The of the normal modes form an arithmetic series •

ν1, 2ν1, 3ν1, . · · ·

The fundamental ν1 is thus the difference between the fre- • quencies of two adjacent modes, i.e.

ν1 = ∆ν = ν +1 ν . (3.26) n − n

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 of 5.0 mm. Calculate both the of the standing and tension in the string.

Solution: so the string, fixed at both ends, form standing . Recall that the wavelength of the third (n = 3) is given by

2L 2L 2.60 λ = λ3 = = = 0.867 m . n n −→ 3 3

If the speed of the waves in the string is v = ν3λ3, then

−1 v = ν3λ3 = (175) (0.867) = 151.7 m s .

Of course, the speed of the waves is also given by v = Ts/µ, and so the tension, Ts, is p

m 0.005 T = µv2 = v2 = (151.7)2 = 88.5 N . s L 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 mir- rors that reflect light back and forth. The mirrors are boundaries, analogous to the clamped/fixed points act- • ing as boundaries at the end of the string.

Two such facing mirrors (figure 5647) form a cavity.

Figure 56: Standing light waves in a laser cavity.

The boundary conditions are that the light must have a 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. • ≈ 47Knight, 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

2L (2.0) (0.30) m = = 106 , (3.27) λ ≈ (600 10−9) × i.e. a standing light wave inside a laser cavity has approximately one million antinodes.

3.6.2 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. For • example, in musical instruments,

– Open both ends – flute.

– Open one end – .

Recall that at a discontinuity, some of the energy of the wave is partially • transmitted and partially reflected.

48Knight, Figures 21.14(a),(b), page 656

104 Figure 57: Physical representation of a standing .

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 in both types of tube are the same as those of the string tied at both ends.

For closed–closed and open–open tubes, •

2L nν λ = , ν = = nν1 , where n = 1.2.3. . (3.28) n n n 2L · · · 49Knight, Figure 21.15(a), page 657 50Knight, Figure 21.15(b), page 657 51Knight, Figure 21.15(c), page 657

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 ν1 is half that for the closed–closed and • open–open tubes of the same length.

For open–open tubes, •

4L nν λ = , ν = = nν1 , where n = 1, 3, 5, , (3.29) n n n 4L · · ·

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 po- • sition 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 displace- • ment.

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 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

52Knight, 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 di- rections.

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 u1 and u2 are given by

u1 (x1, t) = a sin (kx1 ωt + φ10) , (3.30) − and

u2 (x2, t) = a sin (kx2 ωt + φ20) , (3.31) − where φ10 and φ20 are the constants of the waves.

Note: φ10 and φ20 are characteristics of the sources, not the medium.

Figures 63, 64 and 6553 show snapshot graphs at t = 0 for waves emitted • by three sources with phase constants φ0 = 0 rad, φ0 = π/2 rad and

φ0 = π rad.

It can be seen that the phase constant tells us what the source is • doing at t = 0.

– For example, a at its centre position, and moving back-

ward at t = 0 has φ0 = 0 rad.

Identical sources have the same phase constants. •

Let us consider the superposition of u1 and u2 graphically (figures 66 and 6754).

53Knight, Figure 21.17, page 660 54Knight, Figure 21.18, page 661

112 Figure 63: Phase constant φ0 = 0 rad.

Figure 64: Phase constant φ0 = 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 φ0 = 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. u1(x) = u2(x). Thus, • they must have the same phase (i.e. be in phase),

(kx ωt + φ10) = (kx ωt + φ20) , (3.32) − − or more precisely,

(kx ωt + φ10) = (kx ω20 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 u1(x) = u2(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 of the waves are not equal (but their wavelengths are), then we can still get destructive interference, but it is not perfect.

116