# Quick pendulum question

1. Oct 22, 2004

### ptrainerjoe

How would the period of a pendulum change if A)the string's mass was negligible and b) the stings mass had to be accounted for. I can think of arguments for both but can't find an equation or definate answer for either one.
I think the period will increase since I=I(string)+ml^2 and the period equals T=2pi*sqrt(I/mgl)

Last edited: Oct 22, 2004
2. Oct 22, 2004

### HallsofIvy

Staff Emeritus
If the string's mass was not negligible, in comparison to the "bob", where would the center of mass be? How does the period depend upon the length of a pendulum?

3. Oct 22, 2004

### Staff: Mentor

The period of a physical pendulum is given by:
$$T = 2 \pi \sqrt{I/mgl_{cm}}$$
If we include the mass of the string: I increases, of course, but so does m; but the length (from pivot to center of mass) decreases. To find out which effect dominates, you'll have to plug in expressions for I, m, and l and then compare the period to that of a simple pendulum without the string's mass.

According to my analysis (do it for yourself), if you include the mass of the string, the period would slightly decrease.