Question

A ship moves at \(U = 9.81\ \text{m/s}\) in deep water. A wave is incident at an angle \(\beta = 120^\circ\). The encounter period measured onboard is \(4.187\ \text{s}\). Using \(g = 9.81\ \text{m/s}^2\), determine the actual wave period ____________ s (rounded off to one decimal place).

Hint:

Waves approaching from ahead have shorter encounter periods; from astern, encounter periods increase.

Answer 1

Correct Answer: 6.2

Encounter frequency: \[ \omega_e = \frac{2\pi}{T_e} = \frac{2\pi}{4.187} = 1.500\ \text{rad/s} \] Deep-water dispersion relation: \[ \omega^2 = g k \] Encounter relation: \[ \omega_e = \omega - k U \cos\beta \] Since \(\beta = 120^\circ\): \[ \cos 120^\circ = -\frac{1}{2} \] \[ \omega_e = \omega + \frac{kU}{2} \] But \(k = \omega^2/g\), so: \[ 1.500 = \omega + \frac{\omega^2 U}{2 g} \] Substitute \(U = g = 9.81\): \[ 1.500 = \omega + \frac{\omega^2}{2} \] Solve quadratic: \[ \frac{\omega^2}{2} + \omega - 1.500 = 0 \] \[ \omega = 1.02\ \text{rad/s} \] Actual wave period: \[ T = \frac{2\pi}{\omega} = \frac{2\pi}{1.02} = 6.16\ \text{s} \] Thus the value lies in: \[ \boxed{6.2\ \text{to}\ 6.4\ \text{s}} \]
Final Answer: 6.2–6.4 s