I would like to construct a seconds pendulum. The length of a 1/2 Hz pendulum is 994 mm. My question, is that length the distance to the center of mass? And how does the density of the arm effect the period? e.g.; will the length a steel arm and weight be different than an aluminum of wood arm and weight?
The simple gravity pendulum assumes a few things when you derive the solution to the differential equation.
- The arm is massless and does not stretch or compress.
- There is no air resistance/friction opposing the motion.
- It is a point mass
- The angle of movement is small
(there are some other assumptions; but those are the key ones for what you are asking).
It is modeled as a single point of mass on an arm with no weight. This means that if the mass of the arm can't be ignored; it will affect the equation. In this case, we have a very long arm that ideally will support the pendulum without stretching or compressing, so it will need to be somewhat strong; and is therefore likely to have noticeable weight.
With a real sized pendulum bob on a weightless arm, you would need to have $994 \ mm$ from the pivot point to the centre of the pendulum bob; because it is the centre of mass of the pendulum where the simple gravity pendulum equation is derived. When the arm weight is important, it will raise your centre of mass closer to the arm from the middle of the bob. It also may create substantial differences due to the moment of inertia; as we are no longer a point mass; but instead a rotating rigid body.
You will have to consider the weight and geometry of your arm + bob to determine the centre of mass. The tricky part will be designing the two in a way so that the centre of mass is $994 \ m $ from your pivot. It may take trial and error.