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...