Simple Pendulum: The gradient of the graph of T^2 against length

[Tangeton]

I've been asked to investigate the value of g.

My graph shows that the gradient is around 4.4 as when the length was 0.27m, the squared time period was 1.18s (since this is a T² against L graph).

My question is, of course, why?

If I am detecting g, which is 9.81 on earth, why did I get such an odd value?

I tried to see how a calculation would work with that...

Since T = 2 pi * sqrt of L/g
g = (4 pi² x L) / T²

I used one of my results, where L = 0.20m, T² = 0.904
After substituting those values into the equation I still got only 8.734 for the value of g.

I need to do a write up on this and of course this is the analysis and conclusion section.

Thank you for all the help you can give me.

 

[vela]

There could be all sorts of reasons. For example, it could be that:

1. Your experimental setup is flawed.
2. Your measurements are flawed.
3. You have large uncertainties in your measurements.
4. You live in an unusual place where g is abnormally small!

There are differences between your theoretical model of the situation and the actual physical setup. What assumptions does your theoretical model rely upon? How well do those assumptions hold up in reality?

 

[Tangeton]

Okay well.. I actually found out that I am right and the gradient is correctly defined as 4(pi)² over g, which is what I got.

I have an issue though I was looking on this website: http://www.schoolphysics.co.uk/age1...rmonic motion/text/Simple_pendulum/index.html

The gradient is the same as mine but it is upside down. I am torn because in my book is says that the gradient of T² against L is indeed 4(pi)²/g. In the above example, the g has been easily derived.

Any help of why I can't determine the g? I need to use T = 2(pi) * sqrt of L/g, but I don't know how to derive it :(

 

[BvU]

I wonder what you mean with 'upside down'. If you mean 'inverted', it could well mean that you plotted T² horizontally versus L vertically. Show your results. ALL results.

 

[Tangeton]

I plotted T² vertically and L horizontally indeed. I worked it out to be T²/L = 4(pi)²/g and from there I got the g, sorry to be a bother when I could get to an answer myself :s