# Help with a tricky equation

• Oct 2nd 2007, 06:08 PM
Twilight
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.

• Oct 2nd 2007, 06:11 PM
Jhevon
Quote:

Originally Posted by Twilight
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.

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

$\Rightarrow g = \frac {m}{\left( \frac T{2 \pi} \right)^2} = \frac {4 m \pi^2}{T^2}$
• Oct 2nd 2007, 06:14 PM
DivideBy0
$t=2\pi\sqrt{\frac{m}{g}}$

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:

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