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Thread: [SOLVED] Time elapsed (SHM)?

  1. #1
    Super Member fardeen_gen's Avatar
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    [SOLVED] Time elapsed (SHM)?

    A particle executes SHM with period T about a point O. It passes through a point P with velocity V along OP. Show that the time that elapses when it again comes to P is given by: $\displaystyle t = \frac{T}{\pi}\cdot \arctan \left(\frac{TV}{2\pi\cdot OP}\right)$
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  2. #2
    Member
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    Here's how to do it:

    Spoiler:

    It is well known that the solution for SHM with ($\displaystyle x=0$ when $\displaystyle t=0$) is:

    $\displaystyle x = A \sin (\omega t)$

    with

    $\displaystyle \dot{x} = A \omega \cos (\omega t)$

    where $\displaystyle A$ is the amplitude of oscillations and $\displaystyle \omega = \frac{2 \pi}{T}$.

    Now in the first instance when the particle is at $\displaystyle P$ let the time be $\displaystyle t_0$ and let $\displaystyle d = |\vec{\mathbf{OP}}|$ so then

    $\displaystyle d = A \sin (\omega t_0 )$ and $\displaystyle V = A \omega \cos (\omega t_0)$

    which after dividing gives us:

    $\displaystyle \boxed{\frac{\omega d}{V} = \tan (\omega t_0 )}$ .

    Similarly in the second instance when the particle is at $\displaystyle P$ let the time be $\displaystyle t_1$ and so then

    $\displaystyle d = A \sin (\omega t_1 )$ and $\displaystyle -V = A \omega \cos (\omega t_1)$

    which after dividing gives us:

    $\displaystyle \boxed{-\frac{\omega d}{V} = \tan (\omega t_1 )}$. (Note this is in 2nd quadrant).

    Thus (taking into account the quadrants) we have:

    $\displaystyle \omega (t_1 - t_0 ) = \arctan \left( \frac{\omega d}{-V} \right) - \arctan \left( \frac{\omega d}{V} \right)$
    $\displaystyle = \pi - 2 \arctan \left( \frac{\omega d}{V} \right)$

    so

    $\displaystyle \frac{\omega d}{V} = \tan \left( \frac{\pi}{2} - \frac{\omega}{2} (t_1 - t_0 ) \right)$

    $\displaystyle = \cot \left( \frac{\omega}{2} (t_1 - t_0 ) \right)$

    $\displaystyle = \cot \left( \frac{\pi}{T} (t_1 - t_0 ) \right)$

    $\displaystyle \Leftrightarrow \tan \left( \frac{\pi}{T} (t_1 - t_0 ) \right) = \frac{V}{\omega d}$

    Hence

    $\displaystyle {\color[rgb]{0,0,1} \boxed{t = t_1 - t_0 = \frac{T}{\pi} \arctan \left( \frac{TV}{2 \pi d} \right)}}$ .

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