Question

An ocean wave of period 8 s and height 2 m is propagating in the Indian Ocean from south to north. According to linear wave theory, for the wave to be considered as a deep-water wave, the minimum water depth should be …………. (rounded off to the nearest integer). 
 

Hint:

For deep-water waves: 1. Use the relationship \( h \geq \frac{\lambda}{2} \) to check the water depth.
2. Calculate \( \lambda \) using the wave period and the formula \( \lambda = \frac{g T^2}{2\pi} \) for accurate results.
3. Deep-water waves occur when the water depth is large relative to the wavelength.

Answer 1

Step 1: Recall the condition for a deep-water wave. 
For a wave to be considered a deep-water wave, the water depth \( h \) must satisfy: \[ h \geq \frac{\lambda}{2}, \] where \( \lambda \) is the wavelength of the wave. 

Step 2: Determine the wavelength (\( \lambda \)). 
Using the wave speed equation: \[ C = \frac{\lambda}{T}, \] where: - \( C \) is the wave speed, - \( T = 8 \, \text{s} \) is the wave period. For deep-water waves, the wave speed \( C \) is given by: \[ C = \sqrt{\frac{g \lambda}{2\pi}}, \] where \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity. Rewriting the equation: \[ \lambda = \frac{g T^2}{2\pi}. \] Substitute the values: \[ \lambda = \frac{9.81 \cdot 8^2}{2\pi} = \frac{9.81 \cdot 64}{6.2832} = \frac{627.84}{6.2832} \approx 100 \, \text{m}. \] Step 3: Calculate the minimum depth (\( h \)). 
Using the deep-water condition: \[ h \geq \frac{\lambda}{2} = \frac{100}{2} = 50 \, \text{m}. \] Conclusion: The minimum water depth for the wave to be considered as a deep-water wave is \( 50 \, \text{m} \).