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SHM Equations of Motion

  1. Jun 20, 2011 #1
    Hi,

    I am learning trigonometric graphs and transformations as I am learning my SHM equations and I have a doubt:

    Firstly, I have a hard time defining angular frequency and that might be one of the sources of my problem. Can anyone help me with that? Is it simply how many full, 2pi rotations it performs in one second?

    Now, let's take this equation:

    x = xo
    cos(ωt)

    ω = 2π / T or 2πf

    Firstly, the reason why we use 2π / T as omega:

    My teacher said it but I can't remember perfectly - this is how he said it more or less:

    2π would be one complete cycle. Therefore when the t, representing time, in the equation equals the time period, T, we must have an answer of 2π as what we are applying the cosine function. So we have to find something, that when multiplied to t will equal to 2pi

    nt = 2π
    n = 2π / T

    Is that it? Something, n, multiplied by time must yield 2π? And 2π / T = ω

    Thanks,
    Peter G.
     
  2. jcsd
  3. Jun 20, 2011 #2

    Pengwuino

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    Gold Member

    For the most part, yes.

    Compare it to linear velocity if you need to. Instead of [itex]2\pi[/itex] being a cycle around a circle, imagine a "cycle" being just something traveling 10 meters in 20 seconds. 10 meters is like your [itex]2\pi[/itex] and the 20 seconds is a period, T. The only difference is that you know of a well known function that incorporates this circular motion very nicely.
     
  4. Jun 20, 2011 #3

    Doc Al

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    Staff: Mentor

    I think you have the right idea. Here's a summary:

    Period (T) = the time for one complete cycle

    Frequency (f) = the number of cycles per second (thus f = 1 cycle/1 period = 1/T)

    Angular frequency (ω) = the angle (in radians) per second (since 1 cycle = 2pi radians, ω = 2pi*f)

    [Edit: Looks like Pengwuino beat me to it.]
     
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