# Homework Help: T^2 vs L determining g?

1. Nov 14, 2006

### 8parks11

Okay this is the last question of the lab and I don't get it
Basically its a lab where we find how amplitude, bob mass, and length affects the period T for each time.

Our finding was that length affects the period a lot. I did a bunch of graphs and tables and answered all the questions but don't know how I derive this

the question is "From your graph of T^2 vs. L
determine a value for g."

T here is the period in seconds and L is the length of the string in cm. Now I do have to graph and it looks pretty normal to me.

The slope is 0.03900 and the r is 0.9947
now I dont think this would help much at all
(unless its r-slope haha...)

I have to find g from this graph and calculate the percent of error
I know how to calculate the percent of error but I don't know how I can find g from this graph

the points are for x which is length

50
60
70
80
90
100

and for T^2 i have

2.2831
2.4586
2.89
3.3562
3.7288

My first guess was i need to find an equation for this relationship and then
use the formula 4pi^2/g x L = T^2 but I don't know what to do?

2. Nov 14, 2006

### nrqed

As you said, the equation is $T^2 = {4 \pi2 \over g} L$ . So if you plot T^2 vs L, you should get a straight line. The slope of this line will be equal to what?

3. Nov 14, 2006

### 8parks11

wait I don't get it a lot.
so for the x-axis I get all the L (from 50~100) and then just leave it.
for the y, I should use $T^2 = {4 \pi2 \over g} L$ and then subsitute all the L from the x data. that would gimme T^2
but I don't think this is right because I'm getting 4.028 as my slope

4. Nov 14, 2006

### nrqed

You measured the period for different values of L, right? Just plot your measured values of T^2 versus L. Then use theory to determine what the slope should be equal to. Theory will give you a relation between the slope of this graph and the value of g. Once you measure the slope on your graph you will be able to determine g from your data.

5. Nov 14, 2006

### 8parks11

ok so the slope of my line of T^2 vs L is 0.0390.

since 4pi^2/g=slope g should be 4pi^2/slope...

so the predicted g is 4pi^2/ 0.0390

and this gives me 1012.267... what is wrong?

Last edited: Nov 14, 2006
6. Nov 14, 2006

### nrqed

What is the uncertainty? this is not far from the expected value of about 981 cm/s^2!

7. Nov 14, 2006

### 8parks11

yes haha thanks i forgot that its in cm haha

just to confirm, tthe % error would be 1- (980/1012.267) right?

Last edited: Nov 14, 2006