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  17  -­‐  18  

The amount of this displacement is the negative of the distance A�B and, the associ-ated transverse velocity is

−A�B∆t= −v

A�BAA�= −v

∆y∆x

In the language of partial derivatives

vy = −v∂y∂x

so the transverse velocity is directly proportional to the slope of the pulse profile atthat point

vy =∂y∂t= −v

∂y∂x= −∂y∂x∂x∂t

Differentiating now w. r. t. t yields

∂y∂t= f �

∂(x − vt)∂t

= −v f �

∂2y∂t2

= (−v)2 f �� = v2 f ��

Comparing both second derivatives gives

∂2y∂t2= v2∂

2y∂x2

The same holds for the case y(x, t) = g(x + vt).

Dispersion. Phase and group velocities

The equation of a progressive sine wave

y(x, t) = A sin�2πλ

(x − vt)�= A sin (kx − ωt)

describes waves of all wavelengths propagating at the same speed

v =

�Tµ

for a given tension T and the linear density µ.

This will certainly fail in physical situations!

Pure sinusoidal waves of high frequency and short wavelength tend to travel withsmaller speed than the longer waves. This variation of wave speed with wavelengthis knows as

DISPERSION

(Recall that for the normal modes for large N the simple mode frequency was shownto be given by νn = 2ν0 sin {nπ/ [2(N + 1)]}.)

Examples can be found in many different kinds of media, e.g. dispersion of whitelight through a prism where vblue > vred implies different refraction angles accordingto the Snell’s law:

sin isin r

= n =cv

Light consequently separates out into its different colors.

Simple case:Two sine waves of slightly different wavelengths traveling in the same direction (atdifferent speed) along a string

y1 = A sin (k1x − ω1t)y2 = A sin (k2x − ω2t)

where k1 = 2π/λ1 and k2 = 2π/λ2 are the wavenumbers and ω1 and ω2 are theangular frequencies.

These two waves have different characteristic sppeds

v1 =ω1k1= ν1λ1 v2 =

ω2k2= ν2λ2

The superposition gives the combined disturbance

y = A�sin (k1x − ω1t) + sin (k2x − ω2t)

�

= 2A cos�k1 − k2

2x − ω1 − ω2

2t�

sin�k1 + k2

2x − ω1 + ω2

2t�

At t = 0 this gives the same expression as we obtained earlier for superposed waves.We will however now consider what happens with the passage of time.

The expression above can be interpreted as a rapidly alternating wave of short wave-length, modulated in amplitude by an envelope of long wavelength. Both these dis-turbances move but they may have different speeds. A place of maximum amplitudemoves at the speed of the envelope.

For the waves of the same amplitude we can simplify our description of the combineddisturbance by putting

k1 − k22

= ∆kω1 − ω2

2= ∆ω

k1 + k22

= kω1 + ω2

2= ω

⇒ y = 2A cos (∆k x − ∆ω t) sin (kx − ωt)

In this expression we can identify two velocities:

(i) the phase velocity vpthat is the speed with which the crest belonging to the average wave number k movesalong

vp =ω

k= νλ (2)

(ii) the group velocity vgthat is the velocity with which the modulating envelope moves (this envelope enclosesa group of the short waves)

vg =∆ω

∆k→ dω

dk(3)

Physicall important facts:

- every wave train has a finite extent - a wave group;

- transport of energy takes place at vg.

To treat these questions effectively we will need to use not just two sine functions buta whole spectrum.

The below animations illustrate the difference between phase and group velocities; the top two waves travel with speeds indicated by the red and green dots, and the phase and group velocities of their sum (the bottom wave) can be seen from the movement of the blue and black dots respectively.

In the first example, the red wave has frequency ω1 = 3π/5 rad.s-1 and the green has ω2=0.9ω1. The speed of the green wave is 9/8 times the speed of the red; the phase velocity of the orange wave is the average of these (17/16 times the red wave's speed), and the group velocity - the speed at which the envelope travels - is half of the red wave's speed.

In this second example, the red wave is the same, but now the green has frequency ω2=0.8ω1and travels at 8/9 times the speed of the red. The phase velocity of the orange wave is now 17/18 times the red wave's speed its group velocity is twice the red wave's speed:

The existence of the dispersion has important implications in the context of analyzingan arbitrary pulse into pure sinusoids:• if these sinusoids have different characteristic speed, the shape of the disturbance

must change with time;• in particular, a highly localized pulse at some instant will be spreading as it moves.

Example: Deep water waves (so called ”gravity” waves):These waves are highly dispersive: vp = C�λ1/2 = Ck−1/2, where C is a constant.Since vp = ω/k we have ω = Ck1/2. Therefore

vg =dωdk=

12

Ck−1/2 =12

vp

so the component wave crest will run rapidly through the group, first growing in am-plitude and then apparently disappearing again.

Sound waves in gas are non-dispersive (fortunately!).

The phenomenon of cut-off

is closely related to dispersion. It ddescribes the inability of a dispersive medium totransmit waves above (or possibly below) a certain critical frequency.

The effect is implicite in the analysis of the normal modes of a line of N masses:

ωn = 2ω0 sin�nπl2L

�

where L = (N + 1)l.

We can imagine increasing L indefinitely while keeping l unchanged. The wavenumber kn = 2π/λn = πn/L thus becomes in effect a continuous variable and thefrequency can be written as

ν(k) = 2ν0 sin�k l2

�

This expression does not permit any value of ν(k) > 2ν0 = ω0/π, thus implying amaximum normal mode frequency.

This maximal frequency corresponds to a wave number km such that km l/2 = π/2 orto a wavelength

λm = 2l

We can still shake one end of such a line of masses with frequency greater than νm.What happens?

To find out we go back to the equation relating amplitudes of successive masses inthe coupled system vibrating at some ν (or ω):

Ap−1 + Ap+1Ap

=−ν2 + 2ν20ν20

⇒ Ap =ν20

2ν20 − ν2(Ap−1 + Ap+1)

What happens for different values of ν?

(a) ν = 0 implies Ap =12�Ap−1 + Ap+1

�, i.e. the amplitude varies linearly with dis-

tance along the line; it is a simple static equilibrium with one end of the line pulledtransversely aside from the normal position. The effective wavelength is infinite.

(b) ν << ν0 implies Ap >12�Ap−1 + Ap+1

�, i.e. any one amplitude is greater than

the average of the adjacent ones (but not too much); thus there is a slight curvature,toward the axis, of a smooth curve joining the particles (sinusoidal form).

(c) ν =√

2ν0 givesAp−1+Ap+1

Ap= 0 (very special case), this implies that Ap+1 =

−Ap−1 but places no requirement on the ratio Ap−1/Ap. The wavelength associatedwith this frequency is 4l where l is the interparticle distance. Indeed for k = π/2l

ν = 2ν0 sinπ

4=√

2ν0.

(d) ν = 2ν0 (maximum frequency νm) implies Ap = −12�Ap−1 + Ap+1

�, i.e. this re-

quires an alternation of positive and negative displacements of the same size. Thewavelength is 2l.

(e) ν > 2ν0 (but not much) requires that Ap has sign opposite toAp−1+Ap+1

2 and also���Ap��� < 1

2���Ap−1 + Ap+1

���; this implies a slight curvature, away from the axis, of thesmooth curves joining alternate particles.

The amplitude alternate in sign and fall off in magnitude in geometric proportion, i.e.exponentially; this is therefore the cut-off phenomenon.

Let us put ν = 2ν0+∆ν and let the ratios Ap−1/Ap, Ap/Ap+1, etc. to be set to − (1 + f )where f is a small fraction. Then

Ap−1 + Ap+1Ap

=−ν2 + 2ν20ν20

⇒Ap−1Ap+

Ap+1Ap

=−ν2ν20+ 2

− (1 + f ) − (1 + f )−1 =−(2ν0 + ∆ν)2

ν20+ 2

−1 − f − (1 − f + f 2...) =−[4ν20 + 4ν0∆ν + (∆ν)2]

ν20+ 2

That is

−2 − f 2 + ... = −2 − 4∆νν0−�∆ν

ν0

�2

Hence approximately

f = 2�∆ν

ν0

�1/2

The further we go above the critical νm, the more drastic is the attenuation as we

proceed along the line.

(f) ν > 2ν0 implies the cut-off and Ap−1/Ap ≈ −ν2/ν20 (nearly correct equality).

For example, for ν = 2νm = 4ν0, it will be almost true that only the first particle in the

line will show any appreciable response to the driving agency. The rest of the line

behaves almost as a rigid structure.

Summary:

From the equation for y(x, t) we have

∂y∂x= −2πνA

vcos�2πν�t − x

v

��

Therfore

Fy =2πνAT

vcos (2πνt)

We can now calculate the work done in any given time t

W =�

Fydy0 =2πνAT

v

�cos (2πνt) d [A sin (2πνt)]

=(2πνA)2T

v

�cos2 (2πνt) dt

=u2

0Tv

�cos2 (2πνt) dt

where we used that 2πνA = u0.

Let us evaluate this work for one complete period of the wave

Wcycle =u2

0T2v

� T=1/ν

0[1 + cos (4πνt)] dt

=u2

0T2vν

where the cosine term contributes nothing in this complete cycle. Since T = µv2 andν = v/λ, we can write

Wcycle =12

(λµ) u20 = 2π2ν2A2λµ

which is twice the value of the kinetic energy and potential energy per wavelength.

Waves in two (2D) and three dimensions (3D)

Consider a 2D membrane subject to a uniform tension S (per unit length). We in-

troduce rectangular coordinates x, y in the plane of the membrane and describe the

transverse displacements in terms of the coordinate z. EOM for 2D waves is:

∂2z∂x2 +

∂2z∂y2 =

1v2∂2z∂t2

where v2 = S/σ and σ is the surface density.

In rectangular case, we can apply EOM at once and get solutions in the form os

straight waves

z(x, y, t) = f (αx + βy − vt)