A uniform rectangular sign of width 47.6 cm, height 25.0 cm, and negligible thickness hangs vertically from supporting hinges attached at its upper edge. Find the period of small-amplitude oscillations of the sign.

To find the period of small-amplitude oscillations of the sign, we can use the formula for the period of a simple pendulum. The sign can be treated as a physical pendulum, where the rotation of the sign is analogous to the back and forth motion of a pendulum.

The formula for the period of a physical pendulum is given by:

T = 2π√(I / mgd)

Where:
T = Period of oscillation (in seconds)
π ≈ 3.14159 (a mathematical constant)
I = Moment of inertia of the sign
m = Mass of the sign
g = Acceleration due to gravity
d = Distance from the hinges to the center of mass of the sign

To find the moment of inertia (I) of the sign, we need to use the dimensions and shape of the sign. Since the sign is rectangular, the moment of inertia can be calculated using the formula for a rectangular plate:

I = (1/12) * m * (w^2 + h^2)

Where:
w = Width of the sign
h = Height of the sign

Now, let's plug in the values given in the problem:

Width (w) = 47.6 cm
Height (h) = 25.0 cm
Thickness (negligible) = Assume it as zero
Distance from the hinges to the center of mass (d) = (h/2)

First, convert the measurements to SI units:

Width (w) = 0.476 meters (1 cm = 0.01 m)
Height (h) = 0.25 meters (1 cm = 0.01 m)

Calculate the moment of inertia (I):

I = (1/12)*(m)*(w^2 + h^2)

Now, we need to determine the mass (m) of the sign. However, the problem statement does not provide the mass. Without the mass, it is not possible to calculate the period of oscillation.

To obtain the period of oscillation, we need to know the mass of the sign. Please provide the mass value so that we can continue the calculation.