Results 1 to 10 of 10

Math Help - Angular displacement from its equilibrium position

  1. #1
    Newbie
    Joined
    Jul 2010
    From
    Qatar
    Posts
    15

    Angular displacement from its equilibrium position

    Hi all. I'm currently stuck on the following question and I was wondering what I need to do. Any guidance and a few pointers in the right direction will be deeply appreciated!



    Angular displacement from its equilibrium position-angular-displacement.jpg
    Last edited by CaptainBlack; August 8th 2010 at 08:32 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Benji123 View Post
    Hi all. I'm currently stuck on the following question and I was wondering what I need to do. Any guidance and a few pointers in the right direction will be deeply appreciated!



    Click image for larger version. 

Name:	Angular displacement.jpg 
Views:	33 
Size:	56.2 KB 
ID:	18477
    This is a perfectly normal second order linear constant coefficient ODE initial value problem. So what do your notes say?

    Try a trial solution of the form e^{kt}, and solve the resulting quadratic for $$ k etc

    CB
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jul 2010
    From
    Qatar
    Posts
    15

    Post

    Quote Originally Posted by CaptainBlack View Post
    This is a perfectly normal second order linear constant coefficient ODE initial value problem. So what do your notes say?

    Try a trial solution of the form e^{kt}, and solve the resulting quadratic for $$ k etc

    CB
    Thanks for the advice! Using the auxiliary equation I obtained: k^2 + 4k + 13 = 0
    Therefore, k = (-4 √16+52)/2
    k1 = 2.213
    k2 = -6.123
    Θ = Ae^2.123t + Be^-6.123t

    I dont think this is correct because I haven't done anything with t=0, Θ=2 and dΘ/dt = 8...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    Use your initial conditions to find your "constants of integration" A and B. You'll get two equations with two unknowns. Just plug and chug. Make sense?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Benji123 View Post
    Thanks for the advice! Using the auxiliary equation I obtained: k^2 + 4k + 13 = 0
    Therefore, k = (-4 √16+52)/2
    k1 = 2.213
    k2 = -6.123
    Θ = Ae^2.123t + Be^-6.123t

    I dont think this is correct because I haven't done anything with t=0, Θ=2 and dΘ/dt = 8...
    No the roots of the characteristic equation has roots -2 \pm 3 \text{i}

    CB
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Jul 2010
    From
    Qatar
    Posts
    15
    Quote Originally Posted by CaptainBlack View Post
    No the roots of the characteristic equation has roots -2 \pm 3 \text{i}

    CB
    So this would be...

    m^2 + 4m + 13 = 0
    m= -23i
    ^
    y = Ae^(-2+3i)t + Be^(-2-3i)t

    y = e^3t (4cos-2t + 13sin-2t)?
    Last edited by Benji123; August 10th 2010 at 02:13 AM. Reason: changed x to t
    Follow Math Help Forum on Facebook and Google+

  7. #7
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    No, your last line is confusing the real and imaginary parts. You should have

    y = e^(2t) (A cos(3t) + B sin(3t)).
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Jul 2010
    From
    Qatar
    Posts
    15
    Ohh, thanks alot for the help buddy! I knew with a bit of help I'd eventually get there
    Follow Math Help Forum on Facebook and Google+

  9. #9
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    Actually, I wrote it wrong. It's

    y = e^(-2t) (A cos(3t) + B sin(3t)).

    Make sure to thank CB. He did more for you than I did!
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Jul 2010
    From
    Qatar
    Posts
    15
    Thanks for the help everyone! I really appreciate it!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: January 26th 2011, 03:11 PM
  2. Replies: 0
    Last Post: January 25th 2011, 03:01 PM
  3. displacement
    Posted in the Calculus Forum
    Replies: 5
    Last Post: October 28th 2010, 06:33 PM
  4. Finding angular displacement.
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: March 7th 2010, 06:48 AM
  5. angular displacement
    Posted in the Math Topics Forum
    Replies: 8
    Last Post: November 24th 2007, 11:32 AM

Search Tags


/mathhelpforum @mathhelpforum