# Pendulum Lab

1. Jan 14, 2010

### pkpiotr517

This is one question that's giving me a bit of trouble to handle.
The Period of a pendulum is given by the following equation:

where T= period of pendulum (seconds)
L= Length of pendulum (meters)
g= acceleration doe to gravity (meters per second2)

Solve this equation in terms of L, T and pi. That means that g sould be by itself on one side of the equal sign and a combination of L, T, and pi should be on the other side of the equal sign.

T=2(pi)*square.root.of(L/g)

My attempt at the problem was:
(T)/(2*pi)= square.root.of(L/g)2
Then I squared the right side of the equation and did the same on the left in order to cancel out the square root on the right.
((T)/(2*pi))2=(L/g)
Afterwords, I tried to factor out the L by multiplying its inverse on the right and got:
((T)/(2*pi))2(1/L)=(1/g)

There I get stuck, because I'm not too sure if he wanted it set equal to 1/g. So if anyone can send some feedback, then it would be greatly appreciated!

2. Jan 14, 2010

### rock.freak667

From here

$$\left( \frac{T}{2 \pi}\right)^2= \frac{L}{g}$$

You can just invert both sides of the equation and then multiply by L.

Or as you've done in the last line of your working, just invert both sides and you'll have 'g' in terms 'L' and 'T' and 'π'

3. Jan 14, 2010

### pkpiotr517

So I would get:

$$\left( \frac{2 \pi }{T }\right)^2 (L)= {g}$$

Is that correct?

Last edited: Jan 14, 2010
4. Jan 14, 2010

### rock.freak667

Yes that looks correct.