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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2019

Paper 2

Take a <shear-horizontal wave> polarized in the $z$ direction,
$$
\mathbf u=w(x,y,t)\mathbf e_z.
$$
Then $\nabla\cdot\mathbf u=0$, and the elastic-wave equation reduces to
$$
w_{tt}=c_S^2(w_{xx}+w_{yy}),
\qquad
c_S=\sqrt{\frac\mu\rho}.
$$
For $w=Y(y)e^{i(kx-\omega t)}$, the rigid boundary gives $Y(0)=0$, while vanishing shear traction at the free surface gives $Y'(h)=0$. Thus
$$
Y_n(y)=A\sin(q_ny),
\qquad
q_nh=\left(n+\frac12\right)\pi,
\qquad n=0,1,2,\ldots.
$$
The dispersion relation of each <guided shear-horizontal mode> is
$$
\boxed{\omega_n^2=c_S^2(k^2+q_n^2),
\qquad q_n=\frac{(n+1/2)\pi}{h}.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 2 / 38A / a / Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 2 / 38A / a / Solution

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Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 2 / 38A / a / Solution

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