Hooke's law and angular freq.

misogynisticfeminist

hmm ok, i was watching the MIT opencourseware video on oscillations and there was a part where it was mentioned that,

the diff. eq. [tex] x''+ \frac {k}{m} x = 0 [/tex] has solution [tex] x= x_0 cos (\omega t + \phi) [/tex] if and only if [tex]\omega= \sqrt{\frac {k}{m}} [/tex]

how do i show that omega is the sqaureoot of k over m? thanks alot.

Tide

Differentiate the solution twice with respect to time and substitute it back into the differential equation - then tell us what you discovered! :)

misogynisticfeminist

ohhh, i see, it can be gotten by verifying the solution. I thought we needed to do something to hooke's law, but verifying the solution works great too.

: )

edit: a tinge of doubt crosses my mind though. When we solve the original differential equation, we get the solution in terms of k, m and x and no omega. While verifying the solution works when the solution is given, and we see that [tex]\omega= \sqrt{\frac {k}{m}} [/tex]. How do we know that [tex]\omega= \sqrt{\frac {k}{m}} [/tex] when we are solving it?

Tide

If you just substitute [itex]x = x_0 \cos(\omega t - \phi)[/itex] into the differential equation the required value of [itex]\omega[/itex] will jump out at you!

SpaceTiger

In this case, [tex]\omega[/tex] is referring to the angular frequency, which is the multiplicative factor in front of the independent variable in the sine or cosine function. It's just a standard substitution that they probably just didn't bother to define.

DaTario

try the following:

in your original ODE, consider [itex] \omega ^2[\itex] just as a mathematically sound way to express the positiveness of the factor [itex] \frac{k}{m} [\itex]. After all the calculations you will realize that this choice has proven itself useful and physically meaningful.