 Solution

ID: past-exam-of-the-elasticity-and-plasticity-course-of-the-university-of-oxford/2010/2/c/solution

Substitution of

w_{i} = f_{i} (y) e^{i (k x - ω t)}

gives

f_{i}^{''} + (\frac{ω ^{2}}{c _{s i}^{2}} - k^{2}) f_{i} = 0.

The oscillatory seam solution requires

m^{2} = ω^{2} / c_{s 1}^{2} - k^{2}

ω^{2} = c_{s 1}^{2} (k^{2} + m^{2}), c_{L} = \frac{ω}{∣ k ∣} = c_{s 1} 1 + m^{2} / k^{2} .

For an exponentially decaying exterior,

f_{2}^{''} = ℓ^{2} f_{2}

with

ℓ > 0

, so the same differential equation instead requires

ℓ^{2} = k^{2} - ω^{2} / c_{s 2}^{2}, ω^{2} = c_{s 2}^{2} (k^{2} - ℓ^{2}) .

The printed exterior plus sign is inconsistent with its decaying ansatz and must be a sign error. The guided mode has

c_{s 1} < c_{L} < c_{s 2}

, so a slower shear-wave seam than its cladding is necessary.

For an even seam mode

f_{1} = A cos (m y)

, upper-interface displacement and traction continuity give

A cos (mh) = A_{2} e^{- ℓ h}, - μ_{1} A m sin (mh) = - μ_{2} ℓ A_{2} e^{- ℓ h} .

The lower interface gives the same equations when

B_{2} = A_{2}

. Eliminating the amplitudes yields

μ_{1} m tan (mh) = μ_{2} ℓ .

For an odd seam mode

f_{1} = B sin (m y)

, the equations are

B sin (mh) = A_{2} e^{- ℓ h}, μ_{1} B m cos (mh) = - μ_{2} ℓ A_{2} e^{- ℓ h} .

With

B_{2} = - A_{2}

the lower condition agrees, and elimination gives

μ_{1} m cot (mh) = - μ_{2} ℓ .

These interface conditions together with the two corrected dispersion identities determine the guided-wave branches.

 New to topics? Read the docs here!