Hsc 2002 Q16 C. (1 Viewer)

[henry08]

I need some help with Question 16, part C. from the 2002 HSC paper. I'd post up the question, but its a long question with a graph and you really need to see it yourself.

I had no trouble with part a or b. With part c I realize that m = rise/run = T^2/l
So after rearranging th given pendulum equation I get g = 4pie^2/(T^2/l) (i.e. g = 4pie^2/slope).

However maybe I'm just stupid but whatever values I get from the graph and sub in, my g seems wrong. I keep getting a value of g of between 11 and 14 m/s, which is obviously wrong. So some help would be much appreciated.

 

[henry08]

selfish bump

 

[tommykins]

T²/l = 4pi²/g

Rearrangement you get -
g = 4pi²l/T², subbing in the values of L = 0.16 and T² = 0.81 you get g = 9.63m.s/s

g = 9.6m.s/s

 

[henry08]

Yeah, but we'd you get 0.81 from? Shouldn't you take the y vlue at the x value, i.e. at l = 0.16, should y = something smaller (haven't got paper in front of me at moment)? Or if you're going to take 0.81, shouldn't you take the l value of something higher (rise/run)?

 

[tommykins]

You take it from the table they provide you.

 

[henry08]

Crap, theres a table lol? I was trying to read rom the graph. Runs off to check paper...

 

[henry08]

Oh its so obvious now!!! I feel stupid lol.