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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2023

Paper 4

The dynamic condition reduces to
$$
F'(t)+gH(t)=-\frac{p_0}{\rho}\cos(\omega t).
$$
Differentiate $H'=kF$ and define the natural frequency
$$
\Omega^2=gk=\frac{\pi g\sqrt5}{L}.
$$
Then the surface amplitude obeys the <forced harmonic oscillator>
$$
\boxed{H''+\Omega^2H
=-\frac{kp_0}{\rho}\cos(\omega t)}.
$$
For $\omega\ne\Omega$, its general solution and the corresponding potential amplitude are
$$
H=C\cos(\Omega t)+D\sin(\Omega t)
+\frac{kp_0}{\rho(\omega^2-\Omega^2)}\cos(\omega t),
$$
$$
F=\frac{H'}k.
$$

If the fluid is initially quiescent, $H(0)=F(0)=0$, so
$$
\boxed{H(t)=\frac{kp_0}{\rho(\omega^2-\Omega^2)}
\bigl(\cos(\omega t)-\cos(\Omega t)\bigr)},
$$
$$
\boxed{F(t)=\frac{p_0}{\rho(\omega^2-\Omega^2)}
\bigl(\Omega\sin(\Omega t)-\omega\sin(\omega t)\bigr)}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 4 / 16C / b / v / Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 4 / 16C / b / v / Solution

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Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 4 / 16C / b / v / Solution

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