# Damped Oscillator homework

## Homework Statement

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.

## Homework 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##

## 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##.

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##
$$x(t)=e^{-\gamma t}(c_{1}+c_{2}t)+\frac{mg}{k}$$

vela
Staff Emeritus
Homework Helper
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.

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
$$x(t) = [x_0 + (v_0+\gamma x_0)t] e^{-\gamma t}.$$

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.

vela
Staff Emeritus
Homework Helper
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.

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

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.

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

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.

vela
Staff Emeritus
Homework Helper
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.