Choosing a Complex Trial Solution for a Forced Vibration Problem and Expressing The Driving Force in Complex Terms

Question

When solving forced vibration problems I would always choose a trial solution for the particular (stead-state) case in the form $x(t) = A \cos{\omega t} + B \sin{\omega t} $, but reading some books I got to know the complex form for the trial solution. At first I found it more intuitive, but after trying to use it I got some questions that I could not find answers to.

Let us have a equation of motion of the form

$$ m \ddot{x} + c \dot{x} + k x = F_0 \sin{\omega t}$$

What is the difference of choosing a trial solution in the complex form $z = A~e^{i(wt-\phi)}$ and in the $z = A~e^{i(wt+\phi)}$. What should I expect from this change in the phase's sign? What is a more 'visual' interpretation of this change?

Also, when applying these complex forms, one usually has to change the harmonic excitation into a complex term too. The problem is: it seems that regardless of the excitation being a sine or cosine function, books always make

$$F_0 \sin{\omega t} \quad \text{or} \quad F_0 \cos{\omega t} \quad = \quad F_0~e^{i\omega t}$$

and, at the end, they just choose the imaginary or real part of the solution. Does it mean that $A$ and $\phi$ will not change despite we have a sine or cosine excitation?

Answer 1

In this approach, the idea is to test a solution. It might not work, but with experience (and mathematical knowledge), you usually know where you're going of course.

When choosing a solution like $z(t)=A\cos(\omega t)+B\sin(\omega t)$, you're choosing to look for a periodic, harmonic solution. Take note that $\omega$ is assumed to be positive at this point. It'll be important later.

That sort of solution lives in a 2-dimensional space, as can be seen with the presence of two free paramteters $A$ and $B$. In other words, you're trying to write the solution in the basis $(\cos(\omega t),\sin(\omega t))$.

As in any vector space, you're free to take linear combinations of elements, for example to change to another basis. Using for example: $$\cos(\omega t)=\frac{e^{i\omega t}+e^{-i\omega t}}{2} \quad\text{and}\quad \sin(\omega t)=\frac{e^{i\omega t}-e^{-i\omega t}}{2i}$$ you can rewrite the solution in the basis $(e^{i\omega t},e^{-i\omega t})$: $$z(t)=Ce^{i\omega t}+De^{-i\omega t}$$ Restarting from the first form of the solution, you can also rewrite as $z(t)=E\cos(\omega t+\varphi)$ or $z(t)=E\cos(\omega t-\varphi)$ with a bit of trigonometry.

When an author says that they're testing a solution like $e^{i\omega t}$, as long as $\omega$ is free to be positive or negative, they're going to get both terms written $e^{i\omega t}$ or $e^{-i\omega t}$ above. So they're also going to get both terms written as $\cos(\omega t)$ or $\sin(\omega t)$ above.

As for phase $\phi$, it's a phase shift between the excitation and the response. Depending on the sign you chose in the solution ($\omega t+\varphi$ or $\omega t-\varphi$) it'll be a delay or an advance in time.

Answer 2

How you obtain the particular solution with complex ansatz

lets start with this equation:

$$ m \ddot{x} + c \dot{x} + k x = F_0 \cos{\omega t}$$

or

$$ \ddot{x} + \delta \dot{x} + \omega_0^2 x = \frac{F_0}{m} \cos{\omega t}\tag 1$$

we are looking for the particular solution $~x_p(t)~$

with $$ \cos(\omega\,t)=\Re\left(e^{i\,\omega\,t}\right)$$

$\Rightarrow~$

$$\ddot{x}_p + \delta \dot{x}_p + \omega_0^2 x_p =\frac{F_0}{m}\,\,e^{i\,\omega\,t}\tag 2$$

to solve Eq. (2) you make the ansatz $~x_p(t)=A\,e^{i\,\omega\,t}$

thus Eq. (2)

$${{\rm e}^{i\omega\,t}} \left( -{\omega}^{2}+{\omega_{{0}}}^{2}+i\delta \,\omega \right) A={\frac {F_0{{\rm e}^{i\omega\,t}}}{{m}^{2}}}\quad\Rightarrow\\ A={\frac {F_{{0}}}{{m}^{2} \left( -{\omega}^{2}+{\omega_{{0}}}^{2}+i \delta\,\omega \right) }} $$

and the solution $$x_p(t)=\Re(A\,e^{i\,\omega\,t})={\frac {F_{{0}} \left( -{\omega}^{2}+{\omega_{{0}}}^{2} \right) \cos \left( \omega\,t \right) }{{m}^{2} \left( \left( -{\omega}^{2}+{ \omega_{{0}}}^{2} \right) ^{2}+{\delta}^{2}{\omega}^{2} \right) }}+{ \frac {F_{{0}}\delta\,\omega\,\sin \left( \omega\,t \right) }{{m}^{2} \left( \left( -{\omega}^{2}+{\omega_{{0}}}^{2} \right) ^{2}+{\delta} ^{2}{\omega}^{2} \right) }} $$

and if the RHS is $~\frac {F_0}{M}\,\sin(\omega\,t)~$ then

with $$ \sin(\omega\,t)=\Re\left(-i\,e^{i\,\omega\,t}\right)$$

you obtain the particular solution

$$x_p(t)=\Re(-A\,i\,e^{i\,\omega\,t})= -{\frac {F_{{0}}\delta\,\omega\,\cos \left( \omega\,t \right) }{{m}^{2 } \left( \left( -{\omega}^{2}+{\omega_{{0}}}^{2} \right) ^{2}+{\delta }^{2}{\omega}^{2} \right) }}+{\frac {F_{{0}} \left( -{\omega}^{2}+{ \omega_{{0}}}^{2} \right) \sin \left( \omega\,t \right) }{{m}^{2} \left( \left( -{\omega}^{2}+{\omega_{{0}}}^{2} \right) ^{2}+{\delta} ^{2}{\omega}^{2} \right) }} $$