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

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

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

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

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})}