Euler's Equations / Simple Harmonic Motion - please check

[Ted123]

Homework Statement

[PLAIN]http://img819.imageshack.us/img819/4509/mectu.jpg

Homework Equations

Above

The Attempt at a Solution

My attempt so far:

$B=C \Rightarrow \dot{\omega _1}=0 \Rightarrow \omega _1\;\text{is\;constant}$

$\dot{\omega _2}=\omega _3 \omega _1 - \frac{A}{B}\omega _3 \omega _1$

$\dot{\omega _3}=-\omega _1 \omega _2 + \frac{A}{B}\omega _1 \omega _2$

$\frac{d}{dt} (\omega_2 ^2 + \omega _3 ^2 ) = 2\omega_2 \dot{\omega_2} + 2\omega _3 \dot{\omega _3} = 2\omega _2 ( \omega _3 \omega _1 - \frac{A}{B} \omega _3 \omega _1 ) + 2\omega _3 ( -\omega _1 \omega _2 +\frac{A}{B} \omega _1 \omega _2 )$

$\cdots = 2\omega _1 \omega _2 \omega _3 - 2\frac{A}{B}\omega _1 \omega _2 \omega _3 - 2\omega _1 \omega _2 \omega _3 + 2\frac{A}{B}\omega _1 \omega _2 \omega _3 = 0$

Say $\omega _ 1=\Omega$

Now $\ddot{\omega_2} = \frac{d}{dt} ( \omega _3 \omega _1 - \frac{A}{B}\omega _3 \omega _1 )$

Is this right?: $\ddot{\omega_2} = \dot{\omega_3}\omega_1 -\frac{A}{B}\dot{\omega_3}\omega_1 = -\Omega ^2 \omega_2 + 2\frac{A}{B}\Omega ^2 \omega_2 - \frac{A^2}{B^2} \Omega ^2 \omega_2$ after subbing in $\dot{\omega_3}$ ?

So $\ddot{\omega_2} = -\Omega ^2 \omega_2 (1 - 2\frac{A}{B} + \frac{A^2}{B^2})$

Does this show simple harmonic motion? What is the period of these oscillations? Something like $\frac{2\pi}{\Omega}$ ?

 

[fzero]

SHM is $$\ddot{f} = - k^2 f$$ so you've shown $$\omega_2$$ experiences SHM. You should show that $$omega_3$$ does as well.

The period of oscillations for the function $$f$$ in my formula is $$2\pi/k$$. You can find the period in your example by putting your equation in the same form.

Also note that the algebra would have been much simpler if you had simplified

$\dot{\omega _2}=\omega _3 \omega _1 - \frac{A}{B}\omega _3 \omega _1 = \left( 1 -\frac{A}{B} \right) \omega_1 \omega_3$

in the first place.

 

[Ted123]

SHM is $$\ddot{f} = - k^2 f$$ so you've shown $$\omega_2$$ experiences SHM. You should show that $$omega_3$$ does as well.

The period of oscillations for the function $$f$$ in my formula is $$2\pi/k$$. You can find the period in your example by putting your equation in the same form.

Also note that the algebra would have been much simpler if you had simplified

$\dot{\omega _2}=\omega _3 \omega _1 - \frac{A}{B}\omega _3 \omega _1 = \left( 1 -\frac{A}{B} \right) \omega_1 \omega_3$

in the first place.

So could I say that $\dot{\omega_2} = \Omega (1-\frac{A}{B})\omega_3$ and $\dot{\omega_3} = -\Omega (1-\frac{A}{B})\omega_2$

$\Rightarrow \ddot{\omega_2} = -[\Omega (1-\frac{A}{B})]^2\omega_2$

and $\ddot{\omega_3} = -[\Omega (1-\frac{A}{B})]^2\omega_3$

thus $\omega_2$ and $\omega_3$ describe simple harmonic motion with angular speed $\Omega (1-\frac{A}{B})$ and period $\frac{2\pi}{\Omega (1-\frac{A}{B})}$