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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- Oct 2nd 2007, 06:08 PMTwilightHelp 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. - Oct 2nd 2007, 06:11 PMJhevon
do you mean $\displaystyle T = 2 \pi \sqrt {\frac mg}$ ?

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

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

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

$\displaystyle \Rightarrow g = \frac {m}{\left( \frac T{2 \pi} \right)^2} = \frac {4 m \pi^2}{T^2}$ - Oct 2nd 2007, 06:14 PMDivideBy0
$\displaystyle t=2\pi\sqrt{\frac{m}{g}}$

Dividing both sides by $\displaystyle 2\pi$:

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

Squaring both sides:

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

Reciprocating both sides:

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

Multiplying both sides by m:

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

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

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

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