- #1

kelly0303

- 580

- 33

$$\ddot{x}+\gamma\dot{x}+\omega_0^2x=\frac{F}{m}\sin{\omega_dt}$$

where ##\gamma## is the damping, ##\omega_0## is the resonant frequency, ##m## is the mass of the particle, ##F## is the applied force and ##\omega_d## is the frequency of the applying force. Assuming ##F=0## (and ##\gamma<<\omega_0##, which is my case) the homogeneous solution is:

$$x(t)=Ae^{-\gamma t/2}\sin{\omega_0t+\phi}$$

If we include the force, the final solution is:

$$x(t) = \frac{F}{m}\left[\frac{\omega_0^2-\omega_d^2}{(\omega_0^2-\omega_d^2)^2+(\gamma\omega_d)^2}\sin{\omega_dt} + \frac{\gamma\omega_d}{(\omega_0^2-\omega_d^2)^2+(\gamma\omega_d)^2}\cos{\omega_dt} \right]$$

So the final solution is the sum of the 2 expressions above. In my situation, in the ideal case, I want to start with a particle at the very center and with zero velocity and by applying this force I want to take it to a fixed amplitude, ##A_d## then let it evolve freely. In my case I also have that ##\omega_0 = \omega_d## and ##\gamma## is basically zero. Thus I can assume that the homogeneous solution is identically zero, while for the driven part (for the right value of ##F_d##) I can ignore the second term, given that ##\gamma \sim 0## and end up with:

$$x(t) = A_d\sin{\omega_dt} =A_d\sin{\omega_0t}$$

Thus I will need to simply apply this force for ##T_0 = \frac{\pi}{2\omega_0}##.

**(Is this right?)**

Now, let's assume that instead of starting with zero velocity and position, the particle has a random initial position and velocity. I know that the possible amplitude of this motion, ##A##, is much smaller than ##A_d## i.e. ##A<<A_d##. What I want to know is the uncertainty in the final amplitude in this case. I assume that in this case, the homogeneous solution can be in general parameterized as:

$$x(t)=A\sin{(\omega_0t+\phi)}$$

and under the same assumption as above, the general solution is given by:

$$x(t)=A\sin{(\omega_0t+\phi)}+A_d\sin{\omega_0t}$$

**(Is this right?)**

The applied force will be for the same amount as before (as I don't know the initial parameters), so I end up with:

$$x(T_0)=A\sin{(\pi/2+\phi)}+A_d$$

$$\dot{x}(T_0)=A\omega_0\cos{(\pi/2+\phi)}$$

For ##\phi=0##, the uncertainty in the amplitude would simply be ##x(T_0)-A_d = A##. In general, the new amplitude would be defined as the point where the initial velocity is zero, which means ##\pi/2+\phi+\omega_0t = 0## or ##\pi/2+\phi+\omega_0t = 3\pi/2## (depending on the sign of ##\phi##). Which gives an initial position of ##\pm A + A_d##, so even in the general case, the uncertainty in the amplitude is given by ##A##.

**(Is this right?)**

Thank you!