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Math Help - Help with a tricky equation

  1. #1
    Newbie Twilight's Avatar
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    Help with a tricky equation

    I'm completly stumped by this one. Any help would be deeply apreciated.

    T=2*pi*the square root of m/g
    the problem asks you to solve for g.

    Thanks in advance.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Twilight View Post
    I'm completly stumped by this one. Any help would be deeply apreciated.

    T=2*pi*the square root of m/g
    the problem asks you to solve for g.

    Thanks in advance.
    do you mean T = 2 \pi \sqrt {\frac mg} ?

    if so, T = 2 \pi \sqrt {\frac mg} ..........divide both sides by 2 \pi

    \Rightarrow \frac T{2 \pi} = \sqrt {\frac mg} ............square both sides

    \Rightarrow \left( \frac T{2 \pi} \right)^2 = \frac mg ..........flip both sides and multiply by m

    \Rightarrow g = \frac {m}{\left( \frac T{2 \pi} \right)^2} = \frac {4 m \pi^2}{T^2}
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  3. #3
    Senior Member DivideBy0's Avatar
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    t=2\pi\sqrt{\frac{m}{g}}

    Dividing both sides by 2\pi:

    \frac{t}{2\pi}=\sqrt{\frac{m}{g}}

    Squaring both sides:

    \frac{t^2}{4\pi^2}=\frac{m}{g}

    Reciprocating both sides:

    \frac{4\pi^2}{t^2}=\frac{g}{m}

    Multiplying both sides by m:

    \frac{4\pi^2m}{t^2}=g

    g=\frac{4\pi^2m}{t^2}

    We can make this more specific by recognising in the original equation that t cannot be < 0, since 2\pi\sqrt{\frac{m}{g}} cannot be negative. So we have:

    g=\frac{4\pi^2m}{t^2} \ \text{ and } \ t\geq 0
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