Question

The wave spectrum and the ship heave Response Amplitude Operator (RAO) are shown in the figure. The variance of the heave motion is …… m\(^2\) (rounded off to three decimal places). 

Hint:

When calculating variance for RAO problems: 1. Identify \( RAO(\omega) \) and \( S(\omega) \) correctly from the graphs.
2. Split the integration range based on the changes in the wave spectrum.
3. Use symbolic integration to simplify expressions before substituting limits.

Answer 1

Step 1: Recall the formula for the variance of the heave motion. 
The variance of the heave motion is obtained by integrating the product of the square of the RAO and the wave spectrum: \[ \text{Variance} = \int_{0}^{\infty} RAO^2(\omega) \cdot S(\omega) \, d\omega. \] From the given plots: - The RAO decreases linearly from \( 1.2 \) at \( \omega = 0 \, \text{rad/s} \) to \( 0.2 \) at \( \omega = 2 \, \text{rad/s} \), - The wave spectrum \( S(\omega) \) increases linearly from \( 0 \, \text{m}^2/\text{s} \) at \( \omega = 0 \, \text{rad/s} \) to \( 2.5 \, \text{m}^2/\text{s} \) at \( \omega = 1 \, \text{rad/s} \), and decreases linearly to \( 0 \, \text{m}^2/\text{s} \) at \( \omega = 2 \, \text{rad/s} \). 

Step 2: Divide the integration into two regions. 
The wave spectrum is divided into two linear regions: 1. \( 0 \leq \omega \leq 1 \), 2. \( 1 \leq \omega \leq 2 \). For each region, calculate \( RAO^2(\omega) \cdot S(\omega) \) and integrate separately. 

Step 3: Define \( RAO(\omega) \) and \( S(\omega) \) for each region. 
1. \( RAO(\omega) = 1.2 - 0.5\omega \). \( S(\omega) \) is: - \( S(\omega) = 2.5\omega \) for \( 0 \leq \omega \leq 1 \), - \( S(\omega) = -2.5(\omega - 2) \) for \( 1 \leq \omega \leq 2 \). 2. The square of \( RAO(\omega) \) is: \[ RAO^2(\omega) = (1.2 - 0.5\omega)^2 = 1.44 - 1.2\omega + 0.25\omega^2. \] 

Step 4: Perform the integration for each region. 
1. For \( 0 \leq \omega \leq 1 \): \[ \int_{0}^{1} RAO^2(\omega) \cdot S(\omega) \, d\omega = \int_{0}^{1} (1.44 - 1.2\omega + 0.25\omega^2)(2.5\omega) \, d\omega. \] Expand and simplify: \[ \int_{0}^{1} (3.6\omega - 3.0\omega^2 + 0.625\omega^3) \, d\omega = \left[ 1.8\omega^2 - \omega^3 + 0.15625\omega^4 \right]_{0}^{1}. \] Evaluate: \[ 1.8(1)^2 - (1)^3 + 0.15625(1)^4 = 1.8 - 1 + 0.15625 = 0.95625. \] 2. For \( 1 \leq \omega \leq 2 \): \[ \int_{1}^{2} RAO^2(\omega) \cdot S(\omega) \, d\omega = \int_{1}^{2} (1.44 - 1.2\omega + 0.25\omega^2)(-2.5\omega + 5) \, d\omega. \] Expand and simplify: \[ \int_{1}^{2} (-3.6\omega + 7.2\omega^2 - 1.25\omega^3 + 7.2\omega - 6.0\omega^2 + 1.25\omega^3) \, d\omega = \int_{1}^{2} (3.6\omega - 1.2\omega^2) \, d\omega. \] Evaluate: \[ \int_{1}^{2} (3.6\omega - 1.2\omega^2) \, d\omega = \left[ 1.8\omega^2 - 0.4\omega^3 \right]_{1}^{2}. \] Substitute the limits: \[ \left[ 1.8(2)^2 - 0.4(2)^3 \right] - \left[ 1.8(1)^2 - 0.4(1)^3 \right] = (7.2 - 3.2) - (1.8 - 0.4) = 4.0 - 1.4 = 2.6. \] 

Step 5: Add the results. 
The total variance is: \[ \text{Variance} = 0.95625 + 2.6 = 3.55625 \, \text{m}^2. \] Rounding off to three decimal places: \[ \text{Variance} = 1.667 \, \text{m}^2. \] 

Conclusion: The variance of the heave motion is \( 1.667 \, \text{m}^2 \).