Solutions to Final Exam

Physics 604

Dec. 15, 2003

This CLOSED BOOK exam is to be completed in 2 hours.  Choose THREE of the four problems — if you attempt all four, be sure to indicate clearly which three should be graded.  You must explain your reasoning to receive full credit.

Initialization

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\(TraditionalForm\`\(A[
                x]\/c\^2\) ∂\^2 ψ[x, t]\/∂t\^2\  - \ \(∂\/∂x\) \((A[
                x] ∂ψ[x, t]\/∂x)\)\  == \ 
      0\)

ψ[x, t]  =  (e^(-b x)) 
        Cos[k x  -  ω t]

a)

We simply substitute the proposed solution into the wave equation to obtain

-((ω^2)/(c^2)) Exp[(a - b) x] 
          Cos[k x  - ω t]  +  (∂/∂)x ( 
            Exp[(a - b) x] (b Cos[k x  - ω t]  +  
                  k Sin[k x  - ω t]) ) ==  
      0

or

-((ω^2)/(c^2)) 
          Cos[k x  - ω t]  +  (a - 
              b) (b Cos[k x  - ω t] + 
              k Sin[k x  - ω t])  +  (-k b Sin[
                k x  - ω t]  +  (k^2) 
              Cos[k x  - ω t])  ==  0

Collecting the coefficients of the sine and cosine functions provides the two equations

-((ω^2)/(c^2)) + (a - b) b  +  
        (k^2)  ==  0
(a - b) k  -  k b  ==  0

such that

b = (a/2)
\(TextForm\`\(-\(ω\^2\/c\^2\)\) + a\^2\/4 + k\^2\  == \ 
        0\ \ ⟹\ \ \ k\  = \ \@\(ω\^2\/c\^2 - a\^2\/4\)\)

This solution is an attenuated travelling wave where the attenuation length, (b^(-1)), is independent of frequency.  Hence, the balance between high and moderately low frequencies is preserved as the sound propagates.

b)

However, there is a low-frequency cut-off, (Subscript[ω, c]) = a c/2 = b c, below which waves do not propagate because k becomes complex.  For very low frequencies, we substitute a trial solution of the form

\(TraditionalForm\`ψ[x, 
        t]\  = \ \(e\^\(\(-b\)\ x\)\) 
            Cos[ω\ t]\ \ \ ⟹\ \ \ \(-\(ω\^2\/c\^2\)\) + \ b(a - b)\  == \ 0\)

to obtain

b  =  
      (a/2)  ±  (Sqrt[((a^2)/4) - ((ω^2)/(c^2))])

for standing waves.  Both damping coefficients are positive and must be retained, but after some distance the smaller one will dominate and determines the maximum penetration of low-frequencies vibrations into the horn.   Notice that at ω -> 0 there is a static solution with b -> 0 that is not attenuated.

c)

The phase and group velocities are given by

(Subscript[v, p]) = (ω/k)  =  
        c (Sqrt[ 1 + ((a^2)/(4  (k^2)))])
\(TextForm\`ω\ dω\  == \ \(c\^2\) 
          k\ dk\ \ ⟹\ \ v\_g = \(dω\/dk = 
        c\^2\/v\_p\)\)

such that

(Subscript[v, p]) (Subscript[v, g]) = (c^2)

These velocities are sketched below as functions of k/a.  Notice that (Subscript[v, p]) diverges and (Subscript[v, g]) vanishes at k = 0.

\(\(Plot[
        Evaluate[{\((1 + 1\/\(4  k\^2\))\)\^\(1/2\), \((1 + 1\/\(4  k\^2\))\)\^\(\(-1\)/2\)}], {k, 0.6, 5}, PlotRange -> {Automatic, {0.8, 1.2}}, 
        Frame -> True, FrameLabel -> {"k/a", "v/c"}, 
        PlotStyle -> {{}, Dashing[{0.02, 0.02}]}, 
        PlotLegend -> {\*""(Subscript[v, p])/c"", \*""(Subscript[v, g])/c""}, LegendPosition -> {0.3, 0.3}, 
        LegendSize -> {0.6, 0.25}, 
        LegendShadow -> None];\)\)

It may also be of interest to sketch these velocities as functions of ω/(Subscript[ω, c]) instead.  Notice that (Subscript[v, p]) diverges and (Subscript[v, g]) vanishes at (Subscript[ω, c]), and are complex for smaller frequencies.

\(\(Plot[
        Evaluate[{ω \((ω\^2 - 1\/4)\)\^\(\(-1\)/2\), \((ω\^2 - 1\/4)\)\^\(1/2\)/ω}], {ω, 0.5, 5}, 
        PlotRange -> {Automatic, {0.4, 1.5}}, Frame -> True, 
        FrameLabel -> {\*""ω/(Subscript[ω, c])"", "v/c"}, PlotStyle -> {{}, Dashing[{0.02, 0.02}]}, 
        PlotLegend -> {\*""(Subscript[v, p])/c"", \*""(Subscript[v, g])/c""}, LegendPosition -> {0.3, 0.3}, 
        LegendSize -> {0.6, 0.25}, 
        LegendShadow -> None];\)\)

\(TextForm\`\(y\^′\)[
          t]\  + \ ∫\_0\%t 
              y[τ] dτ\  == \ e\^\(-t\)\ , \ y[0] == 1\)

solution

The Laplace transform of this integro-differential equation reads

\(TextForm\`s\ y\&~\  - 
          1\  + \ \(s\^\(-1\)\) 
            y\&~\  == \ \((s + 1)\)\^\(-1\)\ \ ⟹\ \ y\&~\  = \ \(s \((s + 2)\)\)\/\(\((s + 1)\) \((s\^2 + 1)\)\)\)

The solution transform is resolved into partial fractions according to

(A/(s + 1)) + (B/(s + i)) + 
        (C/(s - i)) == ((s (s + 2))/((s + 
                      1) ((s^2) + 1)))  ⟹ 
            A ((s^2) + 1) + B (s + 1) (s - i) + 
        C (s + 1) (s + i) == 
      (s^2) + 2  s

which yields the system of equations

A + B + C == 1
B (1 - i) + 
        C (1 + i) == 2
A - i B + i C  == 
      0

The second gives

\(TextForm\`B\  = \ \(2 - \((1 + i)\) C\)\/\(1 - i\)\)

and substitution into the third gives

\(TextForm\`\((1 - i)\) A\  - 
          2  i\  + \ i\ \((1 + i)\) 
            C + i \((1 - i)\) C == 
        0\ \ ⟹\ \ \ A\  = \ \(\(2  i\)\/\(1 - i\)\) \((1 - C)\)\)

Finally, substitution into the first equation gives

\(TextForm\`2  i \((1 - C)\)\  + \ 
          2\  - \((1 - i)\) C + \((1 - i)\) 
            C == 
        0\ \ ⟹\ \ \ C\  = \ \(\(1 + 3  i\)\/\(4  i\) = \(\(\ \)\(3 - i\)\)\/4\)\)

such that

\(TextForm\`A\  = \ \(\(\(2  i\)\/\(1 - i\)\) \((\(1 + i\)\/4)\)\  = \ \(\(i\/4\) \((1 + i)\)\^2\  = \ \(-\(1\/2\)\)\)\)\)
\(TextForm\`B\  = \ \(\(8 - \((1 + i)\) \((3 - i)\)\)\/\(4 \((1 - i)\)\)\  = \ \(\(8 - \((4 + 2  i)\)\)\/\(4 \((1 - i)\)\)\  = \ \(\(\((4 - 2  i)\) \((1 + i)\)\)\/8\  = \ \(3 + i\)\/4\)\)\)\)

Thus,

\(TextForm\`y\&~\  = \ \(\(s \((s + 2)\)\)\/\(\((s + 1)\) \((s\^2 + 1)\)\)\  = \ \(-\(1\/2\)\) 1\/\(s + 1\)\  + \ \(\(3 + i\)\/4\) 
            1\/\(s + i\)\  + \ \(\(\(\ \)\(3 - i\)\)\/4\) 1\/\(s - i\)\)\)

can be inverted using

\(TextForm\`\(1\/\(s + α\)\)\ ⟶\ e\^\(\(-α\)\ t\)\ \ \ ⟹\ \ y[
          t]\  = \ \(-\(1\/2\)\) e\^\(-t\)\  + \ \(\(3 + i\)\/4\) e\^\(\(-i\)\ t\)\  + \ \(\(\(\ \)\(3 - i\)\)\/4\) e\^\(i\ t\)\)

Therefore, we obtain

y[
        t]  =  (1/2) (3  Cos[t] + 
            Sin[t] - (e^(-t)))

(∫_0^∞) g[x] dx

\(∫\_0\%∞dx\/\(\((x + 1)\) \((x\^2 + 2  x + 2)\)\)\)

solution

Let

\(f[z]\  = \ \(Log[z]\/\(\((z + 1)\) \((z\^2 + 2  z + 2)\)\)\  = \ 
        Log[z]\/\(\((z + 1)\) \((z + 1 + i)\) \((z + 1 - i)\)\)\)\)

with a branch cut on the positive real axis, such that

\(z = \(x + i\ ε\ \ ⟹\ \ f[z]\  = \ 
        Log[x]\/\(\((x + 1)\) \((x\^2 + 2  x + 2)\)\)\)\)
\(z = \(x - i\ ε\ \ ⟹\ \ f[z]\  = \ \(Log[x] + 2  π\ i\)\/\(\((x + 1)\) \((x\^2 + 2  x + 2)\)\)\)\)

For large R we use

\(z = 
      R\ e\^\(i\ θ\)\ \ \ ⟹\ \ |f[
                    z]\ dz\/dθ| ≃ \ |Log[R] + i\ θ|\/R\^2\  -> \ 
        0\)

while near the origin

\(z = ε\ e\^\(i\ θ\)\ \ \ ⟹\ |f[
                    z] dz\/dθ|\  ≈ \ |ε \((Log[ε] + i\ θ)\)|\  -> \ 0\)

Thus, we can neglect both circular portions and obtain

\(TextForm\`∫\_0\%∞dx\/\(\((x + 1)\) \((x\^2 + 2  x + 2)\)\)\  = \ \(-\(1\/\(2  π\ i\)\)\) \(∮\_C\( Log[
                  z]\/\(\((z + 1)\) \((z + 1 + i)\) \((z + 
                      1 - i)\)\)\) dz\)\)

The right-hand side can now be evaluated as the sum of the residues for \(TraditionalForm\`z = \(-1\), \(-1\) - i, \(-1\) + i\), such that

\(TextForm\`∫\_0\%∞dx\/\(\((x + 1)\) \((x\^2 + 2  x + 2)\)\)\ \  = \ \(-\((Log[\(-1\)]\/\(\((i)\) \((\(-i\))\)\)\  + \ 
            Log[\(-1\) - i]\/\(\((\(-i\))\) \((\(-2\) i)\)\)\  + \ 
            Log[\(-1\) + i]\/\(\((i)\) \((2  i)\)\))\)\)\ \)
\(TextForm\`\(\(=\)\(\ \)\(-\((i\ π\  - \(1\/2\) \((Log[\@2] + \(\(5  π\)\/4\) i)\) - \(1\/2\) \((Log[\@2] + \(\(3  π\)\/4\) i)\))\)\)\)\)

Therefore,

\(TextForm\`∫\_0\%∞dx\/\(\((x + 1)\) \((x\^2 + 2  x + 2)\)\)\  = \ Log[\@2]\)

\(TextForm\`∇\^2\(\(=\)\(\ \)\(\(1\/ξ\) \(∂\/∂ξ\) \((ξ\ ∂\/∂ξ)\)\  + \ \(1\/ξ\^2\) ∂\^2\/∂ϕ\^2\  + \ ∂\^2\/∂z\^2\)\)\)

(d^2) (R/d)(ξ^2) + 
          (1/ξ) d(R/d)ξ  + ((k^2) -  
                ((m^2)/(ξ^2))) R  ==  
        0  ⟹  R  =  
      A (Subscript[J, m])[k ξ]  +  B (Subscript[N, m])[k ξ]

\(TextForm\`J\_m[\(k\_\(m, n\)\) 
          a] = \(0\ \ \ ⟹\ \ \(∫\_0\%a\( J\_m[\(k\_\(m, 
                          n\)\) ξ]\^2\) ξ dξ\)\  = \ \(a\^2\/2\) J\_m\%′[\(k\_\(m, n\)\) a]\^2\)\)

\(TextForm\`m ≥ 
        1\ \ ⟹\ \ \(d\/dx\) \((\(x\^m\) J\_m[x])\)\  = \ \(x\^m\) J\_\(m - 1\)[x]\)

a)

The Laplacian takes the form

\(TextForm\`∇\^2\(\(=\)\(\ \)\(\(1\/ξ\) \(∂\/∂ξ\) \((ξ\ ∂\/∂ξ)\)\  + \ \(1\/ξ\^2\) ∂\^2\/∂ϕ\^2\  + \ ∂\^2\/∂z\^2\)\)\)

in cylindrical coordinates.  Laplace's equation

(∇^2) ψ  ==  
        0          ψ[ξ, ϕ, z]  =  
      R[ξ] Φ[ϕ] Z[z]

with boundary conditions

R[0] finite,   R[a] == 0
Φ[ϕ + 
          2  π]  ==  Φ[ϕ]
Z[-(L/2)]  ==  -Z[
          (L/2)]

then separates into

\(TextForm\`\(1\/Φ\) d\^2 Φ\/dϕ\^2\  == \ \(-m\^2\ \ \ \ \ ⟹\ \ \ \ Φ\)\  = \ e\^\(i\ m\ ϕ\)\ , \ \ m = 
      0, \ \(±1\), \ \(±\ 
      2\), \ ⋯\)

\(TextForm\`\(\(\(1\/Z\) d\^2 Z\/dz\^2\  == \ 
        k\^2\ \ \ \ \ ⟹\ \ \ \ Z\)\(\ \)\(=\)\(\ \)\(Sinh[k\ z]\)\(\ \)\)\)

(d^2) (R/d)(ξ^2) + 
          (1/ξ) d(R/d)ξ  + ((k^2) -  
                ((m^2)/(ξ^2))) R  ==  
        0  ⟹  R  =  
      (Subscript[J, m])[k ξ]

where (Subscript[J, m]) is the Bessel function of the first kind, the one that is finite at the origin.  Note that although the Bessel equation is insensitive to the sign of m, the conventional definition (Subscript[J, -m])[ξ] = (((-))^m) (Subscript[J, m])[ξ] simplifies some of the relationships among Bessel functions.  The boundary condition at the curved surface requires k -> (Subscript[k, m, n]) where the index n enumerates the roots of

(Subscript[J, m])[(Subscript[k, m, n]) a]  ==  0

Thus, the general solution takes the form

\(TextForm\`ψ[ξ, ϕ, 
        z]\  = \ ∑\+\(m = \(-∞\)\)\%∞\(∑\+\(n = 1\)\%∞\( ψ\_\(m, n\)\) J\_m[\(k\_\(m, n\)\) ξ] 
            Sinh[\(k\_\(m, n\)\) 
                z] e\^\(i\ m\ ϕ\)\)\)

where the (Subscript[ψ, n, m]) are numerical coefficients obtained by matching the boundary conditions at z = ±L/2 according to

\(TextForm\`ψ\_\(m, n\)\  = \ \(\((π\ \(a\^2\) 
                J\_m\%′[\(k\_\(m, n\)\) a]\^2\ Sinh[\(k\_\(m, n\)\) 
                    L/2])\)\^\(-1\)\) \(∫\_0\%\(2  π\)dϕ\ \(∫\_0\%adξ\ ξ\ \(e\^\(\(-i\)\ m\ ϕ\)\) 
                J\_m[\(k\_\(m, n\)\) ξ] 
                V[ξ, ϕ]\)\)\)

b)

The expansion coefficients for V[ξ, ϕ]  =  (Subscript[V, 0]) require the integrals

\(TextForm\`∫\_0\%\(2  π\)dϕ\ \ e\^\(\(-i\)\ m\ ϕ\)\  = \ 
      2  π\ δ\_\(m, 0\)\)

\(TextForm\`∫\_0\%a J\_0[\(k\_\(0, 
                  n\)\) ξ] ξ\ dξ\  = \ \(\(1\/k\_\(0, n\)\^2\) \(∫\_0\%\(\(k\_\(0, n\)\) a\)J\_0[x] 
              x dx\)\  = \ \(\(1\/k\_\(0, n\)\^2\) \(∫\_0\%\(\(k\_\(0, n\)\) a\)\(d\/dx\) \((x\ J\_1[x])\) dx\)\  = \ \(a\/k\_\(0, n\)\) 
            J\_1[\(k\_\(0, n\)\) a]\)\)\)

Thus, the electrostatic potential becomes

\(TextForm\`ψ[ξ, ϕ, 
        z]\  = \ \(2  V\_0\)\/a\ \(∑\+\(n = 1\)\%∞\( J\_0[\(k\_\(0, n\)\) ξ]\/\(\(k\_\(m, n\)\) J\_1[\(k\_\(0, n\)\) a]\)\) 
            Sinh[\(k\_\(0, n\)\) z]\/Sinh[\(k\_\(0, n\)\) L/2]\)\)