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Using a pendulum to determine g using T = 2π√(l/g) 
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#1
Nov3010, 09:25 PM

P: 3

I can do the solution, I do not understand the theory!
Here it is: Using a pendulum to determine g using T = 2π√(l/g) (that little n looking thing is pi) (given l and T) So, then we get T^2 = (4π^2/g) x l This is where I get lost. Supposedly, I am to make the equation T^2 = kl (where k is the group of constants) Then, I am to compare this formula with the general equation for a straight line y=kx. Thus, k = m (of a graph, where vertical T^2 and horizontal l is the axis) Why/how does k = m ? k being (4π^2/g) and m being the gradient of my graph? ______________ Thank you! 


#2
Nov3010, 10:02 PM

P: 59

y = force
x = displacement from center Comparison of straight line to Hooke's law y=mx > F=kx This is because small oscillations about a point obey Hooke's law, which is a linear relationship 


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