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}}
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}}
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