Elastic Waves HW 3 Solution

HW3.1 Consider a string on an elastic foundation with stiffness per length κ and subject to aforce f(t) concentrated at x = 0

[ρ ∂2

∂t 2−T ∂2

∂x2+κ ]ψ(x;t) = f (t) δ(x)

3.1a For f = 0, what is the dispersion relation, ω as a function of (real) q, for a plane waveexp(iωt-iqx) ? What is q as a function of (real) ω? Observe that there is a frequency ωcutoff

below which no plane wave propagates; it only evanesces. Sketch a plot of ω as a function of q.What is the group velocity vg(ω) = dω/dq as a function of ω?

Solution: Substituting ψ = exp(iωt-iqx) gives − ρω2 +Tq2 +κ = 0I.e −ρω2 +Tq2 +κ = 0 which implies ω(q) = Tq2 / ρ +κ / ρ

Alternative, we can solve for q(ω): q(ω) = ρω2 /T −κ /Twhich is imaginary for ω less than the cutoff ωco = √κ/ρ.

Vg is obtained by (with c defined as √T/ρ )d(−ρω2 +Tq2 +κ = 0) => −2ρωdω + 2Tqdq = 0 =>Vg = dω / dq = Tq / ρω =

= T ρω2 /T −κ /T / ρω = c 1−ωco2 /ω2

Hi frequencies go at Vg = c. Intermediatefrequencies go slower. Frequencies below cutoff donot propagate at all.

3.1b Do a double FT on the above PDE for thecase f(t) = δ(t) and solve for %G(q;ω) . Then dothe inverse q-transform by residues ( you mayhave to introduce the -iε to ω) and determine thefrequency domain Greens function for this

system %G(x;ω) Notice that the locations of the

poles are qualitatively different depending on whether ω is above cutoff or not. You may replace xwith |x| if you like. You may assume ω≥ 0 for simplicity, and use

%G(x;−ω)= %G(x;ω)* if you needto determine what goes on at negative ω.

Solution: The double transform turns the PDE into

[−ρω2 +Tq2 +κ ] %G(q;ω) =1 which implies

%G(q;ω) = 1ρ[−ω2 + c2q2 +ωco

2 ]The inverse q transform is

%G(x;ω) = 12πρ

dq∫ exp(−iq | x |)[−(ω − iε)2 + c2q2 +ωco

2 ]

We will close this in the lower half complex q plane, which means we are interested only in polesthat may lie there. If ω is greater than cutoff, there is a pole close to the real axis in the lower

plane, at q = q† = (1 / c) ω2 −ωco2 . If ω is less than cutoff, there is a pole on the negative

imaginary axis in the lower plane, at q = −iη = −i(1 / c) ωco2 −ω2 . Therefore, for ω larger than

cutoff

%G(x;ω) = 12πρ

(−2iπ ) exp(−iq† | x |)

2c2q†= −i exp(−iq

† | x |)2c2ρq†

And for ω below cutoff,

%G(x;ω) = 12πρ

(−2iπ ) exp(−η | x |)2c2 (−iη)

=exp(−η | x |)2ρc2η

3.1c For the case f(t) = exp(iωt), at what time average rate is work being done by the force? Showthat it is zero if |ω| < ωcutoff. (This makes sense, there is no propagation at such frequencies to carrythe energy away.)

Solution: I recall that <Π >t= −

ω2Im %G(x = 0;ω) . So, if ω is above cutoff,

<Π >t=ω

4ρc2q†(ω)=

ω

4ρc ω2 −ωco2

And if ω is below cutoff, then <Π >t= 0… because Im %G(x = 0;ω) is purely real below cutoff.

3.1d Construct an expression for the response in the time domain G(x,t) (i.e., for the delta functionforcing δ(t)) as an inverse FT on

%G(x;ω) . Show that your integral expression for G(x,t) is real (inspite of initial appearances.) To do this, it is best to convert your integral from ω=-∞ to ∞ to anintegral from ω=0 to ∞ by recognizing how the integrand changes when ω changes sign.

Answer: G(x;t) = 1

2πdω exp(iωt) %G(x;ω)

−∞

∞

∫ . The above expressions for %G(x;ω) are

complicated. In particular, it may not be obvious how they ought be interpreted for negativeω. It is perhaps apparent that η must be real and positive, regardless of the sign of ω. It comesfrom a pole that is always in the lower half q plane). It is perhaps less obvious thatq† = (1 / c) ω2 −ωco

2 must change sign when ω changes sign. The cleanest way to sort this out isto recognize that

%G(x;−ω) must be equal to the complex conjugate of %G(x;ω) ( any FT of a real

function must have this property, as can be seen by examination of the Fourier transform ) .Therefore

2πG(x;t) = dω exp(iωt) %G(x;ω)0

∞

∫ + dω exp(iωt) %G(x;ω)−∞

0

∫

= dω exp(iωt) %G(x;ω)0

∞

∫ + − dξ exp(−iξt) %G(x;−ξ)∞

0

∫

= dω exp(iωt) %G(x;ω)0

∞

∫ + dξ exp(−iξt) %G(x;ξ)0

∞

∫ *

= 2Re dω exp(iωt) %G(x;ω)0

∞

∫This last integral does not require us to evaluate %G(x;ω) at negative ω. We can use theexpressions for it from above.

There is still the potential difficulty of the different form that %G(x;ω) takes depending on whetheror not ω is above cutoff. For this reason we write

G(x;t) = (2 / π )Re[ +ωco

∞

∫ dω exp(iωt) %G(x;ω)0

ωco

∫ ]

= (2 / 2ρc2π )Re[ exp(−η | x |)exp(iωt)η

dω0

ωco

∫ + −i exp(−iq† | x |)exp(iωt)q†ωco

∞

∫ dω ]

=1ρcπ

[ exp(− ωco2 −ω2 | x | /c)cos(ωt)ωco2 −ω2

dω0

ωco

∫ +sin(ωt − ω2 −ωco

2 | x | /c)ω2 −ωco

2ωco

∞

∫ dω

3.1e Evaluate this expression and plot it as a function of t for a fixed value of x = 20 (take T =κ = ρ =1) by doing the ω integration. You may have to do it numerically (thus it is a matter ofdoing a numerical integration over ω, and doing so again and again for every time t of interest.You may find that it is convenient to write out separate expressions for the integrations from 0 to 1and from 1 to ∞. You may suppress the high frequencies and the problematic convergence of theintegral by inserting an ad-hoc factor exp(-ω2/Ω2) that may be interpreted as a lo-pass filter; justmake sure Ω >> 1 so you don't lose the interesting physics near and below ω=1. You should beable to observe in your plot of ψ(t) an apparent relation between the time-of-arrival and theoscillation rate: low frequencies arrive late, in accord with the group velocity derived in 3.1a Youshould also be able to observe causality: there should be no signal arriving before the fastest groupvelocity could carry it tearliest arrival= 20/vgmax.

Here is a procedure ( in no particular language and with no particular attention to efficiency orprecision ) for doing the integral

wmax = 50; x=20; dw = 0.002 ( this small dw assures that sin and cosine of wt doesn't vary too fast. )For t = 0 to 150 by 0.1

sum = 0for w = .005 to .995 by dwsum = sum + dw exp( - sqrt(1-w^2) x ) cos(wt) exp(-w^2/500) / sqrt( 1-w^2)end loop over w

for w = 1.005 to wmax by dwsum = sum + dw sin(wt-sqrt(w^2-1) x) exp(-w^2/500) / sqrt(w^2-1)end loop over wintegral = sum / piprint t, integral

end loop over t

The above procedure skips around the problematic singular (but integrably singular) point atomega = 1 ( a better procedure would refine the spacing near that point, or change variables so thatthere is no infinity there.)

The result is plotted here

We notice that there is no response at times before t = 20. This makes sense, because the fastestgroup velocity is 1. We also notice that high frequencies arrive soonest, and lower frequencies takelonger, in accord with expectations based on the group velocity as a function of frequency. Theslowest frequencies ( at the latest times in the figure) have a period of about 6 ( as best we canmeasure off the plot. )… consistent with the cutoff frequency ω = 1, i.e f = 1/2pi, i.e period 2 pi.

We also observe no contributions from frequencies below cutoff. (There actually must be some,but they are so evanescent that they do not show up well at this large distance.)