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SHO stuck on a simple question -.-

  1. Aug 7, 2009 #1
    1. The problem statement, all variables and given/known data

    A horizontal massless spring of spring constant k is attached to a immovable wall at one end and a mass of 0.45kg at the other end. The spring which was not originally under tension, is now extended by 0.18 m by pulling the mass horizontally. The mass was then released by giving it an instantantaneous push in the direction of the extension such that it had an initial velocity of 0.5 ms-1. The mass then undergoes firctionless simple harmonic motion (SHM) at [tex]\omega[/tex] = 2.2 rad s-1. With an explanation of your reasoning:

    (i) deduce the equation for the displacement of the mass as a function of time, (2)

    (ii) show that your equation complies with the general form of SHM

    x(t) = Acos([tex]\omega[/tex]ot + [tex]\phi[/tex]) (2)
     
  2. jcsd
  3. Aug 7, 2009 #2
    woah why did this get moved? lol this is 2nd year physics :p and i saw a fellow forum user ask about a 1st year physics question in the advance section. no fair ;[
     
  4. Aug 7, 2009 #3

    rock.freak667

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    For SHM you can either have x=x0sinwt or x=x0coswt

    So why did you pick [itex]x=x_0cos\omega t[/itex] instead of [itex]x=x_0sin\omega t[/itex] ?
     
  5. Aug 7, 2009 #4
    I did not pick it that is the second part of the question -.-
     
  6. Aug 7, 2009 #5

    rock.freak667

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    then what exactly did you try in this question? (I ask because you did not post any attempt)
     
  7. Aug 7, 2009 #6
    this is not a homework question it is a past exam question and i dont know where to begin. :S
     
  8. Aug 7, 2009 #7

    rock.freak667

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    Well what is the force acting on the spring with extension x? When you get that recall that F=ma and a=d2x/dt2
     
  9. Aug 8, 2009 #8
    force acting on a spring is f = -kx and [tex]\ddot{}x[/tex] = [tex]\omega[/tex]o2x

    i know how to arrive to that equation f = - kx but also f = ma therfore ma = -kx re arrange and you get [tex]\ddot{}x[/tex] = [tex]\omega[/tex]o2x
     
  10. Aug 8, 2009 #9

    ideasrule

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    If ma=-kx, then ma+kx=0 and
    [tex]
    \ddot{}x+\omega^2x=0
    [/tex]

    So you were off by a negative sign.

    Now show that this equation satisfies this condition:

    x(t) = Acos([tex]\omega[/tex]ot + [tex]\phi[/tex])

    In other words, show that the second time derivative of x(t) is indeed equal to [tex]
    -\omega^2x
    [/tex]

    That would show x(t) = Acos([tex]\omega[/tex]ot + [tex]\phi[/tex]) is a solution to the differential equation and therefore describes simple harmonic motion.
     
  11. Aug 8, 2009 #10


    the original question asked for an equation as a function of time,
    [tex]

    \ddot{}x+\omega^2x=0

    [/tex]

    there is no time in that equation o_O
     
  12. Aug 8, 2009 #11
    Yes there is...[tex]x\ddot{}[/tex] is the second derivative of x wrt time
     
  13. Aug 8, 2009 #12
    oh yeah there is
     
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