1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Damped Oscillator homework

  1. Sep 15, 2013 #1
    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
    $$mg=kx$$
    $$m\ddot{x}+b\dot{x}+kx=mg$$
    $$\omega=\sqrt{\frac{k}{m}}$$
    $$\gamma=\frac{b}{2m}$$

    Answers:

    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. jcsd
  3. Sep 15, 2013 #2
    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##
    $$x(t)=e^{-\gamma t}(c_{1}+c_{2}t)+\frac{mg}{k}$$
     
  4. Sep 15, 2013 #3

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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.
     
  5. Sep 15, 2013 #4
    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.
     
  6. Sep 15, 2013 #5

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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.
     
  7. Sep 15, 2013 #6
    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.
     
  8. Sep 15, 2013 #7

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Damped Oscillator homework
  1. Damped oscilator (Replies: 2)

  2. Damped oscillator (Replies: 1)

  3. Damped oscillations (Replies: 3)

  4. Damped Oscillator (Replies: 2)

Loading...