Heave motion of a point energy absorber in the form of a buoy due t incident wave of frequency $w$ and amplitude $a$ is described by
$(m+m_a)y'' + (c_h + c_e)y' +ky = F_i(t)$
$F_i(t)=aF(w)\cos(wt)$ is the hydrodynamic force acting on the buoy due to incident wave.
When the waves first arrive at the buoy there are transient parts to the solution which die down as time progresses until a steady state oscillation is established. The steady state oscillation of the buoy is described
$y(t)=Y\cos(wt+d)$
(i)Find the amplitude Y of the steady state solution.
So i have differentiated $y(t)=Y\cos(wt+d)$ to find $y''$ and $y'$ and I have substituted this into the given equation but I am not really sure what I need to do next nothing seems to cancel. Is this right? Do I need to substitute $F_i(t)$ for $aF(w)\cos(wt)$?
Thanks
$\endgroup$2 Answers
$\begingroup$The differential equation you want to solve is this
$$My'' + Cy' +ky = aF(\omega)\cos(\omega t)$$ where $M=m+m_a$, $C=c_h + c_e$ You're given that you're looking for steady-state solution
$$y(t)=Y\cos(\omega t+d),$$
so the question is what value of the amplitude ($Y$) and the phase ($d$) satisfies this equation. Let's take the alternate form of the solution, $$ y(t) = I\cos(\omega t) + Q\sin(\omega t), $$ which equals our earlier expression if we set $Y=\sqrt{I^2+Q^2}$ and $d=\arctan(Q/I)$. The derivatives are now $$ y'(t) = \omega Q \cos (\omega t)- \omega I \sin (\omega t) $$ $$ y''(t) = -\omega ^2(I \cos(\omega t) + Q \sin (\omega t)) $$ Now plugging in and collecting terms in sine and cosine we have $$ \cos (\omega t) \left(C Q \omega +I k-I M \omega ^2\right)+\sin (\omega t ) \left(-C I \omega +k Q-M Q \omega^2\right) = a F(\omega) \cos(\omega t) $$ From which we see that for the two sides to be equal, we must have $$ C Q \omega +I k-I M \omega ^2 = aF(\omega) $$ and $$ -C I \omega +k Q-M Q \omega^2 = 0 $$ The solution to that system is $$ I = -\frac{a F(\omega) \left(k-M \omega ^2\right)}{-C^2 \omega ^2-\left(k-M \omega ^2\right)^2} $$ $$ Q = \frac{a C F(\omega) \omega }{C^2 \omega ^2+k^2-2 k M \omega ^2+M^2 \omega ^4} $$ And the magnitude and phase $$ Y = \frac{aF(\omega)}{\sqrt{C^2\omega^2+(k-M\omega^2)^2}} $$ $$ d = \arctan\left(\frac{C \omega }{k-M \omega ^2}\right) $$ You can directly verify that the $I$-$Q$ form solves the original differential equation. Doing so with the $Y$-$d$ form directly requires tedious trig identities.
$\endgroup$ $\begingroup$That (finding $y'$ and $y''$ and substituting into the original equation) is a good first step, assuming you did it correctly. You still have some work to do to finish the problem. The next step would be to also substitute the expression for $F_i(t)$ (i.e. $aF(\omega)\cos(\omega t)$) into the original equation. Next, use basic trig identities to rewrite the equation in terms of $\cos(\omega t)$ and $\sin(\omega t)$ (rather than in terms of $\cos(\omega t +d)$ and $\sin(\omega t +d)$) and then group the like terms so that you have an equation of the form $C_1 \cos(\omega t)+C_2\sin(\omega t)=aF(\omega)\cos(\omega t)$, where $C_1$ and $C_2$ are constant coefficients. Since there is no $\sin$ term on the right side of the original equation, the coefficient in front of the $\sin(\omega t)$ term on the left side must be $0$. And, the coefficient in front of the $\cos(\omega t)$ term on the left side of the equation must equal the coefficient in front of the $\cos(\omega t)$ term on the right side of the equation. So now you have two equations and two unknowns, namely $Y$ and $d$. You can solve the system, and then you'll have your answer for the amplitude, $\lvert Y\rvert$.
$\endgroup$
