Question:medium
A ship undergoing harmonic oscillation has an uncoupled heave motion, \(\eta_3(t)\), governed by the following equation
\[ (M + A_{33})\,\ddot{\eta}_3(t) + B_{33}\,\dot{\eta}_3(t) + K_{33}\eta_3(t) = f_3(t) \]
If \(F_3(\omega)\) is the Fourier transform of \(f_3(t)\), then the response in the frequency domain can be written as ______.
A ship undergoing harmonic oscillation has an uncoupled heave motion, \(\eta_3(t)\), governed by the following equation
\[ (M + A_{33})\,\ddot{\eta}_3(t) + B_{33}\,\dot{\eta}_3(t) + K_{33}\eta_3(t) = f_3(t) \]
If \(F_3(\omega)\) is the Fourier transform of \(f_3(t)\), then the response in the frequency domain can be written as ______.
\[ (M + A_{33})\,\ddot{\eta}_3(t) + B_{33}\,\dot{\eta}_3(t) + K_{33}\eta_3(t) = f_3(t) \]
If \(F_3(\omega)\) is the Fourier transform of \(f_3(t)\), then the response in the frequency domain can be written as ______.
Show Hint
Turn each time derivative into a factor of i times omega using the standard Fourier transform property.
Updated On: Jul 28, 2026
- \( \dfrac{F_3(\omega)}{-\omega^2(M + A_{33}) + i\omega B_{33} + K_{33}} \)
- \( \dfrac{iF_3(\omega)}{-\omega^2(M + A_{33}) + i\omega B_{33} + K_{33}} \)
- \( \dfrac{F_3(\omega)}{\omega^2(M + A_{33}) - i\omega B_{33} + K_{33}} \)
- \( \dfrac{F_3(\omega)}{\omega^2(M + A_{33}) + \omega B_{33} + K_{33}} \)
Show Solution
The Correct Option is A
Solution and Explanation
Step 1: Use the Fourier transform property for derivatives.
Taking the Fourier transform is linear, and it turns a time derivative into multiplication by $i\omega$ in the frequency domain. So if $H_3(\omega)$ is the transform of $\eta_3(t)$, then $\dot{\eta}_3(t)$ transforms to $i\omega H_3(\omega)$, and $\ddot{\eta}_3(t)$ transforms to $(i\omega)^2 H_3(\omega) = -\omega^2 H_3(\omega)$.
Step 2: Transform the whole equation term by term.
$(M+A_{33})\ddot{\eta}_3(t)$ becomes $-\omega^2(M+A_{33})H_3(\omega)$, $B_{33}\dot{\eta}_3(t)$ becomes $i\omega B_{33} H_3(\omega)$, and $K_{33}\eta_3(t)$ becomes $K_{33}H_3(\omega)$. The right side $f_3(t)$ becomes $F_3(\omega)$ directly, by definition.
Step 3: Combine and solve for $H_3(\omega)$.
Adding the transformed terms and setting them equal to $F_3(\omega)$ gives $H_3(\omega)\left[-\omega^2(M+A_{33}) + i\omega B_{33} + K_{33}\right] = F_3(\omega)$, the same bracket as a mass, damping and stiffness combination evaluated at frequency $\omega$.
Final Answer:
Solve for the frequency response directly. \[ \boxed{H_3(\omega) = \dfrac{F_3(\omega)}{-\omega^2(M + A_{33}) + i\omega B_{33} + K_{33}}} \]
Taking the Fourier transform is linear, and it turns a time derivative into multiplication by $i\omega$ in the frequency domain. So if $H_3(\omega)$ is the transform of $\eta_3(t)$, then $\dot{\eta}_3(t)$ transforms to $i\omega H_3(\omega)$, and $\ddot{\eta}_3(t)$ transforms to $(i\omega)^2 H_3(\omega) = -\omega^2 H_3(\omega)$.
Step 2: Transform the whole equation term by term.
$(M+A_{33})\ddot{\eta}_3(t)$ becomes $-\omega^2(M+A_{33})H_3(\omega)$, $B_{33}\dot{\eta}_3(t)$ becomes $i\omega B_{33} H_3(\omega)$, and $K_{33}\eta_3(t)$ becomes $K_{33}H_3(\omega)$. The right side $f_3(t)$ becomes $F_3(\omega)$ directly, by definition.
Step 3: Combine and solve for $H_3(\omega)$.
Adding the transformed terms and setting them equal to $F_3(\omega)$ gives $H_3(\omega)\left[-\omega^2(M+A_{33}) + i\omega B_{33} + K_{33}\right] = F_3(\omega)$, the same bracket as a mass, damping and stiffness combination evaluated at frequency $\omega$.
Final Answer:
Solve for the frequency response directly. \[ \boxed{H_3(\omega) = \dfrac{F_3(\omega)}{-\omega^2(M + A_{33}) + i\omega B_{33} + K_{33}}} \]
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