Chapter 6
Standing Waves
Reference: George C. Kings, Vibrations and waves, A John WileyandSons, Ltd., Publication, 2009.
In this chapter we turn our attention to standing waves. These are the kind of waves that occur when
we pluck a guitar string. Indeed musical instruments provide a rich variety of standing waves. Etc.
We will find that a standing wave can be considered as the superposition of two travelling waves of
the same frequency and amplitude travelling in opposite directions.
We will see that standing waves are the normal modes of a vibrating system and that the general
motion of the system is a superposition of these normal modes.
This will give us the energy of a vibrating string. It will also introduce us to the powerful technique of
Fourier analysis.
Outline
6.1 StandingWaves on a String
6.2 StandingWaves as theSuperposition of TwoTravellingWaves
6.3 TheEnergy in a StandingWave
6.4 StandingWaves as Normal Modes of a VibratingString
6.1 STANDING WAVES ON A STRING
We shall explore the physical characteristics of
standing waves by considering transverse waves on a
taut string. The string is stretched between two fixed
points, which we take to be at 𝑥 = 0and 𝑥 = 𝐿,
respectively. The transverse displacement of the string
is in the 𝑦-direction.
An example of such a standing wave is illustrated in
Figure. Snapshots of the string at successive instants of
time are shown in(a)–(e), while (f) shows these
individual snapshots on a singleset of axes. The
displacement 𝑦is always zero at 𝑥 = 0 𝑎𝑛𝑑𝑥 =
𝐿since thestring is held fixed at those points.
However, midway between the fixed ends wecan see
that the displacement of the string is also zero at all
times. This point iscalled a node. Midway between this
node and each end point the wave reaches its
maximum displacement. These points are called
antinodes. The positions of thesemaxima and minima
do not move along the 𝑥 −axis with time and hence
the namestanding or stationary waves. When the string
vibrates, all particles of the stringvibrate at the same
frequency. Moreover they do so in SHM about their
equilibriumpositions, which is the line along which the
string lies when at rest. However, asshown in Figure.
The amplitude of vibration of the particles varies along
thelength of the string. These characteristics suggest
that the displacement 𝑦can berepresented by
𝑦(𝑥, 𝑡) = 𝑓(𝑥) 𝑐𝑜𝑠(𝜔𝑡 + 𝜑)
The function 𝑓(𝑥)describes the variation of the
amplitude of vibration along the𝑥 −axis. The function
𝑐𝑜𝑠(𝜔𝑡 + 𝜑)describes the SHM that each particle of
the stringundergoes. If we choose the maximum
displacements of the particles to occur at𝑡 = 0, then
the phase angle φ is zero and
𝑦(𝑥, 𝑡) = 𝑓(𝑥) 𝑐𝑜𝑠(𝜔𝑡 )
[Imposing the condition φ = 0 is equivalent to saying that initially, at 𝑡 = 0, thestring has zero
𝜕𝑦
velocity, i.e. 𝜕𝑡
= −𝜔𝑠𝑖𝑛𝜑 = 0, implies φ = 0.]
𝑡=0
We now substitute the solution 𝑦(𝑥, 𝑡) = 𝑓 𝑥 𝑐𝑜𝑠 𝜔𝑡 into the one-dimensional wave equation
𝜕2 𝑦
𝜕2 𝑦
2
=
𝑣
𝜕𝑡 2
𝜕𝑥 2
these expressions into the one-dimensional wave equationleads to
𝜕2 𝑓
𝜔2
= − 2 𝑓(𝑥)
𝜕𝑡 2
𝑣
We can compare this result with the equation of SHM:
𝜕2 𝑥
= −𝜔2 𝑥
𝜕𝑡 2
which has the general solution
𝑥 = 𝐴𝑐𝑜𝑠 (𝜔𝑡 + 𝜙)
Thus it follows that the general form of the 𝑓(𝑥)
is:
𝜔
𝑓(𝑥) = 𝐴𝑐𝑜𝑠 ( 𝑥 + 𝜙)
𝑣
where𝐴and 𝜙are constants that are determined by
the boundary conditions. Inthis case the boundary
conditions are 𝑓 𝑥 = 0 𝑎𝑡𝑥 = 0 𝑎𝑛𝑑𝑎𝑡𝑥 =
𝜋
𝐿.The firstcondition gives𝜙 = .. The second
2
condition gives
𝜔
𝜋
𝜔
𝜔
0 = 𝐴𝑐𝑜𝑠
𝐿+
= 𝐴𝑠𝑖𝑛
𝐿 => 𝐿
𝑣
2
𝑣
𝑣
= 𝑛𝜋
Thus, we can write
𝑛𝜋𝑣
,
𝐿
Then solution of the wave equation is given by:
𝑛𝜋
𝑦𝑛 𝑥, 𝑡 = 𝐴𝑛 sin
𝑥 cos 𝜔𝑛 𝑡
𝐿
This equation describes the standing waves on the
string, where each value of𝑛corresponds to a
different standing wave pattern. The standing
wave patternsare alternatively called the modes of
vibration of the string.
𝑛𝜋
The functions 𝑓 𝑥 = 𝐴𝑛 𝑠𝑖𝑛 𝐿 𝑥 for 𝑛 =
1 𝑡𝑜 4 are plotted in Figure, respectively. 𝑛 = 1is
the fundamentalmode or first harmonic of the
string; 𝑛 = 2 corresponds to the second
harmonic,𝑛 = 3 corresponds to the third harmonic, etc. We see that the number of antinodesin the
𝑛𝑡𝑕 harmonic is equal to 𝑛.
The time period 𝑇for a standing wave pattern to exactly toreproduce its shape is given by
2𝜋 2𝐿
𝑇=
=
𝜔𝑛 𝑛𝑣
We again define the wavelength λ of a standing wave as the repeat distance ofthe wave pattern. Since
𝑣 = 𝜈λ and 𝜔 = 2𝜋𝜈, then wavelength is given by
2𝐿
𝜆𝑛 =
𝑛
𝜔𝑛 =
If we write this equation as
𝑛𝜆𝑛
=𝐿
2
we see that we will obtain a standing wave only if an integral number ofhalf-wavelengths fits between
the two fixed ends of the string. The wave number is given by
2𝜋
𝑘𝑛 =
𝜆𝑛
Using this last relationship we can write
𝑦𝑛 𝑥, 𝑡 = 𝐴𝑛 sin 𝑘𝑛 𝑥 cos 𝜔𝑛 𝑡
which is an alternative expression for a standing wave. In order to obtain frequency of a taut string, i.e,
the angular frequency ofthe fundamental, with 𝑛 = 1, is
𝜋𝑣
𝜔1
𝑣
𝜔1 =
𝑎𝑛𝑑𝜈1 =
=
𝐿
2𝜋 2𝐿
Since the velocity of a wave on a taut string is given by
𝑣=
𝑇
𝜇
Then
𝜈1 =
1 𝑇
2𝐿 𝜇
This equation shows how the fundamental frequency of a taut string depends on itslength 𝐿, the
tension 𝑇in the string and its mass per unit length 𝜇.We can readilyrelate these results to stringed
instruments. For example, a guitar has six strings ofthe same length and these are held under
approximately the same tension. However,the strings have different values of mass per unit length and
so their fundamentalfrequencies are different: the larger the mass per unit length the lower the note.
For most vibrating systems this is not the case. These will also vibrate at a series ofhigher frequencies
in addition to the fundamental frequency. These higher frequenciesare called overtones.
We have used the example of a taut string to explore the physical characteristicsof standing waves.
However, standing waves occur in many different physicalsituations and the ideas we have been
discussing are important to a wide range ofphysical phenomena. In a microwave oven, electromagnetic
waves reflect from thewalls of the oven to form standing wave patterns in the oven compartment.
Thismeans that there will inevitably be places in the compartment where the intensity
of the microwave radiation is reduced and the food will not be properly cooked.
6.2 STANDING WAVES AS THE SUPERPOSITION OF TWOTRAVELLING WAVES
In Section 5, we saw that the general solution of the onedimensional waveequation is
𝑦 = 𝑓 (𝑥 − 𝑣𝑡) + 𝑔(𝑥 + 𝑣𝑡).
A specific example is
𝐴
𝑦 = 2 sin
2𝜋
𝜆
𝑥 − 𝑣𝑡
𝐴
+ 2 sin
2𝜋
𝜆
𝑥 + 𝑣𝑡
or, in terms of wavenumber𝑘 = 2𝜋/𝜆and angular
frequency 𝜔 = 𝑘𝑣,
𝐴
𝐴
𝑦 = sin 𝑘𝑥 − 𝜔𝑡 + sin 𝑘𝑥 + 𝜔𝑡
2
2
The first term in the right-hand side of this equation
represents a sinusoidal wave ofamplitude 𝐴/2 travelling
in the positive 𝑥 −direction and the second term
representsa sinusoidal wave of amplitude 𝐴/2 travelling
in the negative 𝑥 − 𝑑irection. Bothwaves have the same
angular frequency. Using the identity
𝑠𝑖𝑛(𝛼 + 𝛽) + 𝑠𝑖𝑛(𝛼 − 𝛽) = 2 𝑠𝑖𝑛𝛼𝑐𝑜𝑠𝛽
we obtain
𝑦 = 𝐴 sin 𝑘𝑥 cos 𝜔𝑡
Hence, we have theimportant result that a standing wave is the superposition of two travelling wavesof
the same frequency and amplitude travelling in opposite directions. This isillustrated in Figure, which
shows the two travelling waves at successive instantsof time separated by 𝑇/8 where 𝑇is the period of
the wave. The wave travelling
towards the right is represented by the thin continuous curve and the wave travellingtowards the left is
represented by the dotted curve. The arrows attached tothesecurves indicate the directions of travel.
Anypoint on the standing wave is described by Equation. 𝑦 = 𝐴𝑠𝑖𝑛𝑘𝑥𝑐𝑜𝑠𝜔𝑡.The transverse
displacement of every point on the standing wave varies with SHMas 𝑐𝑜𝑠𝜔𝑡and the amplitude of this
motion varies as 𝐴𝑠𝑖𝑛𝑘𝑥, i.e. the nodes andantinodes occur at fixed points on the x-axis. The two
travelling sinusoidal waves that we have considered above extend tolarge distances in both directions
(in principle to 𝑥 = ±∞). A string stretchedbetween two rigid walls has a finite length. However, it
can still support standingwaves. In this case it is reflections at the two walls that produce the two
waves
travelling in opposite directions.
6.3 THE ENERGY IN A STANDING WAVE
In Section 5.5 we considered the energy of a travelling wave and found that thisenergy is carried along
with the wave at the wave velocity. The situation for astanding wave is different. As we have seen, a
standing wave is a superposition oftwo waves of the same frequency and amplitude travelling in
opposite directions.
The energies of these two waves are also transported in opposite directions and sothere is no net
transport of energy. Clearly, however, there is energy in a standingwave: a vibrating string is in
motion and it stretches in moving away from itsequilibrium position. Thus the string has both kinetic
and potential energies. InSection 5.5 we obtained a general expression for the total energy E contained
in aportion a ≤ x ≤ b of a string that carries a transverse wave:
𝑏
1
𝜕𝑦 2
𝜕𝑦 2
2
𝐸= 𝜇
+𝑣
𝑑𝑥
2 𝑎
𝜕𝑡
𝜕𝑥
where𝜇is the mass per unit length of the string and v is the wave velocity. Thefirst term in the integral
relates to the kinetic energy of the string and the secondterm to its potential energy. The standing wave
solution for this case is given by
𝑦𝑛 = 𝐴𝑠𝑖𝑛𝑘𝑥𝑐𝑜𝑠𝜔𝑡
And
𝜕𝑦
𝜕𝑦
= −𝜔𝐴𝑠𝑖𝑛𝑘𝑥𝑠𝑖𝑛𝜔𝑡;
= 𝑘𝐴𝑐𝑜𝑠𝑘𝑥𝑠𝑖𝑛𝜔𝑡
𝜕𝑡
𝜕𝑥
Substituting derivatives into integral and integrating over [0,L] we obtain
1
𝐸𝑛 = 𝜇𝐿𝐴2𝑛 𝜔𝑛2
4
This equationshows that the energy of the system flows continuously between kinetic and potential
energies although the total energy remains constant. This is a characteristicfeature of oscillating
systems, as we similarly found for the simple harmonic oscillator. When the string is at its maximum
displacement, the stringis instantaneously at rest and all the energy is in the form of potential energy.
When the string passes through its equilibrium position, all the energy is in theform of kinetic energy.
6.4 STANDING WAVES AS NORMAL MODESOF A VIBRATING STRING
In Chapter 4 we discussed the normal modes of a coupled oscillator. The strikingcharacteristic of a
normal mode is that all the masses move in SHM at the samefrequency: indeed this defined the normal
modes. We also saw that these normalmodes are completely independent of each other and the general
motion of thesystem is a superposition of the normal modes. All of these properties are sharedby
standing waves on a vibrating string; all the particles of the string performSHM with the same
frequency. Indeed the standing waves are the normal modesof the vibrating string and from now on we
shall generally refer to them as normalmodes. So far we have only considered the case in which a
single normal mode ofthe string is excited. We shall begin bydescribing the superposition principle.
6.4.1 The superposition principle
The superposition principle states that, if 𝑦1 (𝑥, 𝑡)and 𝑦2 (𝑥, 𝑡)areany two solutions of the wave
equation, then so is any linear combination
𝑦(𝑥, 𝑡) = 𝐴1 𝑦1 (𝑥, 𝑡) + 𝐴2 𝑦1 (𝑥, 𝑡)
where𝐴1 𝑎𝑛𝑑𝐴2 are arbitrary constants. This result follows at once from thelinearity of the wave
equation:
𝜕 2 𝑦1
𝜕 2 𝑦1
𝜕 2 𝑦2
𝜕 2 𝑦2
2
2
=
𝑣
,
=
𝑣
,
𝜕𝑡 2
𝜕𝑥 2
𝜕𝑡 2
𝜕𝑥 2
The equations can be written as:
𝜕 2 𝑦1
𝜕 2 𝑦2
𝜕 2 𝑦1
𝜕 2 𝑦2
2
2
𝐴1
+
𝐴
=
𝐴
𝑣
+
𝐴
𝑣
,
2
1
2
𝜕𝑡 2
𝜕𝑡 2
𝜕𝑥 2
𝜕𝑥 2
𝜕2
𝜕2
2
(𝐴
𝑦
+
𝐴
𝑦
)
=
𝑣
(𝐴 𝑦 + 𝐴2 𝑦2 )
1
1
2
2
𝜕𝑡 2
𝜕𝑥 2 1 1
it follows that the linear superposition 𝑦(𝑥, 𝑡),wave equation, is also a solutionof the wave equation.
This result clearly generalises to the superposition ofany number of solutions of the wave equation.
These can be any solutions: theydo not have to be normal modes. However, for reasons that will
become clearer inthe course of the following discussions we now choose a general superposition of
normal modes.
6.4.2 The superposition of normal modes
We found the expression for the 𝑛𝑡𝑕 normal mode of a vibratingstring of length 𝐿:
𝑦𝑛 = 𝐴𝑛 𝑠𝑖𝑛𝑘𝑛 𝑥𝑐𝑜𝑠𝜔𝑛 𝑡
In general, the motion of the string will be a superposition of normal modes givenby
𝑛𝜋
𝑛𝜋𝑣
𝑦 𝑥, 𝑡 =
𝐴𝑛 𝑠𝑖𝑛𝑘𝑛 𝑥𝑐𝑜𝑠𝜔𝑛 𝑡 ; 𝑛𝑜𝑡𝑒: 𝑘𝑛 =
; 𝜔𝑛 =
𝐿
𝐿
𝑛
whereωn= nπv/L. An example of this is presented in Figure
which showsthe superposition of the third normal mode with a
relative amplitude of 1.0 andthe thirteenth normal mode with a
relative amplitude of 0.5 (at t=0). (We choose such ahigh normal
mode to demonstrate the superposition of the waves more
clearly.)The third thirteen normal modes are
3𝜋
𝑦3 = 0.1𝑠𝑖𝑛
𝑥𝑐𝑜𝑠𝜔3 𝑡
𝐿
13𝜋
𝑦13 = 0.05𝑠𝑖𝑛
𝑥𝑐𝑜𝑠𝜔13 𝑡
𝐿
Snapshots of these two normal modes at 𝑡 = 0shown in Figure
(a) and (b), respectively. The superposition of the two normal
modes is given by
3𝜋
13𝜋
𝑦 𝑥, 𝑡 = 0.1𝑠𝑖𝑛 𝑥𝑐𝑜𝑠𝜔3 𝑡 + 𝑦13 = 0.05𝑠𝑖𝑛
𝑥𝑐𝑜𝑠𝜔13 𝑡
𝐿
𝐿
and describes the motion of the vibrating string. This is
illustrated in Figure (c)which again is a snapshot of the string at
𝑡 = 0.
Of course, it is impractical to do this and in practice we pluck a
string to causeit to vibrate.
The action of plucking a string is illustrated in Figure. In thisexample the string is displaced a distance
𝑑 at one quarter of its length. Initially,the string has a triangular shape and thisshape clearly does not
match any of theshapes of the normal modes shown in Figure. For one thing the triangle hasa sharp
corner while the sinusoidal shapes of the normal modes varysmoothly.
The remarkable thing is, however, that it is possible to reproduce this triangularshape by adding
together the normal modes of the string with appropriate amplitudes.
This is illustrated by Figure. In Figure the first three
normal modes 𝑦1 (𝑥, 0), 𝑦2 (𝑥, 0) 𝑎𝑛𝑑𝑦3 (𝑥, 0)are shown.
𝐴 𝐴
Their amplitudes are 𝐴, 2 2 , 9 respectively, where
𝐴 = 32𝑑/3𝜋 2 . (The general procedure for finding the
values of these amplitudesis developed in next Section.)
Figure (c) shows the superposition of these threenormal
modes, i.e.
𝑦(𝑥, 0) = 𝑦1 (𝑥, 0) + 𝑦2 (𝑥, 0) + 𝑦3 (𝑥, 0)
and enables a comparison with the initial shape of the
string. Even using just the firstthree normal modes we get a
surprisingly good fit to the triangular shape. By adding
more normal modes, we would achieve even better
agreement, especially withrespect to the sharp corner. The
corresponding frequencies of the normal modesare given
by the usual expression 𝜔𝑛 = (𝑛𝜋𝑣/𝐿). Thus when
wepluck a string we excite many of its normal modes and
the subsequent motion of thestring is given by the
superposition of these normal modes.
6.4.3 The amplitudes of normal modes and Fourier analysis
In previous section we saw that the general motion of a vibrating string is a superpositionof normal
modes, 𝑦𝑛 = 𝐴𝑠𝑖𝑛𝑘𝑥𝑐𝑜𝑠𝜔𝑡. In particular, the initial shape of thestring 𝑓(𝑥), i.e. at 𝑡 = 0, is from
given by
𝑛𝜋
𝑦𝑛 𝑥, 0 =
𝐴𝑛 𝑠𝑖𝑛
𝑥 = 𝑓(𝑥)
𝐿
𝑛
We now state a remarkable result: any shape 𝑓(𝑥)of the string with fixed endpoints [𝑓(0) = 𝑓(𝐿) =
0] can be written as a superposition of these sine functionswith appropriate values for the
coefficients. This result is due to Fourier. The idea that an essentiallyarbitrary function 𝑓(𝑥)can be
expanded in a Fourier series can be generalized andis of great importance in much of theoretical
physics and technology.
The Fourier expansion theorem, involves some difficult mathematicsand we will simply assume its
validity. In contrast, its application in practiceis quite straightforward. Given 𝑓(𝑥), i.e. the shape of the
string, the amplitudes 𝐴𝑛 (𝑛 = 1, 2, . . . ) are easily found. It is this that makes Fourier analysis such a
powerfultool. The determination of the amplitudes depends on two integrals involvingsine functions:
𝐿
𝑛𝜋
𝐿
sin2
𝑥 𝑑𝑥 =
𝐿
2
0
𝐿
𝑛𝜋
𝑚𝜋
𝑠𝑖𝑛
𝑥 𝑠𝑖𝑛
𝑥 𝑑𝑥 = 0; 𝑚 ≠ 𝑛.
𝐿
𝐿
0
where𝑚𝑎𝑛𝑑𝑛are integers throughout. For the secondintegral , we use the trignometric identity
1
𝑠𝑖𝑛𝛼𝑠𝑖𝑛𝛽 = cos 𝛼 − 𝛽 − cos 𝛼 + 𝛽
2
Multiplying 𝑓(𝑥) with 𝑠𝑖𝑛(𝑚𝜋𝑥/𝐿)and integrating the resultingequation with respect to 𝑥over the
range 𝑥 = 0 𝑡𝑜𝑥 = 𝐿we obtain the finalexpression for the Fourier amplitude
2 𝐿
𝑛𝜋
𝐴𝑛 =
f x sin
𝑥 𝑑𝑥
𝐿 0
𝐿
Our final result: a statement of the Fourier theorem.For any specific function 𝑓 (𝑥), i.e. the shape of
the string at 𝑡 = 0, gives us the Fourier amplitudes 𝐴1 , 𝐴2 , . ...
6.4.4 The energy of vibration of a string
We considered a string vibrating in a single normal mode, givenby
𝑦𝑛 = 𝐴𝑛 𝑠𝑖𝑛𝑘𝑛 𝑥𝑐𝑜𝑠𝜔𝑛 𝑡
and we derived the energy 𝐸𝑛 of the string vibrating in this mode:
1
𝐸𝑛 = 𝜇𝐿𝐴2𝑛 𝜔𝑛2
4
We now want to obtain the energy E of the vibrating string when there are severalmodes present. The
general superposition of normal modes is given by
𝑛𝜋
𝑦𝑛 𝑥, 𝑡 =
𝐴𝑛 𝑠𝑖𝑛
𝑥 𝑐𝑜𝑠𝜔𝑛 𝑡
𝐿
𝑛
The most interesting feature of this result is that each normal mode contributes an energy
1
𝐸𝑛 = 𝜇𝐿
𝐴2𝑛 𝜔𝑛2
4
𝑛
quite independently of the other normal modes. This is quite typical of normalmodes as we discussed
in Chapter 4. They are independent of each other and thereis no coupling between them. Consequently
their energies are additive.