Question:hard
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\(^2\).
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\(^2\).
The frequency of the incoming wave is ______ rad/s (answer in integer).
Assume g = 10 m/s\(^2\).
Show Hint
Convert the counted crest period to encounter frequency, then use the head sea encounter relation.
Updated On: Jul 28, 2026
Show Solution
Correct Answer: 1
Solution and Explanation
Step 1: Turn the counted period into an encounter frequency.
The crests the observer counts arrive every $T_e = \pi$ s, so $\omega_e = 2\pi/T_e = 2$ rad/s.
Step 2: Bring in the head sea encounter formula as a quadratic in $\omega$.
For head seas, the ship meets waves faster than they actually travel: $\omega_e = \omega + \omega^2 V/g$. Treating this as an equation in the unknown $\omega$, with $V/g = 10/10 = 1$, it becomes $\omega^2 + \omega - \omega_e = 0$, i.e. $\omega^2 + \omega - 2 = 0$.
Step 3: Solve with the quadratic formula and pick the sensible root.
$\omega = \dfrac{-1 \pm \sqrt{1 + 8}}{2} = \dfrac{-1 \pm 3}{2}$, so $\omega = 1$ or $\omega = -2$. A negative frequency has no physical meaning here, so $\omega = 1$ rad/s.
Final Answer:
The incoming wave frequency is 1 rad/s. \[ \boxed{\omega = 1 \text{ rad/s}} \]
The crests the observer counts arrive every $T_e = \pi$ s, so $\omega_e = 2\pi/T_e = 2$ rad/s.
Step 2: Bring in the head sea encounter formula as a quadratic in $\omega$.
For head seas, the ship meets waves faster than they actually travel: $\omega_e = \omega + \omega^2 V/g$. Treating this as an equation in the unknown $\omega$, with $V/g = 10/10 = 1$, it becomes $\omega^2 + \omega - \omega_e = 0$, i.e. $\omega^2 + \omega - 2 = 0$.
Step 3: Solve with the quadratic formula and pick the sensible root.
$\omega = \dfrac{-1 \pm \sqrt{1 + 8}}{2} = \dfrac{-1 \pm 3}{2}$, so $\omega = 1$ or $\omega = -2$. A negative frequency has no physical meaning here, so $\omega = 1$ rad/s.
Final Answer:
The incoming wave frequency is 1 rad/s. \[ \boxed{\omega = 1 \text{ rad/s}} \]
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