Question

A ship is moving at 10 m/s in head sea condition. A person on board counts the time period between two successive wave crests as \(\pi\) seconds.
The frequency of the incoming wave is ______ rad/s (answer in integer).
Assume g = 10 m/s\(^2\).

Hint:

Convert the counted crest period to encounter frequency, then use the head sea encounter relation.

Answer 1

Correct Answer: 1

Step 1: Identify what the person on board is actually measuring.
Someone standing on a moving ship sees waves pass at the encounter period \(T_e\), not the true wave period. So the encounter frequency is \(\omega_e = 2\pi / T_e = 2\pi / \pi = 2\) rad/s.

Step 2: Relate encounter frequency to true wave frequency in head seas.
In deep water, \(\omega_e = \omega - \dfrac{\omega^2 V}{g}\cos\mu\), where \(\mu\) is measured from the bow. In a head sea \(\mu = 180^{\circ}\), so \(\cos\mu = -1\) and the relation becomes \(\omega_e = \omega + \dfrac{\omega^2 V}{g}\).

Step 3: Substitute and solve the quadratic.
With \(V = 10\) m/s and \(g = 10\) m/s\(^2\), \(V/g = 1\), so \(2 = \omega + \omega^2\), i.e. \(\omega^2 + \omega - 2 = 0\). Factoring, \((\omega - 1)(\omega + 2) = 0\), giving \(\omega = 1\) or \(\omega = -2\). Only the positive root is physical.

Final Answer:
The true wave frequency is \(\omega = 1\) rad/s, matching the official key. \[ \boxed{\omega = 1 \text{ rad/s}} \]