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Could anybody provide some guidance or steps for this physics problem? It would be much appreciated?
A uniform rectangular sign of width 48.6 cm, height 25.6 cm, and negligible thickness hangs vertically from supporting hinges attached at its upper edge. Find the period of small-amplitude oscillations of the sign.
I know that this is a pendulum, and that the formula T = 2pi (L/g)^(1/2) can probably be used. I tried the following:
Exponential 1/2 = square root of L/g
L = 0.5 times (0.256) <-- cm to m
g = 9.8
Neither the width nor the mass matter as they do no appear in the formula
The answer is supposed to be 0.829 sec
Any advice on where I went wrong?
There are two other formulas which may be the correct ones to use: α=−((mgd)/I)*θ and T=2π(I/(mgd))^(1/2)
If either of these are the correct formula, I am not sure how to use them, particularly with I and θ
1 Answer
- billrussell42Lv 72 years agoFavorite Answer
Any swinging rigid body free to rotate about a fixed horizontal axis is called a compound pendulum or physical pendulum. The appropriate equivalent length L for calculating the period of any such pendulum is the distance from the pivot to the center of oscillation. This point is located under the center of mass at a distance from the pivot traditionally called the radius of oscillation, which depends on the mass distribution of the pendulum.
For the special case of a beam of uniform density of length L, the center of oscillation is 2/3 of the length of the uniform beam L from the pivoted end.
equivalent length = (2/3)(0.256) = 0.1707 m
T ≈ 2π√(0.1707/9.8) = 0.829 s