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Simple Oscillator: find x(t)

  1. Oct 18, 2004 #1

    cj

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    At time t = T/2 the speed of a simple oscillator of angular frequency

    [tex]\omega = \omega_0[/tex]

    has maximum amplitude U and positive value.

    Find x(t).

    ----------
    The textbook answer for this problem is:

    [tex]x(t) = -\frac {U}{\omega_0}sin(\omega_0t)[/tex]

    where

    [tex]U = \omega_0A}[/tex]
    ----------
    This is confounding me.

    For a simple sin solution (I'm assuming zero phase shift), wouldn't x at t = T/4 have to be positive, i.e.:

    [tex]x = \frac {U}{\omega_0}[/tex]

    and NOT negative (as given by the alleged solution):

    [tex]x = -\frac {U}{\omega_0}[/tex]

    Comments?
     
  2. jcsd
  3. Oct 18, 2004 #2

    Integral

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    What is the definition of the coordinate system?
     
  4. Oct 18, 2004 #3

    cj

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    Vertical axis = displacement (x), velocity (v) and acceleration (a).

    Horizontal axis = time, t
     
  5. Oct 18, 2004 #4

    Integral

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    I am trying to get you to think about the definition of what is meant by negative and positive. Have you given us the complete problem?
     
  6. Oct 18, 2004 #5

    cj

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    Yes, the posted question and textbook answer (which I guess could be wrong) is verbatim.

    Maybe I don't understand the meaning of the negative sign.

    I'm imagining a displacement-speed-acceleration relationship as given in the graph at the top of the following linked page:

    Sample

    The bottom line is that if x(t) is as given by the textbook, it would clearly have a negative value (relative to the equilibrium position at t = T/4. This seems incorrect since even junior high school students understand that at t = T/4 the value for a sine function is clearly a maximum and clearly positive.

    Right?
     
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