PDA

View Full Version : angular frequency - how is it derived?


UrbanXrisis
Apr10-05, 07:27 AM
How is the formula for angular derived?

\omega=\sqrt{\frac{g}{L}}

my book has these equations:

F= -mg sin\theta = m \frac{d^2s}{dt^2}

\frac{d^2 \theta}{dt^2}=-\frac{g}{L}sin\theta

\frac{d^2 \theta}{dt^2}=-\frac{g}{L}\theta

\omega=\sqrt{\frac{g}{L}}

what exactly is m \frac{d^2s}{dt^2} and \frac{d^2 \theta}{dt^2}

dextercioby
Apr10-05, 09:12 AM
\mbox{s} [/itex] is arc length...Because the movement is on a circle

[tex]s=L\theta

,where \theta is the angle at the center of the circle...

Daniel.

Galileo
Apr10-05, 09:16 AM
s is the position of the particle, so the second derivative wtr time denotes the acceleration 'a' of the particle. (Hence F=ma in the first equation)

Solve the D.E. and check what that you get a periodic function. What is its angular frequency?

dextercioby
Apr10-05, 09:29 AM
Instead of applying Newton's second law for the translation movement \frac{d\vec{\mbox{p}}}{dt}=\sum_{k} \vec{\mbox{F}}_{k} ,try to apply it for the rotation movement \frac{d\vec{\mbox{L}}}{dt}=\sum_{k} \vec{\mbox{M}}_{k}


Daniel.

UrbanXrisis
Apr10-05, 04:01 PM
okay, I understand up to here:
\frac{d^2 \theta}{dt^2}=-\frac{g}{L}\theta


I think the next step is this...
\sqrt{\frac{d^2 \theta}{dt^2}}=\sqrt{-\frac{g}{L}\theta}
\sqrt{\frac{d^2 \theta}{dt^2}}=\frac{d \theta}{dt}= \omega
\omega = \sqrt{\frac{g}{L}

but I dont think I'm correct

dextercioby
Apr10-05, 04:08 PM
Of course not.

It's a definition

\frac{g}{l}=:\omega_{0}^{2}

That's all to it.

Daniel.

UrbanXrisis
Apr10-05, 04:48 PM
but how do I get \frac{d^2 \theta}{dt^2}=-\frac{g}{L}\theta to go to \omega=\sqrt{\frac{g}{L}}

jtbell
Apr10-05, 06:24 PM
How much have you studied about solving differential equations? Here in the USA, very few people study differential equations at the K-12 level.

It turns out that a general solution of the differential equation

\frac {d^2 \theta} {d t^2} = - \frac {g}{l} \theta

is

\theta = A \sin \left(\sqrt {\frac {g}{l}} t + \theta_0\right)

where A and \theta_0 are arbitrary constants. You can verify this by working out the second derivative and plugging it back into the differential equation. A general form of a sinusoidal function is

x = A \sin (\omega t + \theta_0)[/itex]

where \omega is the angular frequency. Matching up the preceding two equations gives you

[tex]\omega = \sqrt {\frac {g}{l}}