Finding the equation of motion based on the diagram and its transfer function

classical-mechanicscontrol theorylaplace transformordinary differential equations

I have a diagram that looks like below:

enter image description here

Description: By applying force p(t), point A is moving with displacement $y = asin𝜔t$. $k_1$ and $k_2$ stand for the coefficients of the corresponding spring, and C stands for the coefficient of the damper. The friction between the surface and the object can be ignored.

The questions have three subquestions:

(1) Find the equation of motion

(2) Find the transfer function with $P(s)$ as output and $Y(s)$ as input

(3) Find the amplitude and phase of $p(t)$ when reach steady state

Now, I would like to find the equation of motion for the diagram and find its transfer function based on the equation as $G(s) =\dfrac{P(s)}{Y(s)}$.

As far as I know, the equation could be like this:

\begin{equation}
m\dfrac{d^2x}{dt^2} + k_1x(t) + k_2x(t) + (\dfrac{1}{k_1} + \dfrac{1}{k_2})x(t) – C(\dfrac{dy}{dt} – \dfrac{dx}{dt}) = p(t)
\end{equation}

The transfer function I am asked to find is to take $Y(s)$ as input and $P(s)$ as output and this seems to contradict what I know about this kind of equation of motion, and I am also confused with the $y = asin𝜔t$ at point A.

I have tried to write out the equation I believe to be, but it does not make sense if I want to find the transfer function with the defined input and output. Actually, I think the $y = asin𝜔t$ should also be listed in the equation, but I have no clue or idea about how should I put it into it.

Can anyone help me solve this equation of motion or point out what I missed?

Thank you for your help and advice.

(Note: To me, based on the instruction, I believe that p(t) should be input, however, the question demands that the p(t) be output and y(t) be input to find the transfer function of $G(s) = \dfrac{P(s)}{Y(s)}$ which looks a bit strange if I follow from the description of the question.)

Best Answer

We have

$$ \cases{ (\dot y-\dot x)C = p\\ m \ddot x = -k x + (\dot y-\dot x)C } $$

with $k = \frac{(k_1+k_2)^2+k_1k_2}{k_1+k_2}$

then

$$ \cases{ s(\hat y(s)-\hat x(s))C = \hat p(s)\\ \left(ms^2+Cs+k\right)\hat x(s) = s C \hat y(s) } $$

now as

$$ \hat x(s) = \frac{s C}{ms^2+Cs+k}\hat y(s) $$

we have

$$ \frac{\hat p(s)}{\hat y(s)} = s\left(1-\frac{s C}{ms^2+Cs+k}\right)C $$