Finding Displacements for forced oscillations

[Jacobpm64]

Homework Statement

A particle of mass $$m$$ is at rest at the end of a spring (force constant $$= k$$) hanging from a fixed support. At $$t = 0$$, a constant downward force $$F$$ is applied to the mass and acts for a time $$t_0$$. Show that, after the force is removed, the displacement of the mass from its equilibrium position ($$x = x_0$$, where $$x$$ is down) is

$$x - x_0 = \frac{F}{k}[cos \omega_0(t-t_0) - cos \omega_0 t]$$​

where $$\omega_0^2 = \frac{k}{m}$$.

Homework Equations

Newton's 2nd law $$F = ma$$ and knowing how to solve second order linear ODEs.

The Attempt at a Solution

Okay, so I have no idea if what I'm doing is right, but I've been working on this (trying different things) for about 6 hours now.. It's painful. So, any assistance would be appreciated.

First, I split this thing into two parts.. one after the force is removed and one while the force is applied.

After force is removed: ($$t \geq t_0$$ and $$t = 0$$)
$$mg - kx = m \ddot{x}$$
$$\ddot{x} + \omega_0^2 x = g$$
We know $$x(t) = Acos(\omega_0 t - \phi ) + \frac{g}{\omega_0^2}$$.
We also know that at $$t = 0, x = x_0,$$ so:
$$x_0 = Acos\phi + \frac{g}{\omega_0^2}$$
Now, the displacement from the equilibrium is:
$$x - x_0 = Acos(\omega_0 t - \phi ) - Acos\phi$$
Simplifying, $$x - x_0 = A[cos(\omega_0 t - \phi ) - cos\phi]$$

While the force is applied: ($$0 < t \leq t_0$$)
$$F + mg - kx = m \ddot{x}$$
$$\ddot{x} + \omega_0^2 x = \frac{F + mg}{m}$$
So, $$x_2 (t) = Acos(\omega_0 t - \phi) + \frac{F+mg}{m\omega_0^2}$$



Now, at $$t = t_0$$, we know $$x = x_2$$, so,
$$Acos(\omega_0 t_0 - \phi ) + \frac{g}{w_0^2} = Acos(\omega_0 t_0 - \phi) + \frac{F+mg}{m\omega_0^2}$$
But this gives me $$F = 0$$ which is of course true, because $$t = t_0$$ is the time when the force is removed. I just do not know where to go and if this is helpful at all. Eventually, i want to have $$A = \mbox{something}$$ so that i can plug that into my equation for displacement from the equilibrium $$x - x_0$$ above.

What am I doing wrong? What should I do? Where should I go?

 

[gabbagabbahey]

If the force is applied at t=0, for a time $t_0$, doesn't that mean

$$F+mg-kx=m\ddot{x}, \;\;\;\;\; 0\leq t\leq t_0$$

$$mg-kx=m\ddot{x}, \;\;\;\;\; t> t_0$$ so that your equilibrium position is actually $x_0=A\cos(\phi) + \frac{F+mg}{m\omega_0^2}$?

 

[Jacobpm64]

okay, so, using your advice.. I have, as you said..

$$x_0 = A\cos(\phi) + \frac{F+mg}{m\omega_0^2}$$

Therefore, our displacement from equilibrium is:

$$x - x_0 = A[cos(\omega_0 t - \phi) - cos\phi] - \frac{F}{m\omega_0^2}$$

Still though, when we have time $$t = t_0$$, $$x = x_2$$. This still cancels the $$Acos(\omega_0 t_0 - \phi)$$ from both sides and leaves $$F = 0$$. Again, I don't know where to go from here to clean up the expression for $$x - x_0$$.

 

[gabbagabbahey]

I think you are confusing yourself by using the same variables to represent the amplitude and phase of the oscillation during both intervals...

During the interval $0\leq t\leq t_0$ you have:

$$x (t) = A_1\cos(\omega_0 t - \phi_1) + \frac{F+mg}{m\omega_0^2}$$

and during the interval $t\geq t_0$, you have:

$$x (t) = A_2\cos(\omega_0 t - \phi_2) + \frac{g}{\omega_0^2}$$

Use the fact that $x(0)=x_0$ and $x(t)$ and $x('t)$ must be continuous at $t_0$...