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Using a pendulum to determine g using T = 2π√(l/g)

by TaraMarshall
Tags: acceleration gravity, gravity, pendulum, physics
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TaraMarshall
#1
Nov30-10, 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!
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Schr0d1ng3r
#2
Nov30-10, 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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