Integration ##\ddot\phi = -\omega^2\phi##

[happyparticle]

Homework Statement

integrate angular acceleration

Relevant Equations

## \phi(t) = A sin(\omega t) + B cos(\omega t)##
##\ddot\phi = -\omega^2\phi##

Hi,
I'm wondering how can I get ## \phi(t) = A sin(\omega t) + B cos(\omega t)##
I know I have to integrate 2 times ##\ddot\phi = -\omega^2\phi##. However, I don't have any more explanation in my book.

I know A and B are the constants of integration.

 

[haruspex]

Homework Statement:: integrate angular acceleration
Relevant Equations:: ## \phi(t) = A sin(\omega t) + B cos(\omega t)##
##\ddot\phi = -\omega^2\phi##

Hi,
I'm wondering how can I get ## \phi(t) = A sin(\omega t) + B cos(\omega t)##
I know I have to integrate 2 times ##\ddot\phi = -\omega^2\phi##. However, I don't have any more explanation in my book.

I know A and B are the constants of integration.

One way is simply to run it backwards: differentiate the answer twice.
Another is to multiply through by ##\dot\phi##. That makes all terms integrable, but then you get an equation for ##\dot\phi## equal to a square root. So the next step is to make a trig substitution.

 

[Steve4Physics]

Hi.

Solving differential equations is more of an art than a science. They key here is to recognise that we want a function which, when differentiated twice, gives itself multiplied by a constant. This is quite a common problem.

One approach is to recognise that a solution could be of the form [itex]\phi = e^{rt}[/itex]

Taking this approach gives solutions as complex exponentials. These then reduce to sin and cos. Here is a video which shows this: Another approach is to notice [itex] \phi = sin(\omega t + \alpha)[/itex] has the required behavior (when differentiated twice). Then use the fact that [itex]sin(\omega t + \alpha) [/itex] can be expressed in the form [itex]Asin(\omega t)+ Bcos(\omega t)[/itex].

 

[rude man]

Another approach is via Laplace transform:
## \ddot\phi = -\omega^2\phi ##
## s^2\Phi - s\phi(0) - \dot\phi(0) + \omega^2 \Phi = 0 ##
where ##\Phi(s) ## is the transform of ## \phi(t) ##.

$$ \Phi = \frac {s\phi(0) + \dot\phi(0)} {s^2 + \omega^2} $$
Let ## A = phi(0) ##
## B = \dot\phi(0)/\omega ##
$$ \Phi = \frac {As + B\omega} {s^2+\omega^2} $$
and from tables,
## \phi(t) =A cos(\omega t) + B sin(\omega t) ##

If this is unfamiliar now, if you're going into electrical eng. you will soon learn it. The only way to solve linear ODE's with constant coefficients! No "guessing" and initial conditions included!

 

[PeroK]

Or, you could use a power series expansion for ##\phi##.