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    • Thread Starter
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    HI!
    For Simple Harmonic Motion, what does the graph of ACCELERATION against Velocity look like?

    I've googled long and hard and can't find this!

    Any help would be much appreciated.
    Thanks.

    Buzzie. {No, not a vibrator as someone once though..}.
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    I think it may look like tanx
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    sorry, complete ********. I'll think about it.
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    I think I have the answer:

    If you take x = A sin (ωt)
    Then v = Aω cos (ωt)
    And a = -Aω² sin (ωt) through differentiation

    If you plot a against v, then

    y = a = -Aω² sin (ωt)
    y/(-Aω²) = sin (ωt)
    y²/A²ω4 = sin² (ωt)

    x = v = Aω cos (ωt)
    x/Aω = cos (ωt)
    x²/A²ω² = cos² (ωt)

    Now, there is a trigonometric identity that states
    sin² θ + cos² θ ≡ 1

    So by adding our two previous equations:
    y²/A²ω4 + x²/A²ω² = sin² (ωt) + cos² (ωt)
    (y/Aω²)² + (x/Aω)² = 1
    And this just happens to be the Cartesian equation of an ellipse, centre the origin, which crosses the y-axis at (0, ±Aω²) and crosses the x-axis at (±Aω, 0)

    I think that's correct, hope it helps!
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    And this just happens to be the Cartesian equation of an ellipse
    that' REALLY great!
    i'm sure that that is right, because : when a is max, v is min, and when v is max, a is min, and as v sometimes have the same direction as a, and other times opposite direction so it's circular ( ellipse to be more specific )
    you answer is REALLY great!
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    (Original post by Mikemania)
    I think I have the answer:

    If you take x = A sin (ωt)
    Then v = Aω cos (ωt)
    And a = -Aω² sin (ωt) through differentiation

    If you plot a against v, then

    y = a = -Aω² sin (ωt)
    y/(-Aω²) = sin (ωt)
    y²/A²ω4 = sin² (ωt)

    x = v = Aω cos (ωt)
    x/Aω = cos (ωt)
    x²/A²ω² = cos² (ωt)

    Now, there is a trigonometric identity that states
    sin² θ + cos² θ ≡ 1

    So by adding our two previous equations:
    y²/A²ω4 + x²/A²ω² = sin² (ωt) + cos² (ωt)
    (y/Aω²)² + (x/Aω)² = 1
    And this just happens to be the Cartesian equation of an ellipse, centre the origin, which crosses the y-axis at (0, ±Aω²) and crosses the x-axis at (±Aω, 0)

    I think that's correct, hope it helps!

    WHOA i think we got a new fermat ....jkzzzz
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    Many thanks for that, Mikemania.
    Cheers.
    .
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    Sure! I just happened to have done this kind of graph in maths, so it was lucky really.
    Was this on an actual physics paper? Seems a bit tricky
 
 
 
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