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Zeppelin
Guest
How exactly does one use the line of best fit to find g? It says in the marking guidelines not to substitute in points. I know you need to find the gradient of the line, but then what?
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smallcattle
Member
read number 6, Calculate g using the relationship T = 2pi * sqr(L/g)
Z
Zeppelin
Guest
Huh? I don't get what you mean.
T = 2pi * sqr(L/g)
T<sup>2</sup> = 4pi<sup>2</sup>(l/g)
g = 4pi<sup>2</sup>l/T<sup>2</sup>
Now 4pi<sup>2</sup> is a constant and l/T<sup>2</sup> is the reciprocal of the gradient of the line of best fit (T<sup>2</sup>/l)
T<sup>2</sup> = 4pi<sup>2</sup>(l/g)
g = 4pi<sup>2</sup>l/T<sup>2</sup>
Now 4pi<sup>2</sup> is a constant and l/T<sup>2</sup> is the reciprocal of the gradient of the line of best fit (T<sup>2</sup>/l)
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