1. Not finding help here? Sign up for a free 30min 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!

Critical Damping Question

  1. Jan 25, 2016 #1

    RJLiberator

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    A spring balance consists of a pan that hangs from a spring. A damping force F_d = -bv is applied to the balance so that when an object is placed in the pan it comes to rest in the minimum time without overshoot. Determine the required value of b for an object of mass 2.5 kg that extends the spring by 0.06m.


    2. Relevant equations

    (ω_0)^2 = k/m = ϒ^2/4
    b = 2*sqrt(k*m)

    3. The attempt at a solution

    So if we find k, the spring constant, we find b which is what we are looking for.
    We know the mass is 2.5 kg
    We know that the change in x is 0.06m.
    We know that there is critical damping.

    I need to use the change in x information to find b and/or k.

    The general solution for crit. damping is x = Ae^(-ϒt/2)+Bte^(-ϒt/2)
    So the derivative, which is the rate of change is equal to
    dx/dt = e^(-ϒt/2)(B-ϒBt/2-ϒA/2)

    max displacement occurs when dx/dt is equal to 0.

    0 = e^(-ϒt/2)(B-ϒBt/2-ϒA/2)
    0.06 = Ae^(-ϒt/2)+Bte^(-ϒt/2)

    Am I heading in the right direction here? Is there anything that I am missing in my steps that might lead me in the right direction?
     
  2. jcsd
  3. Jan 25, 2016 #2
  4. Jan 25, 2016 #3

    RJLiberator

    User Avatar
    Gold Member

    I learned here that
    A = x(0) and we can define our initial starting spot as x =0 so A =0.
    B = dx/dt + w_0 * x(0)
    so B = dx/dt

    We have x(t) = Bte^(-w_0*t)
    dx/dt = B(e^(-w_0t)+t*(-w_0)e^(-w_0*t))
     
  5. Jan 25, 2016 #4

    RJLiberator

    User Avatar
    Gold Member

    I'm not sure how gravity is playing a part here.

    The two pieces of information that I am not using are change in x = 0.06m and gravity = 9.81 m/s^2.


    x(t) = v_0 *t*e^(-w_0*t)
    v(t) = v_0*w_0*t*e^(-w_0*t)
     
  6. Jan 25, 2016 #5

    RJLiberator

    User Avatar
    Gold Member

    Perhaps I have thought of something

    Displacement is maximized when dx/dt = 0.
    So set v(t) = 0.
    We see then that 1/t = w_0

    If I can find w_0 then I can find k.

    We can find t through
    t = sqrt(2*distance/g)

    distance = 0.06
    g = 9.81

    t = 0.111

    which means 1/t = 9.04 = w_0.

    Which thus means w_0^2 *m = k = 204.375

    And finally, b = 2*sqrt(2.5*204.375) = 45.2

    Correct? :D
     
  7. Jan 26, 2016 #6

    RJLiberator

    User Avatar
    Gold Member

    Sorry to bump this up, but any last minute help/confirmation on my logic here?
     
  8. Jan 26, 2016 #7
    When you plot the resulting solution, how does it compare with the under damped and over damped cases?
     
  9. Jan 26, 2016 #8

    gneill

    User Avatar

    Staff: Mentor

    You can find k simply by examining the force balance for the extended spring when the mass is hanging at rest. That's basic Hooke's law. The damping plays no role when the mass is not moving, as the damping force is -bv.

    Sketch a schematic of the system and an FBD for the mass. Can you write the differential equation from the FBD? It's handy to use the x, x', x'' shorthand to represent the displacement x and its derivatives w.r.t. time.

    If you can do the above then you will be able to write the characteristic equation for the differential equation by inspection. The discriminant of its roots holds the key to the behavior of the system (damped, critically damped, underdamped). Do you know what that relationship is?
     
  10. Jan 26, 2016 #9

    RJLiberator

    User Avatar
    Gold Member

    @qneill
    What you are suggesting is that I use Hooke's law, F = kx
    where F = m*a = 2.5kg*9.81m/s^2
    and x = 0.06m
    so F/x = k = 408.75

    Then we simply use b = 2*sqrt(m*k) to find a b value of 63.93

    I suppose this makes sense as we are using all of the giving information.
     
  11. Jan 26, 2016 #10

    gneill

    User Avatar

    Staff: Mentor

    Yes, but you've slipped a decimal point in your calculation of k. Re-do the calculation.

    I was also implying that you don't have to rely on canned formulas for every system, but can quickly derive the necessary formulas from scratch if you can recall the general principles. This can be a life-saver in an exam situation :wink:

    Your formula for the critical value of b falls out of the discriminant of the characteristic equation when you know that it must be zero for critical damping.
     
  12. Jan 26, 2016 #11

    gneill

    User Avatar

    Staff: Mentor

    Oops! Forget that! I mistakenly read the mass to be 0.25 kg rather than 2.5 kg. o:)
     
  13. Jan 26, 2016 #12
    You can also stand to take more care with converting principles into equations.

    The equilibrium condition is that Fnet = 0 at rest, which for the vertical component gives:

    Fnet = F(spring) + F(gravity) = 0 (sum over vertical components)

    = +kd - mg = 0 (substituting in for spring force and gravitational force. Note g rather than a. acceleration is zero at equilibrium)

    leading to kd = mg

    and then k = mg/d (take care with the units)
     
  14. Jan 26, 2016 #13

    RJLiberator

    User Avatar
    Gold Member

    Yes, currently I'm struggling mightily in classical mechanics type questions. I do great with mathematics and even my quantum mechanics class, but give me a problem like this one that should be solved in 2 easy steps as you proposed and I'm lost.

    I just don't understand what I'm missing when it comes to classical mechanics. Perhaps experience.
     
  15. Jan 26, 2016 #14

    RJLiberator

    User Avatar
    Gold Member

    Yes, I was just in the shower and also thinking to my self how when the mass is at rest, the damping force Fd must be 0 as velocity = 0. Fd = -bv.
    Therefore, I could have saw this and immediately used Hooke's law as stated.
    Which would have allowed me to solve this problem in 5 minutes instead of the 5 hours it took me.
    Sigh.
     
  16. Jan 26, 2016 #15

    gneill

    User Avatar

    Staff: Mentor

    Usually these things boil down to sketching the system, drawing relevant FBD's, then writing the equations of motion from them.

    With the equations and a general knowledge of the behavior of differential equations it becomes a math problem, and the DE's of many systems in physics have the same form. Knowing about how the discriminant of the characteristic equation effects the nature of the response of a second order DE is one of the things that is handy to recall, rather than trying to memorize a bunch of formulas for different types of systems or their configurations. That's more "math smarts" than "physics smarts", so perhaps that will help? At any rate, experience definitely makes a difference. Once you've spent several hours on a problem and then come across a shortcut, the associated "Aha!" moment and relief usually sticks it firmly in your memory, ready for the next similar situation.

    If you wish to go into more detail on a solution approach for this problem let me know. Otherwise, shall we declare the problem Solved? (I'm blatantly hinting about the new thread-marking "Solved" feature of PF :smile:)
     
  17. Jan 26, 2016 #16

    RJLiberator

    User Avatar
    Gold Member

    Yes this problem is solved. Thank you kindly.
     
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: Critical Damping Question
Loading...