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Math Help - Varaition!!

  1. #1
    corpse_bride
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    Varaition!!

    OK, I have a question here that's part of a homework task I have due tomorrow:

    The time taken, t, for a pendulum to swing varies as the sqare root of its length, L. If one swing of a pendulum 98cm long takes 2 seconds, find the time taken for one swing of a pendulum 32cm long.

    I know of varaition, direct variation etc, but what do I do in this case? I can't work it out.

    Any help is greatly appreciated
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  2. #2
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    Hello, corpse_bride!

    The time t for a pendulum to swing varies as the sqare root of its length L.
    If one swing of a pendulum 98cm long takes 2 seconds,
    find the time taken for one swing of a pendulum 32cm long.
    This is direct variation . . .

    Write the equation! . t \;=\;k\sqrt{L}

    We are told: when L = 98,\;t = 2
    . . So we have: . 2 \;=\;k\sqrt{98}\quad\Rightarrow\quad k \:=\:\frac{2}{\sqrt{98}} \:=\:\frac{2}{7\sqrt{2}} \:=\:\frac{\sqrt{2}}{7}
    Hence: . t \;=\;\frac{\sqrt{2}}{7}\sqrt{L}

    Now find t when L = 32.

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  3. #3
    MHF Contributor kalagota's Avatar
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    Taguig City, Philippines
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    Quote Originally Posted by corpse_bride View Post
    OK, I have a question here that's part of a homework task I have due tomorrow:

    The time taken, t, for a pendulum to swing varies as the sqare root of its length, L. If one swing of a pendulum 98cm long takes 2 seconds, find the time taken for one swing of a pendulum 32cm long.

    I know of varaition, direct variation etc, but what do I do in this case? I can't work it out.

    Any help is greatly appreciated
    we have t = k\sqrt L from the given, we have 2 = k \sqrt {98} or k = \frac{2}{\sqrt {98}}

    using this for the second pendulum, we have t = \frac{2\sqrt {32}}{\sqrt {98}} = 2 \sqrt{\frac{32}{98}} = (2) \left( {\frac{4}{7}} \right) \left( \sqrt 1 \right) = \frac{8}{7} \approx 1.142857.
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