Download Phys 6303 Final Exam Solutions December 19, 2012 You may NOT

Phys 6303 Final Exam Solutions
December 19, 2012
You may NOT use any book or notes other than supplied with this test. You will have
3 hours to finish. DO YOUR OWN WORK. Express your answers clearly and concisely
so that appropriate credit can be assigned for each problem. There are 6 problems. You
must do 5 for full credit. TURN IN ONLY 5 PROBLEMS - I WILL GRADE ONLY
THE FIRST 5 PROBLEMS YOU SUBMIT. Full credit for each problem is 25 points.
1)
Use the technique of Lagrange multipliers to find the ratio of the radius, r, to height, h, of
a right circular cylinder of fixed volume which minimizes the surface area.
Solution
The Volume of the cylinder is;
V = Constant = πr 2 h
The surface area is;
A = 2πr 2 + (2πrh)
Form the function, J, using the Lagrange multiplier, λ.
J = [2πr 2 + 2πrh] + λ[πr 2 h − V ]
Set the derivatives with respect to r and h to zero, obtaining 2 equations. Remove the
Lagrange parameter λ
∂J = 4πr + 2πh + λ 2πrh = 0
∂r
∂J = 2πr + λ 2πr 2 = 0
∂h
2r = h
2)
The equation;
1
∂ρ
∂2ρ
= a2 2 − κ ρ
∂t
∂x
with κ and a constants, represents neutron density, ρ, in a material of thickness, L. Assume
neutrons are reflected at the surface so ρ = 0 at x = 0 and x = L. Solve the equation by a
finite sine expansion and a Laplace transform with ρ = C, a constant, when t = 0.
Solution
As the solution for ρ(x, t) must vanish at x = 0, L it will have the form,
ρ(x, t) =
P
An (t) sin(nπx/L)
n=1
This leaves the time dependent equation for A(t).
P dAn (t)
= −a2 (nπ/L)2 An (t) − κ An (t)
dt
n
Apply a Laplace transform such that;
An (p) =
R∞
dt An (t) e−pt
0
Multiply each term in the equation by e−pt and integrate from 0 to ∞. The derivative term
must be integrated by parts.
R∞
0
dt dAn = An e−pt |∞
0 + p
dt
R∞
0
dt An e−pt
Transform the boundary condition;
A(p)[t=0] =
RL
0
dx C sin(nπx/L) = D = 2CL
nπ n even
D + pAn (p) = −[(nπa/L)2 + κ] An (p)
An (p) = −
D
p + (nπa/L)2 + κ
The inverse transform then gives;
1
An (t) = − 2πi
γ+i∞
R
γ−i∞
dp
D ept
p + (nπa/L)2 + κ
This integral is completed in the complex plane closing the integral around a loop to the left
of the imaginary axis. There is one pole;
2
p = −[(nπa/L)2 + κ]
The evaluation of An is;
An (t) = D e−αt
α = (nπa/L)2 + κ
Finally substitute this into the sum over n above to get, ρ(x, t)
3)
A long wire with vanishingly small radius, carries a charge per unit length, λ. It is placed
a distance, b, from the center of a long, perfectly-conducting, grounded cylinder with radius
a < b. Find the electric potential in the space for a < r < ∞ using the Green’s function
method. [hint; separate variables in Laplace’s equation as Bessel functions are not solutions]
Solution
The Green’s function is a soution to the equation;
∇2 V = δ(θ − θ′ )δ(ρ − ρ′ )
We seek a solution in cylinrical coordinates and it must be independent of the variable, z.
Thus using separation of variables the solution for the homogeneous equation has the form
V (ρ, θ) = R(ρ)Θ(θ). The separated homogeneous equation is;
2
(1/R)[ρ d (ρ dR )] = − d Θ
= ω2
dρ dρ
dθ2
In the above ω is the separation constant. It is selected so that Θ is continuous as θ →
2π → 0. Since the solution depends on 2 variables, there will be one eigenvalue solution, Θ,
and one dual solution for R. Build the Green’s function solution from these functions. The
eigenfunction solutions of Θ which satisfy the boundary conditions are; sin(nθ) with ω = n.
Thus expand δ(θ − θ′ ) in the eigenfunctions in these eigenfunctions to obtain;
δ(θ − θ′ ) = (1/π)
P
sin(nθ) sin(nθ′ )
n
The homogeneous radial equation is;
ρ d (ρ dR )] = n2 R
dρ dρ
The two solutions are;
3
R =
Aρn
Bρ−n
Require that the Green’s function solution be finite at ρ = 0 and ρ = ∞. To match the
boundary conditions;
A[(ρ/a)n ) − (ρ/a)−n ] ρ < b
R =
Bρ−n
ρ>b
The solution is continuous at ρ = ρ′ . However, the derivative of the solution is discontinuous. This results in the two equations;
A [(b/a)n − (b/a)−n ] = B (ρ/a)−n
b+ǫ
R
b−ǫ
b+ǫ
R
dρ ρ d (ρ dR ) =
dρ ρ δ(ρ − ρ′ )/ρ
dρ dρ
b−ǫ
A[(b/a)n − (b/n)−n ] + B(b/a)−n ] = 1/n
(b/a)n
n[(b/a) + (b/a)n + (b/a)−n + 1]
[(b/a)n − (b/a)−n ]
B =
2n
n[(b/a) + (b/a)n + (b/a)−n + 1]
A =
2n
The solution for the Green’s function is;
G(ρ, ρ ) =
′
P
sin(nθ) sin(nθ )
n
′
A [(ρ/a)n − (ρ/a)−n ]
B ρ−n
ρ < ρ′
ρ > ρ′
Then use the Green’s function to find the potential, V .
1 R ρ dρ dθ Gλ
∇2 V = − 4πǫ
An alternative method to obtain the Green’s function is the method of images replacing the
cylindrical surface with an image line charge below the surface. This is not developed and
was not expected to be attempted.
4)
Using the Lagrangian, find the position where a mass, M, sliding without friction along on
a hemispherical surface of radius, a, leaves the surface. The mass starts from rest at the top
of the hemisphere.
4
Solution
The Lagrangian is developed from the Kinetic and Potential energy. Use polar coordinates.
T = (M/2)(ṙ 2 + (r θ̇)2 )
V = mgr cos(θ)
The constraint is;
f = r −a = 0
The Lagrangian is;
L = (M/2)(ṙ 2 + (r θ̇)2 ) − mgr cos(θ)
Add in the Lagrange multiplier constraint, λf
The Euler-Lagrange equations become;
∂L − d [ ∂L ] + λ ∂f = 0
∂r
dt ∂ ṙ
∂r
∂L − d [ ∂L ] + λ ∂f = 0
∂θ
dt ∂ θ̇
∂θ
This gives the equations;
Mr θ̇2 − Mg cos(θ) − M r̈ + λ = 0
mgr sin(θ) − mr 2 θ̈ − 2mr ṙθ̇ = 0
Now r is constant, so the derivatives of r vanish. Solving the equations;
θ̈ = (g/a) sin(θ)
θ̇2 /2 = (−g/a) cos(θ) + g/a
λ = 0 = mg(3 cos(θ0 ) − 2)
θ0 = cos−1 (2/3)
5)
5
Find the Fourier components of a square pulse of width, d, and height, h.
Solution
The pulse has the form;


0
x < d/2
f (x) =  h d/2 < x < d/2 
0
x > d/2
P
f (x) =
An cos(nπx/d)
n odd
d/2
R
dx f (x) cos(nπx/d) = (d/2)An
−d/2
h sin(nπ/2)
An = nπ
However, for a truely single pulse use the following Fourier transform.
A(k) =
p
1/2π
R∞
dx f (x) e−ikx
−∞
2h sin(kd/2)
√
A(k) =
2π k
∞
ikx
R
h dk e sin(kd/2)
f (x) = π
k
∞
6)
Show that;
δ(x) = (1/π 2 )
R∞
dt
−∞
1
t(t − x)
represents a delta function by evaluating;
I =
R∞
dx f (x) δ(x)
−∞
Solution
Use the Principle Value integration. When f (x) is analytic and has appropriate limits at
± ∞, then;
−f (x0 ) = (1/iπ)
R∞
−∞
dx
f (x)
(x0 − x)
6
Apply this twice to evaluate each integral (note that using the principal value of x0 − x
introduces a negative sign for this integral);
J =
R∞
−∞
dx (f (x)/π 2 )
R∞
−∞
ds
1
t(t − x)
7