How does the mass of a string affect the period of a pendulum?

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The discussion centers on how the mass of a string affects the period of a pendulum. When the string's mass is negligible, the period is determined solely by the mass of the bob and the length of the pendulum. However, when the string's mass is significant, the period can be calculated using the formula T = 2π√(I/mglcm), where I is the moment of inertia, m is the total mass, g is the acceleration due to gravity, and lcm is the distance from the pivot to the center of mass. The analysis concludes that including the mass of the string results in a slight decrease in the period of the pendulum.

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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 definite 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)
Help please!
 
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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?
 
The period of a physical pendulum is given by:
[tex]T = 2 \pi \sqrt{I/mgl_{cm}}[/tex]
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.
 

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