How Does Hooke's Law Relate to Angular Frequency?

[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. $$x''+ \frac {k}{m} x = 0$$ has solution $$x= x_0 cos (\omega t + \phi)$$ if and only if $$\omega= \sqrt{\frac {k}{m}}$$

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

 

[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 $$\omega= \sqrt{\frac {k}{m}}$$. How do we know that $$\omega= \sqrt{\frac {k}{m}}$$ when we are solving it?

 

[Tide]

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

 

[SpaceTiger]

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 $$\omega= \sqrt{\frac {k}{m}}$$. How do we know that $$\omega= \sqrt{\frac {k}{m}}$$ when we are solving it?

In this case, $$\omega$$ 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 $\frac{k}{m} [\itex]. After all the calculations you will realize that this choice has proven itself useful and physically meaningful.$[/itex]