Results 1 to 5 of 5

Thread: Mass on a spring problem

  1. #1
    Member
    Joined
    Nov 2008
    Posts
    103

    Mass on a spring problem


    Can someone explain how to solve this? Cheers.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Mar 2009
    Posts
    378
    You will need to solve the homogeneous problem
    $\displaystyle \frac{d^2x}{dt^2} + w^2x = 0$
    Then you will find a particular solution you can do this using the method of undetermined coefficients. Then you add the two solutions together.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2008
    Posts
    103
    Can you explain how to find the C.F.? Apparently its x=cos(wt)+sin(wt) ? I don't know how they got that answer though.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Mar 2009
    Posts
    378
    Okay, I will go through the particular solution first. I will let $\displaystyle x_p$ denote the particular solution. Since the nonhomogeneous part is a trigonometric function we will let the particular solution be of the form $\displaystyle x_p = A\sin \gamma t + B\cos \gamma t$. Now we must determine the coefficients $\displaystyle A,B$. To do this we will plug $\displaystyle x_p$ back into the original DEQ.
    $\displaystyle x''_p = -\gamma^2 A \sin \gamma t - \gamma^2 B \cos \gamma t$
    and so
    $\displaystyle -\gamma^2 A \sin \gamma t - \gamma^2 B \cos \gamma t + w^2 \cdot \left( A\sin \gamma t + B\cos \gamma t \right) = \cos \gamma t$
    $\displaystyle \left(w^2 -\gamma^2\right) A \sin \gamma t + \left(w^2 - \gamma^2\right) B \cos \gamma t = \cos \gamma t$.
    Thus we have
    $\displaystyle \left(w^2 - \gamma^2\right)A \sin \gamma t = 0$
    and
    $\displaystyle \left(w^2 - \gamma^2\right)B \cos\gamma t = \cos \gamma t$.
    Therefore, we have
    $\displaystyle \left(w^2 - \gamma^2\right)A = 0$
    and
    $\displaystyle \left(w^2 - \gamma^2\right)B = 1$,
    so
    $\displaystyle A = 0$
    $\displaystyle B = \frac{1}{w^2 - \gamma^2}$.
    Finally, we see that our particular solution is
    $\displaystyle x_p = \frac{1}{w^2 - \gamma^2} \cos \gamma t$.

    Deleted mistake see following post.
    Last edited by lvleph; Dec 5th 2009 at 08:16 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Mar 2009
    Posts
    378
    Sorry I made a mistake. The homogeneous solution is wrong. The solution of the characteristic equation is not $\displaystyle \lambda = \pm w$. The solution is $\displaystyle \lambda = \pm w i$.
    Therefore, the homogeneous solution is of the form
    $\displaystyle x_h = C_1\cos w t + C_2 \sin w t$.
    Finally, the general solution should be
    $\displaystyle x(t) = C_1\cos w t + C_2 \sin w t + \frac{1}{w^2 - \gamma^2} \cos \gamma t$.
    To determine the constants $\displaystyle C_1,C_2$ we must have initial conditions.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Spring mass problems
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: Mar 23rd 2011, 10:13 AM
  2. mass-spring system
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: Feb 6th 2010, 08:27 AM
  3. Mass-spring system
    Posted in the Differential Equations Forum
    Replies: 9
    Last Post: Dec 20th 2009, 10:16 AM
  4. Mass-Spring System
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: Dec 19th 2009, 04:46 AM
  5. spring mass-undamped
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: Oct 9th 2009, 05:22 AM

Search Tags


/mathhelpforum @mathhelpforum