No I don't I'm afraid as I only used 3 masses, so the period was 0.7, 0.75 and 0.8.
I think you're right though because when I added a theoretical higher point that fitted with the equation the gradient improved, so I'll go back and repeat the experiment with smaller and greater masses...
Thanks :)
All SI units so kg, seconds etc.
I didn't calculate the slope myself, I put it into excel - I think it made the points (0.64,0.49) and (0.1,0.06) I've sorted that so that's okay now.
BUT... now the gradient = 3.75 now... so k = 10.5. This is still double what it should be, and double...
(the right value for k is about 4 or 5)
mass 0.06
period 0.7
period^2 0.49
calculated value for k 4.834091952
mass 0.1
period 0.8
period^2 0.64
calculated value for k 6.168502751
graph of period^2 (y) and mass (x) gives gradient 1.2558, value for k from this is 31.4...
halp...
Homework Statement
Using experimental values for mass and period of oscillations, I'm trying to graph period^{2} (y-axis) against mass (x-axis) to get a gradient = 4(pi^{2})/k so I can find k (spring constant).
The problem is that when I use each pair of results in the equation below, I...