• The frequencies of the normal modes form an arithmetic series

ν1, 2ν1, 3ν1, · · · .

• The fundamental frequency ν1 is thus the difference between the fre-

quencies of two adjacent modes, i.e.

ν1 = ∆ν = νn+1 − νn . (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

λn =2L

n−→ λ3 =

2L

3=

2.60

3= 0.867 m .

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

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

Of course, the speed of the waves is also given by v =√

Ts/µ, and so the

tension, Ts, is

Ts = µv2 =m

Lv2 =

(

0.005

1.30

)

(151.7)2 = 88.5 N .

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 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.

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

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. For

example, in musical instruments,

– 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.

48Knight, 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,

λn =2L

n, νn =

nν

2L= nν1 , where n = 1.2.3. · · · . (3.28)

49Knight, Figure 21.15(a), page 65750Knight, Figure 21.15(b), page 65751Knight, 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 fundamental frequency ν1 is half that for the closed–closed and

open–open tubes of the same length.

• For open–open tubes,

λn =4L

n, νn =

nν

4L= nν1 , 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 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 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

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 phase 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 loudspeaker 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 66054Knight, 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 amplitudes of the waves are not equal (but their wavelengths

are), then we can still get destructive interference, but it is not perfect.

116