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Damped Oscillator homework

  • #1
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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


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

Answers:

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##.
 

Answers and Replies

  • #2
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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
vela
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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
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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.
 
  • #5
vela
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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
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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.
 
  • #7
vela
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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.
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.

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.
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.
 

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