# Damped Oscillator homework

1. Sep 15, 2013

### jbrussell93

1. The problem statement, all variables and given/known data
A mass of 1000 kg drops from a height of 10 m on a platform of negligible mass. It is desired to design a spring and dashpot on which to mount the platform so that the platform will settle to a new equilibrium position 0.2 m below its original position as quickly as possible after the impact without overshooting.

a)Find the spring constant k and the damping constant b of the dashpot.
b)Find the time required for the platform to settle within 1 mm of its final position.

2. Relevant equations
$$mg=kx$$
$$m\ddot{x}+b\dot{x}+kx=mg$$
$$\omega=\sqrt{\frac{k}{m}}$$
$$\gamma=\frac{b}{2m}$$

a) $k=49000kgm/s^{2}$, $b=70700 kgm/s$

b)$0.076sec$

3. The attempt at a solution

I am having trouble finding the value of $b$. I assumed that since the question asks for the quickest amount of time without overshooting, this would be a critical damping problem with $\omega=\gamma$ but when I solve there is overshooting. The only way that I am able to get the answer for part b) is to take away from forcing function $mg$.

2. Sep 15, 2013

### jbrussell93

Additionally, when $\gamma>\omega$ my solution is of the form
$$x(t)=e^{-\gamma t}(c_{1}e^{\sqrt{\gamma^{2}-\omega^{2}}t}+c_{2}e^{-\sqrt{\gamma^{2}-\omega^{2}}t})+\frac{mg}{k}$$

Instead of the usual form for critical damping when $\omega=\gamma$
$$x(t)=e^{-\gamma t}(c_{1}+c_{2}t)+\frac{mg}{k}$$

3. Sep 15, 2013

### vela

Staff Emeritus
I don't see how they got their value of $b$ either.

Given the initial conditions and parameters, it doesn't look like there's a way to not have the platform overshoot the new equilibrium position. For a critically damped oscillator, you have
$$x(t) = [x_0 + (v_0+\gamma x_0)t] e^{-\gamma t}.$$ With the assumption that the upward direction is positive, to have no overshoot, the coefficient of the linear term has to be positive. If not, $t$ will eventually get large enough so that x(t) becomes negative, which means that the mass has overshot the new equilibrium point.

4. Sep 15, 2013

### jbrussell93

That is what I ended up with for the situation with $F(t)=0$ but wouldn't $F(t)=mg$? Assuming downward positive and $x=0$ at the equilibrium position, I took

$x_0=-.2 m$
$v_0=14 m/s$
$\gamma=70 s^{-1}$

which causes $(v_0+\gamma x_0)t$ to equal 0. There would be no overshooting in this case, just a smooth decay to zero. By doing this, I got the correct answer for part b) but I still believe that there should be a forcing function somewhere in there.

5. Sep 15, 2013

### vela

Staff Emeritus
F(t)=mg only serves to shift the equilibrium point downward.

How'd you get γ=70 1/s? I get that it's equal to 7 1/s.

6. Sep 15, 2013

### jbrussell93

Hmm this explains why I can't figure out how to get the mass to settle to x=0 with the forcing function included... I asked my professor about this one and he told me that there should be a forcing function but I don't see how because all other terms go to zero leaving only that constant. It should be zero.

By solving $$(v_0+\gamma x_0)t=0$$ but that is partly why I'm so confused. I was under the impression that in order to use that solution x(t), $\omega$ had to equal $\gamma$ in which case $\gamma=7$ and the mass overshoots.

7. Sep 15, 2013

### vela

Staff Emeritus
The solution with the forcing function is $x(t) = mg/k + (c_1+c_2 t)e^{-\gamma t}$. As $t \to \infty$, $x(t) \to mg/k$. What you can do is define $x' = x - mg/k$, and x'(t) will satisfy the homogeneous equation. The solution I wrote above is really for x'(t), not x(t).

With your sign convention, if you use x(t), the initial conditions are x(0)=0 and $\dot{x}(0) = 14\text{ m/s}$, and the equilibrium position is at x=0.2 m. If you use x'(t), the initial conditions are x'(0)=-0.2 m and $\dot{x}'(0) = 14\text{ m/s}$, and equilibrium is at x'=0.

You're right. The numbers given in the problem don't work out, so I wouldn't worry about it if I were you. Whoever wrote the problem made an error, perhaps using the wrong value of $\gamma$ when solving it.