MathematicalPhysics

MP205 Vibrations and Waves

Lecturer:              Dr.  Jiří  Vala  Office:                Room  1.9,  Mathema<cal  Physics,    

               Science  Building,  North  Campus  Phone:              (1)  708  3553  E-­‐Mail:              jiri.vala@nuim.ie  

Lecture  15  -­‐  16  

FOURIER ANALYSIS

A string of length L fixed at its two ends should be able (subject to certain assump-tions) to vibrate in any of an infinite number of normal modes. We shall put

yn(x, t) = An sin�nπx

L

�cos (ωnt − δn)

We can imagine that all these modes are permitted to be present, so that the motionof the string is specified by the equation

y(x, t) =∞�

n=1An sin

�nπxL

�cos (ωnt − δn)

For a fixed specified time t0, the quantities cos�ωnt0 − δn

�can be treated as a set of

fixed numbers, and the displacement of the string at any x is given as

y(x) =∞�

n=1Bn sin

�nπxL

�(1)

where

Bn = An cos�ωnt0 − δn

�

We make the following assertions (Fourier 1807):

It is possible to take any form of the profile of the string described by y as a functionof x between x = 0 and x = L (subject tot he conditions y = 0 at x = 0 and x = L andanalyze it into an infinite series of sine functions as given by Eq. 1.

We will consider the continuous string as the limit for N → ∞ of a row of N connectedparticles (We are translating our problem from the world of physics to the world ofmathematics).

Before proceeding, we point another result in our solution for vibrating system:

Consider the general transverse motion of the continuous string:1) the frequencies ωn = nω1, where n is a positive integer;2) on a particular fixed value of x, we can write An sin(nπx/L) = Cn which is a con-stant and thus have

y(t) =∞�

n=1Cn cos (ωnt − δn)

This states:

that any possible motion of any point on the string is periodic in the time 2π/ω1

and that

this periodic motion can be written as a combination, with suitable amplitudes andphases of pure sinusoidal vibrations comprising all possible harmonics of ω1,that is as

a Fourier analysis in time(rather than space).

Fourier analysis in action

We must be able to determine the coefficients of the component sine and cosinefunctions, that is, we must perform harmonic analysis.

Consider the expansion for y(x)

y(x) =∞�

n=1Bn sin

�nπxL

�

Suppose we want the amplitude associated with a particular value of n, let’s say n1:

to find it we multiply both sides of the equation by sin (n1πx/L) and integrate w.r.t.x over the range from 0 to L:

� L

0y(x) sin

�n1πxL

�dx =

∞�

n=1Bn

� L

0sin�nπx

L

�sin�n1πx

L

�dx

for any two angles θ and φ, we have

cos(θ − φ) = cos θ cos φ + sin θ sin φcos(θ + φ) = cos θ cos φ − sin θ sin φ

and therefore

sin θ sin φ =12�cos(θ − φ) − cos(θ + φ)

�

Hence we can put

sin�nπx

L

�sin�n1πx

L

�=

12

�cos�(n − n1)πx

L

�− cos

�(n + n1)πx

L

��

Therefore�

sin�nπx

L

�sin�n1πx

L

�dx =

L2π(n − n1)

sin�(n − n1)πx

L

�− L

2π(n + n1)sin�(n + n1)πx

L

�

If we insert the limits x = 0 and x = L, the values of sin(n ± n1)πx/L are all zero, butthe quantity (n − n1) appears in the denominator of one of the integrals, i.e. leadingto 0

0 for n = n1.

For n = n1, the integral that is nonzero is� L

0sin2�n1πx

L

�dx =

12

� L

0

�1 − cos

�2n1πx

L

��dx

The cosine term contributes nothing but the other part gives L/2, thus we arrive at� L

0y(x) sin

�nπxL

�dx =

L2

Bn

that is

Bn =2L

� L

0y(x) sin

�nπxL

�dx

This equation determines the amplitude Bn associated with any given value of n inthe harmonic analysis of y(x).

If y(x) is a purely empirical curve, the evaluation of the Fourier coefficients Bn is donenumerically but if y(x) is given as an analytical function, Bn’s can be obtained in aform of a general formula.Example:

Bn =2L

� L

0kx sin

�nπxL

�dx

=2kL

�− L

nπ

�x cos

�nπxL

��L0+

Lnπ

� L

0cos�nπx

L

�dx�

=2knπ

�−�x cos

�nπxL

��L0+

Lnπ

�sin�nπx

L

��L0

�

= −2kLπ

cos nπn

Bn falls into two categories, according to whether n is odd or even (cos(nπ) alternatesbetween +1 and −1):n odd:

Bn =2kLnπ

n even:

Bn = −2kLnπ

and by combining both expressions, we get

Bn = (−1)n+1 2kLnπ

It is easy to tabulate the various amplitudes.

n Bn/kL

1 2π = 0.636

2 −1π = −0.318

3 23π = 0.212

4 − 12π = −0.159

5 25π = 0.127

Our description of the triangular profile becomes

y(x) =2kLπ

�sin�πx

L

�− 1

2sin�2πxL

�+

13

sin�3πxL

�...

�

The sine curves in terms of which the Fourier analysis is made represent an exampleof orthogonal functions. This orthogonality is described by

� L

0sin�n1πx

L

�sin�n2πx

L

�dx = 0

for n1 � n2.

What is the connection with orthogonality we know from the context of geometry?

Consider vectors �A and �B. The condition that they are orthogonal to each other,�A ⊥ �B, is that their scalar product is zero:

AxBx + AyBy + AzBz =3�

p=1ApBp = 0

Replacing, the continuous integral above by sum

LN

N�

p=1sin�n1πp

N

�sin�n2πp

N

�= 0

for n1 � n2, reveals that in purely formal sense, the difference between both expres-sions is merely that one of them involves quantities described by three componentsand the other has n components (and in the limit, infinitely many components).

Analyzing an arbitrary function in terms of a set of orthogonal functions is one ofthe most important and widely used techniques in physics.

Normal modes and orthogonal functions

A few remarks on actual physical systems:(i) normal modes, described by orthogonal functions, exist independently of all theothers;(ii) they are orthogonal (independent) also dynamically: the total energy of a stringvibrating in a superposition of its normal modes is just the sum of the energies forthe modes individually;

the sum of kinetic and potential energy for a small segment of the string at somex consists of two terms, involving:(1) square of sines and cosines of the same argument (mode);(2) cross terms from different modes which all yield zero due to orthogonality.

orthogonality = independence = normal (mode)

Fourier series are very useful tools indeed, but they have their limits when applied to functions which are discontinuous. Consider a square wave of period 1 which, in the interval [0,1], is 1 for 1/4 < x < 3/4 and 0 elsewhere. Below is an animation which shows the sequential addition of the first 100 nonzero terms in the Fourier expansion:

The sum gets closer and closer to the actual square wave everywhere except at the discontinuities x=1/4 and x=3/4, where there is always an over/undershoot. This is unavoidable, and persists no matter how many terms you include in the Fourier expansion. This is called the Gibbs phenomenon (after the American physicist Josiah Willard Gibbs).

VII. PROGRESSIVE WAVES

A condition of some kind is transmitted from one place to another by means of amedium, but the medium itself is not transported.

All material media – solids, liquids, and gases – can carry energy and information bymeans of waves.

Our focus will be on a simple system: the stretched string.

Normal modes and traveling waves

How do we excite a particular normal mode?(i) set up the shape of the string for a particular mode at the maximum amplitude andrelease it.

(ii) drive the string from one end by harmonic motion.

Formal analysis

We start with the normal modes of a stretched string fixed at both ends. For thestring of the length L, the tension T and the linear density µ we get for a given modenumber n = 1, 2, 3 , etc. the displacement

yn(x, t) = An sin�nπx

L

�cosωnt

ωn =nπL

�Tµ

�1/2

Using some elementary mathematics:

sin (θ + φ) + sin (θ − φ) = 2 sin θ cos φ

⇒ sin θ cos φ =12�sin (θ + φ) + sin (θ − φ)�

Does the new expression for yn(x, t) describe two traveling waves going in oppositedirections?

We focus on the first of the two terms:

y(x, t) = A sin�2πλ

(x − vt)�

λ =2Ln

At any instant of time this is a sine wave with a wavelength λ.

Where will a value of y, corresponding to certain values of x and t, be found at aslightly later instant t + ∆t?

If the appropriate location is x + ∆x, we have

y(x, t) = y(x + ∆x, t + ∆t)

sin�2πλ

(x − vt)�= sin

�2πλ

[(x + ∆x) − v (t + ∆t)]�

⇒ ∆x − v∆t = 0 i.e.∆x∆t= v

thus the wave moves in positive direction.

In the expression for the displacement of the n-th normal mode

yn(x, t) =12

An sin�2πλ

(x − vt)�+

12

An sin�2πλ

(x + vt)�

the first term corresponds to a wave moving in positive direction with the velocityv and the second to a wave moving in negative direction with the velocity −v.

Suppose we take

y(x, t) = A sin�2πλ

(x + vt)�

⇒ ∂y∂x=

2πλ

A cos�2πλ

(x + vt)�

∂y∂t=

2πvλ

A cos�2πλ

(x + vt)�

and then we get ∂y∂x = +1v∂y∂t .

However by forming the second derivatives, we arrive at EOM for (transverse) sinewaves traveling in either direction (the equation we studied in the context of ”Normalmodes of continuous systems”):

∂2y∂x2 =

1v2∂2y∂t2

(1)

Comments: Regarding v =�γp/ρ?

The general gas equation for a mass of an effectively ideal gas of molecular mass Mis

pV =mM

RT

where R is the gas constant and T is the absolute temperature. Since m/V = ρ then

v =

�γRTM

Thus the velocity of sound would be(a) independent of pressure or density,(b) proportional to

√T , and

(c) inversely proportional to√

M.Results (a) and (b) are correct for any given gas over a wide range of p and T , and (c)is borne out if we compare various gases of the same molecular type, e.g. diatomic.

Since both waves have the same velocity v, the combined disturbance moves like astructure of unchanging shape. The shape of the combination is easily consideredat t = 0

y = A�sin

2πxλ1+ sin

2πxλ2

�

Indeed, this is a beat phenomenon, though the modulation of the amplitude is herea function of x (instead of t).

Now it is extremely convenient to introduce the reciprocal of the wavelength, calledthe wave number, which corresponds to the number of complete wavelengths perunit distance

k =2πλ

In terms of wave numbers, the equation for the superposed wave form is

y = A�sin (k1x) + sin (k2x)

�

= 2A cos�k1 − k2

2x�

sin�k1 + k2

2x�

The distance from peak to peak of the modulating factor is defined by the change ofx corresponding to an increase of π in the quantity (k1 − k2) x/2:

D =2π

k1 − k2=λ1λ2λ2 − λ1

If λ1 and λ2 are almost equal, i.e. λ, λ + ∆λ, then

D ≈ λ2

∆λThat is, a number of wavelengths given approximately by λ/∆λ is contained betweensuccessive zeroes of the modulation amplitude.