Question

For a regular sinusoidal wave propagating in deep water having wave height of 3.5 m and wave period of 9 s, the wave steepness is .................... (round off to three decimal places)

Hint:

Remember the deep water approximation for wavelength: \( L \approx 1.56 T^2 \) (using \(g = 9.81\) m/s\(^2\) and \(2\pi \approx 6.28\)). This is a quick way to estimate the wavelength from the period. For this problem: \(1.56 \times 9^2 = 1.56 \times 81 = 126.36\) m, which is very close.

Answer 1

Step 1: Understanding the Concept:
Wave steepness is a dimensionless parameter that characterizes waves, defined as the ratio of the wave height to the wavelength. To find the steepness, we first need to calculate the wavelength from the given wave period using the dispersion relation for deep water waves.
Step 2: Key Formula or Approach:
1. Wave Steepness (\(S\)): \( S = \frac{H}{L} \), where \(H\) is the wave height and \(L\) is the wavelength.
2. Dispersion Relation for Deep Water: In deep water (where the water depth is greater than half the wavelength), the relationship between wavelength (\(L\)) and wave period (\(T\)) is given by:
\[ L = \frac{g T^2}{2\pi} \] where \(g\) is the acceleration due to gravity (approx. 9.81 m/s\(^2\)).
Step 3: Detailed Calculation:
Given values:
- Wave height, \(H = 3.5\) m.
- Wave period, \(T = 9\) s.
- Acceleration due to gravity, \(g \approx 9.81\) m/s\(^2\).
1. Calculate the wavelength (L): \[ L = \frac{g T^2}{2\pi} = \frac{9.81 \times (9)^2}{2\pi} = \frac{9.81 \times 81}{2\pi} \] \[ L = \frac{794.61}{2\pi} \approx 126.49 \text{ m} \] 2. Calculate the wave steepness (S): \[ S = \frac{H}{L} = \frac{3.5 \text{ m}}{126.49 \text{ m}} \] \[ S \approx 0.02767 \] 3. Round to three decimal places: \[ S \approx 0.028 \] Step 4: Final Answer:
The wave steepness is 0.028.