Physics 229& 100 Homework #9 Name: Nasser Abbasi

à 1. Use dimensional reasoning and the tools described in Dimensional Analysis to determine the order of magnitude of several physical quantities. a) determine the pressure at the center of the earth.

Remove@"Global` ∗" D

dimanal A9ρ ikjj Kilogram �������������������������

Meter 3 y {zz, d HMeter L, ReduceUnits @ToSymbolsUnits @GDD=,

8Kilogram, Meter, Second

dimanal @params, 8Kilogram, Meter, Second

dimanal @params, 8Kilogram, Meter, Second

à d) Determine the relation between the speed of waves in deep water ( like the ocean) and the wavelength of the waves. (Hint: what is the restoring force for big waves in the ocean?)

Speed of a wave C = Wave Length (l) * frequency (f)

i.e. C = l f

But 2 pf = w

And if we model the wave going up and down as a mass/spring system, with stiffness K, then this leads to the

standard formula that w="#######KÅÅÅÅÅÅm where K is the stiffness of the wave as it goes up and down, and M is the mass of each wave.

and we know that from the mass/spring model that

restoring Force = K * displacement

Here the displacement is the average wave hight.

For deep water waves, it is gravity that causes a wave to fall down again after it goes up. Hence the restoring Force is

the weight of the wave

M g = K * wave height

K= M gÅÅÅÅÅÅÅÅÅÅh hence w= "########kÅÅÅÅÅÅÅM ="######gÅÅÅÅh

So f = wÅÅÅÅÅÅÅÅ2 p = "######gÅÅÅÅh 1ÅÅÅÅÅÅÅÅ2 p

so C = l f

C= l "######gÅÅÅÅh 1ÅÅÅÅÅÅÅÅ2 p

But 2ph = l (since one full cycle over the circle of radius h gives the length of the circumeference, which is the wave

length).

So C = "########glÅÅÅÅÅÅÅÅ2 p

AbbasiN93_graded_FINAL.nb 5

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à e) Black holes have a characteristic scale known as the event horizon. Use dimensional reasoning to estimate its size.

AbbasiN93_graded_FINAL.nb 6

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What does event horizon depends on? The event horizon is the distance from the black hole where light does not

escape (escape velocity equals the speed of light). Mass of the black hole must be involved, and the universal

gravtional constant as well. Therefore, I expect that the size must involve G, Mass, and c

params =

9mHKilogram L, G ik jjjj Meter

2 Newton ������������������������������������

Kilogram 2 y { zzzz, c HMeter êSecond L= êê ReduceUnits

9Kilogram m, G Meter 3

�������������������������������������������� Kilogram Second 2

, c Meter �������������������� Second

=

AbbasiN93_graded_FINAL.nb 7

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dimanal @params, 8Kilogram, Meter, Second

ode = − —2

��������� 2 m

ψ '' @xD + m g x ψ@xD En ψ@xD

g m xψ@xD − — 2 ψ′′@xD

���������������������� 2 m

En ψ@xD

The parameters involved in the ODE are m,g, Ñ and E

condp = 8m Kilogram, g , —< êê ToSymbolsUnits êê ReduceUnits

9Kilogram m, g Meter��������������������� Second 2

, Kilogram Meter 2 — ����������������������������������������������

Second =

Find what is the length x is propertional to (this is the independent variable)

dimanal @condp, 8Kilogram, Meter, Second, Coulomb

dimanal @condp, 8Kilogram, Meter, Second, Coulomb

ode2 = Simplify @ode2D 2 Hx r − λL ψr @x r D ψr ′′@x r D

The above represents the ODE in its dimensionless form. l is now the eigenvalue of the ODE

AbbasiN93_graded_FINAL.nb 11

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à b) Use finite differences to estimate the first few eigenvalues, and plot the first 4 eigenfunctions. ( use finitedifEVP from the Boundary Value Problems notebook; either copy and paste into this notebook, or execute the relevant lines in the browser. ) Explain how you deal with the boundary condition at •.

Needs@"LinearAlgebra`MatrixManipulation`" D; finitedifEVP @de_, bc_, xRange_, lam_, yvar_, xvar_,

npts_, verbose_ D : = Block @8fdsub, bcrules, n, evals, evecs, sortvals, mat, fe, fs, h, elimb1, elimb2, B1, B2

If @verbose, 8matall, rhs < = LinearEquationsToMatrices @ Append@Prepend @interioreqs, bceqs @@1DDD, bceqs @@2DDD, Table @fdyvar @i D, 8i, 1, npts

xv = Table @xRange@@1DD + Hi − 1L h, 8i, 1, npts

evc êê TableForm 33.3342

66.6676

à c) Solve the same equation using the shooting method. The shooting method also requires a finite interval. Recall that the shooting method can be unstable for large values of xr, so you may need to experiment with the appropriate form of the boundary conditon at the right hand boundary.

ψp0 = 1;

ode2

2 Hx r − λL ψr @x r D ψr ′′@x r D

Solve using shooting method by looking for solution that makes y goes to zero far away. I used x=100 value to

represent far away or infinity.

AbbasiN93_graded_FINAL.nb 15

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ψAtFar @ψp0_ ?NumericQ, eigenValue_ ?NumericQ D : = ψr @1D ê. NDSolve @8ode2 ê. λ → eigenValue, ψr @0D 0, ψr ' @0D ψp0

FindRoot @ψAtFar @ψp0, λD, 8λ, 80 •, so cannot be part of the solution. Then find the values of er which satisfy the boundary condition at xr=0.

sol = ψr @x r D ê. Flatten @DSolve @ode2, ψr @x r D, x r DD

AiryAi A 2 x r − 2 λ����������������������� 22ê3

E C@1D + AiryBi A 2 x r − 2 λ����������������������� 22ê3

E C@2D

Now I have 2 solutions being added togother to give a general solution. From the boundary condition that tells us

Ÿ0¶yy* „ x = 1 it implies that solution is zero at zero, and also zero at far away. So I plot each of the above solutions, then see which solution blow up as x gets large. Then I discard this solution since it would not lead to the boundary

condition given.

Plot @AiryAi @xD, 8x, 0, 10

Plot @AiryBi @xD, 8x, 0, 10

Plot @sol, 8λ, 0, 10

FindRoot @sol, 8λ, 6.5

Plot @fMain @xD, 8x, 0, 3

Plot @fOnePeriod @xD, 8x, 0, 3

Do@Plot @Evaluate @8xseries @x, n D, fMain @xD

0.2 0.4 0.6 0.8 1

0.1

0.2

0.3

0.4

0.5 n=3

0.2 0.4 0.6 0.8 1

0.1

0.2

0.3

0.4

0.5 n=4

0.2 0.4 0.6 0.8 1

0.1

0.2

0.3

0.4

0.5 n=5

AbbasiN93_graded_FINAL.nb 24

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0.2 0.4 0.6 0.8 1

0.1

0.2

0.3

0.4

0.5 n=6

0.2 0.4 0.6 0.8 1

0.1

0.2

0.3

0.4

0.5 n=7

0.2 0.4 0.6 0.8 1

0.1

0.2

0.3

0.4

0.5 n=8

AbbasiN93_graded_FINAL.nb 25

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0.2 0.4 0.6 0.8 1

0.1

0.2

0.3

0.4

0.5 n=9

Now plot the approximation over many periods

Do@Plot @Evaluate @8xseries @x, n D, fMain @xD

0.5 1 1.5 2 2.5 3

0.2

0.4

0.6

0.8

1 n=2

0.5 1 1.5 2 2.5 3

0.2

0.4

0.6

0.8

1 n=3

0.5 1 1.5 2 2.5 3

0.2

0.4

0.6

0.8

1 n=4

0.5 1 1.5 2 2.5 3

0.2

0.4

0.6

0.8

1 n=5

AbbasiN93_graded_FINAL.nb 27

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0.5 1 1.5 2 2.5 3

0.2

0.4

0.6

0.8

1 n=6

0.5 1 1.5 2 2.5 3

0.2

0.4

0.6

0.8

1 n=7

0.5 1 1.5 2 2.5 3

0.2

0.4

0.6

0.8

1 n=8

AbbasiN93_graded_FINAL.nb 28

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0.5 1 1.5 2 2.5 3

0.2

0.4

0.6

0.8

1 n=9

Function for animation

AnimateToClosedGroup @graphicslist_, animationdisplaytime_ D : = HNotebookWrite @EvaluationNotebook @D,

CellGroupData @Table @Cell @GraphicsData @"PostScript", DisplayString @graphicslist @@i DDDD, "Graphics" D,

8i, Length @graphicslist D

0.2 0.4 0.6 0.8

0.1

0.2

0.3

0.4

0.5

n=1

Below is animation of the whole function

plotlist = Table @Show@Plot @Evaluate @8xseries @x, n D, fMain @xD

à 4. A triangle function is zero for »x»>a, and has magnitude L at x=0 as shown in the figure.

à Calculate the Fou