Question

An ocean wave is propagating from deep to shallow water. The wave is approaching the coast at \( 45^\circ \) counterclockwise from the shore normal with an initial phase speed of \( 12.5 \, {m/s} \). After entering the shallow water, the wave direction becomes \( 30^\circ \) counterclockwise from the shore normal. The phase speed of the wave at the shallow water is ________ m/s (rounded off to one decimal place).

Hint:

For wave refraction between deep and shallow waters, use Snell's Law: \[ \frac{\sin \theta_1}{C_1} = \frac{\sin \theta_2}{C_2} \] Here, \( \theta \) is the angle from the shore normal, and \( C \) is the phase speed.

Answer 1

Step 1: Use Snell's law for wave propagation in water. 
Snell's Law for water waves relates the phase speed \( C \) and angle of wave approach \( \theta \): \[ \frac{\sin \theta_1}{C_1} = \frac{\sin \theta_2}{C_2} \] Given: \[ C_1 = 12.5 \, {m/s}, \quad \theta_1 = 45^\circ, \quad \theta_2 = 30^\circ \] Step 2: Substitute into Snell's Law: \[ \frac{\sin 45^\circ}{12.5} = \frac{\sin 30^\circ}{C_2} \] \[ \frac{0.7071}{12.5} = \frac{0.5}{C_2} \quad \Rightarrow \quad C_2 = \frac{0.5 \times 12.5}{0.7071} \approx \frac{6.25}{0.7071} \approx 8.83 \, {m/s} \] Rounded to one decimal place: \[ C_2 = \boxed{8.8 \, {m/s}} \]