Results 1 to 2 of 2

Math Help - Damped Oscillations & Sequences

  1. #1
    Junior Member
    Joined
    Sep 2007
    Posts
    51

    Damped Oscillations & Sequences

    1) Spring of mass 1kg and its damping constant c = 10. The spring starts from it equilibrium position with a velocity of 1 m/s. graph the position function for the following values of the spring constant k: 10,20,25,30,40. What type of damping occurs in each case?

    so m = 1kg, c =10, x(0) = 0 and x'(0) = 1

    1r^2 + 10r + kx = 0.
    do i calculate each one by using the given k values? is it necessary to use 10 pages?


    2)Can someone show me how to figure out the formula mathematically?
    {5,1,5,1,5,1...}
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,929
    Thanks
    333
    Awards
    1
    Quote Originally Posted by xfyz View Post
    1) Spring of mass 1kg and its damping constant c = 10. The spring starts from it equilibrium position with a velocity of 1 m/s. graph the position function for the following values of the spring constant k: 10,20,25,30,40. What type of damping occurs in each case?

    so m = 1kg, c =10, x(0) = 0 and x'(0) = 1

    1r^2 + 10r + kx = 0.
    do i calculate each one by using the given k values? is it necessary to use 10 pages?
    The equation for a damped spring is
    \frac{d^x}{dt^2} + \frac{c}{m} \cdot \frac{dx}{dt} + \frac{k}{m} \cdot x = 0
    Where the damping force is given as F = -c \frac{dx}{dt}.

    This is a homogeneous linear differential equation so the auxiliary equation is
    r^2 + \frac{c}{m} \cdot r + \frac{k}{m} = 0

    The solution to this equation is (using the quadratic formula)
    r = \frac{ -\frac{c}{m} \pm \sqrt{ \frac{c^2}{m^2} - 4(1) \cdot \frac{k}{m}}}{2}

    r = \frac{-c \pm \sqrt{c^2 - 4km}}{2m}

    So the general solution for the differential equation is
    x(t) = A \cdot exp \left [ x \left ( \frac{-c + \sqrt{c^2 - 4km}}{2m} \right ) \right ] + B \cdot exp \left [ x \left ( \frac{-c - \sqrt{c^2 - 4km}}{2m} \right ) \right ]

    So the nature of the discriminant will determine the nature of the solution.

    Your problem specifies m = 1 kg and c = 10 Ns/m. (You didn't specify the units for this, so I'm presuming they are MKS units.)

    So your solution comes out to be:
    x(t) = \left ( \frac{1}{2\sqrt{25 - k}} \right ) \cdot   exp \left [ x \left ( \frac{-10 + \sqrt{100 - 4k}}{2m} \right ) \right ]  - \left ( \frac{1}{2\sqrt{25 - k}} \right ) \cdot   exp \left [ x \left ( \frac{-10 - \sqrt{100 - 4k}}{2m} \right ) \right ]

    The nature of the solution changes depending on the sign of the discriminant. If the discriminant is negative we have damped harmonic motion. If the discriminant is 0 we have critically damped harmonic motion. If the discriminant if positive we have overdamped harmonic motion.

    -Dn
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Damped Forced Oscillations and Beats
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: December 8th 2010, 12:15 PM
  2. Nonlinear Oscillations
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: January 10th 2010, 02:46 AM
  3. Forced, Damped Oscillations
    Posted in the Advanced Applied Math Forum
    Replies: 7
    Last Post: April 7th 2008, 02:16 PM
  4. Vertical Oscillations
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: February 13th 2008, 12:39 AM
  5. Damped Oscillations
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 25th 2007, 09:45 AM

Search Tags


/mathhelpforum @mathhelpforum