Considering the following motion equation :

\begin{equation} \ddot x + \frac{a^2 b^2}{c^2} x = -V \frac{a b}{c^2} \end{equation}

where $a$, $b$, $c$ and $V$ are all constant. One can identify the period as being

\begin{equation} \omega = \frac{a b}{c} \end{equation}

so that the motion equation becomes

\begin{equation} \ddot x + \omega^2 x = \frac{-V \omega}{c} \end{equation}

I understand that if the motion equation was $\ddot x + \omega^2 x = 0$ instead, the general solution would simply be $A\cos(\omega t) + B\sin(\omega t)$ where $A$ and $B$ would be identified using the motion initial conditions.

But how does this general solution changes in the presented case where the right hand side is non-null?

  • $\begingroup$ I think you should ask this question in math SE. $\endgroup$ – lucas Jun 29 '16 at 4:58

To expand on Andrei's answer a bit, start with your equation:

$$ \ddot x + \omega^2 x = C $$

and rewrite it as:

$$ \ddot x + \omega^2 \left(x - \frac{C}{\omega^2}\right) = 0 $$

Then define a new variable $y$ by:

$$ y = x - \frac{C}{\omega^2} $$

and differentiate twice to get:

$$ \ddot{y} = \ddot{x} $$

Finally substitute into your original equation to get:

$$ \ddot y + \omega^2 y = 0 $$

And this is just the usual SHO equation with the solution:

$$ y = A\cos(\omega t) + B\sin(\omega t) $$

The last step is simply to substitute for $y$ to get:

$$ x - \frac{C}{\omega^2} = A\cos(\omega t) + B\sin(\omega t) $$


$$ x = A\cos(\omega t) + B\sin(\omega t) + \frac{C}{\omega^2} $$

which is the equation that Andrei finished with.


Just add a constant say $C$ to your solution. $x=A\cos(\omega t)+B\sin(\omega t)+C$ Taking the second derivative, the term with $C$ will be 0. But you still have $\omega^2C=-V\omega/c$ so $C=-V/(\omega c)$. Therefore $x=A\cos(\omega t)+B\sin(\omega t)-V/(\omega c)$


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