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Simple Harmonic Motion Solving for the phase and applying it.

  1. Apr 8, 2010 #1
    1. The problem statement, all variables and given/known data
    I am having a hard time solving this problem and this is my last chance to solve it before it is due and I have to do this for a test tomorrow.

    A simple harmonic oscillator consists of a block of mass 3.90 kg attached to a spring of spring constant 330 N/m. When t = 0.570 s, the position and velocity of the block are x = 0.123 m and v = 3.510 m/s. (a) What is the amplitude of the oscillations? What were the (b) position and (c) velocity of the block at t = 0 s?


    2. Relevant equations

    [tex]\omega=\sqrt{\frac{k}{m}}[/tex]
    [tex]x=x_{m}cos(\omega t + \phi)[/tex]
    [tex]v=-\omega x_{m}cos(\omega t + \phi)[/tex]

    3. The attempt at a solution

    Okay so I tried to solve for the phase constant by dividing the velocity function by the position function.

    This gave me the result of [tex]\phi=tan^{-1}(\frac{v}{x \omega})[/tex]

    I plugged in [tex]\phi[/tex] to both the position functions and my answers are wrong. So I am not sure as to what I need to do or why.
     
    Last edited: Apr 8, 2010
  2. jcsd
  3. Apr 8, 2010 #2

    rock.freak667

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    Your expression for v should be v=-ωxmsin(ωt+φ)


    Also consider what v/x gives and then substitute the conditions to get φ.
     
  4. Apr 8, 2010 #3
    Yes the formula for velocity was wrong. I divided the two which is how I got [tex]\phi[/tex] is inverse tangen times v divided by the quantity of position times the angular frequency. I must be doing something right because I am able to solve for the amplitude of the function, but I cannot get the position at t=0 and the velocity at t=0.
     
  5. Apr 8, 2010 #4

    rock.freak667

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    Could you please post your work on how you found all of your answers?
     
  6. Apr 8, 2010 #5
    [tex]x=x_{m}cos(\omega t + \phi)[/tex]
    [tex]v=-\omega x_{m}sin(\omega t + \phi)[/tex]

    So I divided these two functions which gave me the result of......


    [tex]\omega t + \phi = tan^{-1}(\frac{v}{-x \omega})[/tex]

    Since this is equal to all of the stuff in the position function and the velocity function I put it into the those functions after solving for the amplitude. This cannot lead to a correct answer because the velocity and position in the above mentioned expression are not at time = 0,

    P.S. I solved for the amplitude by dividing x at t=.570s by the [tex]cos(\omega t + \phi)[/tex]

    which came to...... approximately 0.400912 meters
     
  7. Apr 8, 2010 #6

    rock.freak667

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    What did you get for φ?
     
  8. Apr 8, 2010 #7
    For [tex]\phi[/tex] I got approximately.... 0.240112 radians
     
  9. Apr 8, 2010 #8

    rock.freak667

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    Recheck it, I get -1.25 radians.
     
  10. Apr 8, 2010 #9
    Yes I get -1.25897 radians when I solve for [tex]\omega t + \phi[/tex]
     
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