0
$\begingroup$

Consider that we have the following forced vibration with an input frequency $ω(t)$ variable in time. $$m\ddot{x}+c\dot{x}+kx = F_0 \sin{(\omega(t) t)}$$

Assuming that the solution must be a harmonic form but with lag some phase angle $ψ$ we consider a solution to be:

$$x(t)=X_0 \sin{(\omega(t) t-ψ)}$$

Firstly, Is it true that the lag phase $ψ$ is also variable with time and so the particular solution must be as follows? $$x(t)=X_0 \sin{(\omega(t) t-ψ(t))}$$

Secondly, If the first question holds, how is the general solution formulated because the calculations gets really complicated?

$\endgroup$
1
$\begingroup$

$\psi$ is frequency-dependent so if frequency is time-dependent, $\psi$ will be time dependent. That's assuming that $\omega$ is varying slowly enough [$|\frac{d\omega}{dt}|\ll\omega^2$] for the solution you give to be approximately correct. In that case you can treat $\psi$ as a constant when solving the equation, and put its dependency on $t$ in afterwards! But if $\omega$ can vary as rapidly as you like, then the solution will not be of the form you have quoted anyway!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.