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vibrations question (differential equations etc)

  • 07-12-2009 12:17am
    #1
    imported_guy
    Closed Accounts Posts: 1,141 ✭✭✭


    hi dunno if i should post this here or the physics section or the engineering section, so move it if appropriate please, sorry.

    this is the question.

    the mass m=9kg and the spring constant k=81N/m. the spring is unstretched when x=0, at t=0, x=0 and the mass has a velocity of 0.7m/s towards the right, what is the value of x at t=2s

    answer should be
    -0.065

    but im doing it and getting to nearly the last step (checked the detailed solution from the past paper, still dont get it)

    so

    omega = root(k/m)
    i get omega = 3
    differential equation is d2x/dt2 + omegaX = 0

    its solution is

    x = Asin(omega)(t) + Bcos(omega)(t) where A and B are constants

    when i differentiate i get

    dx/dt = A(omega)cos(omega)(t) - B(omega)sin(omega)(t)

    and filling in the corresponding values i get 0.7/3 = A
    and B = 0

    but if i fill it in x = .233sin(3)(2) i dont get the answer (see the spoiler)

    i think that bolded step is what im doing wrong, anyone care to explain?

    thanks a bunch


Comments

  • #2
    henbane
    Registered Users, Registered Users 2 Posts: 1,636 ✭✭✭henbane


    It's very possible I've missed something but [latex]0.223*sin(2*3)=-0.065[/latex]


  • #3
    imported_guy
    Closed Accounts Posts: 1,141 ✭✭✭imported_guy


    henbane wrote: »
    It's very possible I've missed something but [latex]0.223*sin(2*3)=-0.065[/latex]
    somehow im not getting that on my calculator.... it has been pee'ing me off FFS. do i have to be in radians or something? or did i miss out on some thing?

    im getting .104 for sin6, multi plying by 0.223 gives me 0.0233

    edit

    oh yeah lol owned by calculator set to degree mode :/ sorry for the trouble fellas.


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