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Diff EQ spring question

  1. Apr 29, 2008 #1
    [SOLVED] Diff EQ spring question

    1. The problem statement, all variables and given/known data

    a)IF a mass of 0.5kg is attached to a spring with a spring constant of 5(nt/m) and then receives a blow to dislodge it from its equilibrium position, then what is the resistive force coefficient (gamma) if the system is critically damped?
    b)what is the general solution for the position u(t) of the mass?


    2. Relevant equations

    mx''=-kx-ax'+f(t)
    x''+ (a/m)x' + (k/m)x = F(t)/m

    m=.5kg
    k=5N
    gamma=? (a/m)x' ?

    3. The attempt at a solution
    I don't understand what "critically damped" means?

    x"+(a/.5)x'+10x=0
    I really don't know how top start this, any help would be appreciated
     
  2. jcsd
  3. Apr 29, 2008 #2

    HallsofIvy

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    The resistive force is the coefficient of x'.

    The spring is "critically damped" if and only if it is on the "border" between purely exponential and repetitive motion: Since the characteristic equation for this d.e. is quadratic, you would have "repetitive motion" if the characteristic roots are complex, purely exponential motion if the characteristic roots are real. It is "critically damped" if and only if the discriminant of the quadratic equation is 0.
     
  4. Apr 29, 2008 #3
    huh?
    so by saying the system is critically damped the spring will continue to oscillate?
    I need to find the roots- if the characteristic equation contains complex or exponential terms we would have repetitive motion?
    What is the discriminant of the quadratic equation?
    Is my initial 2nd order d.eq set up correctly?
     
  5. Apr 29, 2008 #4

    tiny-tim

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    Hi hils0005! :smile:

    The discriminant of a quadratic expression ax² + bx + c is b² - 4ac.

    If the discriminant of a quadratic differential equation is negative, then the general solution is of the form Acos(kt) + Bsin(kt) … so it oscillates (repetitive motion).

    If the discriminant is positive, then the general solution is of the form Ae^kt + Be^-kt … so it gradually reduces to zero or increases to infinity (non-repetitive motion).

    But if the discriminant is zero, then the general solution is of the form (A + Bt)e^kt … the extra t makes the deceleration (or acceleration) faster. :smile:
     
  6. Apr 29, 2008 #5
    Thanks Tiny Tim!

    because mysystem is "critically damped" I need the discriminant to equal 0.
    If my initial equation is set up correctly.....
    x'' + (a/m)x' + 10x = 0
    r^2 + (a/m)r + 10

    (a/m)^2 - 4(1)(10)=0

    a^2/.5^2=40

    a^2=10, a=sqrt10 ?

    to get the general solution into the form (A+ Bt)e^kt or y=Ae^kt + Bte^kt, don't my roots need to be equal?
     
  7. Apr 29, 2008 #6

    Vid

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    The roots are equal if the discriminant is 0.
     
  8. Apr 29, 2008 #7
    OK, then my quadratic equation must be set up incorrectly??? any insight on that would be helpful
     
  9. Apr 29, 2008 #8

    tiny-tim

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    o hils0005! :smile:

    the discriminant is b² - 4ac

    the roots are -b/2a ±√(b² - 4ac)/2a

    so if the discriminant is zero, the two roots are both -b/2a

    you know that! :rolleyes:
     
  10. Apr 29, 2008 #9
    wow, why is this problem giving me so much trouble!!!

    so I determined that for the discriminant to equal zero
    (a/.5)^2-4(1)(10)=0
    a=sqrt10

    so b= sqrt10/.5

    roots = -(sqrt10/.5)/2

    general solution: y=C(1)e^(-sqrt10/.5/2)t + C(2)e^(-sqrt10/.5/2)t

    is this correct?
     
  11. Apr 30, 2008 #10

    tiny-tim

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    hi hils0005! :smile:

    Yes, but …
    why so complicated?

    :cry: enough with the fractions! :cry:

    Let's start again …

    x"+(a/.5)x'+10x=0

    Rewrite that as x"+ 2ax' + 10x=0.

    Then, as you said earlier, a = √10, so it's x"+ (2√10)x'+10x=0.

    and so on … :smile:
     
  12. Apr 30, 2008 #11
    Thanks for all your help-much appreciated
     
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