What is the proportionality constant for a pendulum's period equation?

[Draygon_Phly]

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.

 

[Janitor]

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.

 

[Doc Al]

Perhaps you are looking for the equation for the period of a simple pendulum:
T = 2 \pi \sqrt \frac{l}{g}

 

[Draygon_Phly]

Yes that's probibly what I'm looking at thank you.

 

[krab]

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 high school: 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.

 

[Janitor]

"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.