and 67 54 ). 53 Knight, Figure 21.17, page 660 54 Knight, Figure 21.18, page 661 112
Figure 63: Phase constant φ 0 = 0 rad. Figure 64: Phase constant φ 0 = π/2 rad. In figure 66, the crests of two waves (or alternatively the troughs) are aligned as they travel along the +x-axis. Because of this crest crest or trough trough alignment, the waves are said to be in phase (i.e. in step ) with each other. In figure 67, the crests of one wave are slightly displaced from those of the other wave. 113
Figure 65: Phase constant φ 0 = π rad. Because of this misalignment, the waves are then said to be out of phase (i.e. out of step ) with each other. Note: in figures 66 and 67, the graphs and wave fronts are slightly displaced from each other simply so you can see what each wave is doing. In reality, of course, the waves are on top of each other. In figure 66: The displacements are the same at every point, i.e. u 1 (x) = u 2 (x). Thus, they must have the same phase (i.e. be in phase), or more precisely, (kx ωt + φ 10 ) = (kx ωt + φ 20 ), (3.32) (kx ωt + φ 10 ) = (kx ω 20 ± 2πn), where n = 0, 1, 2,, (3.33) n being an integer. 114
Figure 66: Constructive interference. Recalling the principle of superposition, combining waves that are in phase gives a net displacement at each point which is twice the displacement of each individual wave. This superposition to create a wave with an amplitude larger than either individual wave is called constructive interference. When waves are exactly in phase, giving A = 2a, we have maximum constructive interference. In figure 67: The crests of one wave align with the troughs of the other so the waves are out-of-phase with u 1 (x) = u 2 (x) at every point. 115
Figure 67: Destructive interference. Two waves aligned crest-to-trough are 180 out-of-phase. A superposition of two waves that creates a wave with amplitude smaller than either individual wave is called destructive interference. In this 180 out-of-phase case, the net displacement is zero at every point along the axis. The combination of two waves that cancel each other out to give no wave is called perfect destructive interference. Note: if the amplitudes of the waves are not equal (but their wavelengths are), then we can still get destructive interference, but it is not perfect. 116