 Solution

ID: past-exam-of-the-year-2-of-the-earth-sciences-geology-course-of-the-university-of-oxford/2024/paper-3/5/b/solution

For a nonzero, time-independent vector

\hat{u}

, substitute

u (t) = \hat{u} e^{iω t}

. Taking two time derivatives gives

\ddot{u} = (iω)^{2} \hat{u} e^{iω t} = - ω^{2} \hat{u} e^{iω t} .

Cancel the nonzero exponential to obtain the eigenvalue problem

A \hat{u} = λ \hat{u}, λ = - ω^{2}, \hat{u} \neq = 0.

Thus an eigenvector specifies a normal mode shape and its negative eigenvalue specifies the square of its angular frequency,

ω = - λ

. Real displacements are obtained by taking the real part, or by combining the two temporal solutions as

\hat{u} (a cos ω t + b sin ω t)

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