View Full Version : Principle of Pendulum
Draygon_Phly
Apr5-04, 05:31 PM
[FONT=Comic Sans MS] In the equation to find out the period of the pendulum (T=k*square root of l) they give me the length (l),or period (T) but no proportionality constant (k). Without the proportionality constant I cannot figure out the equation.
So what I'm asking for is the proportionality constant for this equation.
Thank you. :cool: :frown:
Aside from the issue of what k is, I find it bizarre to see the square root of length in your formula. I think what goes in the square root is going to turn out to have units of time squared. For instance, maybe l/g, which has units of time squared, since l is length and acceleration of gravity g has units of length/time^2.
Perhaps you are looking for the equation for the period of a simple pendulum:
T = 2 \pi \sqrt \frac{l}{g}
Draygon_Phly
Apr6-04, 10:30 AM
Yes thats probibly what I'm looking at thank you. :rolleyes:
Aside from the issue of what k is, I find it bizarre to see the square root of length in your formula. I think what goes in the square root is going to turn out to have units of time squared. For instance, maybe l/g, which has units of time squared, since l is length and acceleration of gravity g has units of length/time^2.
There's nothing wrong with saying T= k\sqrt{L}. It just means that k has units of time per root length. It reminds me of the first experiment I did in highschool: measure the pendulum period as a function of length. I found T in seconds is about 0.32 times the square root of length when length measured in inches. This was all I could conclude: I had no way of finding the dependence of T on gravity, since of course I could not vary it.
"There's nothing wrong with saying..."
Fair enough. In situations where the pendulum is hanging from a point fixed in a gravitational field at a particular value of g, there is no harm in absorbing the reciprocal square root of g into your k constant.
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