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 harmonic oscillator.
The formula for the period (T) of a simple harmonic oscillator is given by:
T = 2π√(m/k)
Where:
T = Period of oscillation
m = Mass of the object in the oscillator
k = Spring constant of the oscillator
Since the sign is hanging vertically and oscillating, we can consider it as a simple pendulum and use the formula for the period of a simple pendulum.
The formula for the period (T) of a simple pendulum is given by:
T = 2π√(L/g)
Where:
T = Period of oscillation
L = Length of the pendulum
g = Acceleration due to gravity (approximately 9.8 m/s²)
In this case, we can consider the height of the sign (25.0 cm) as the length of the pendulum (L).
Therefore, substituting the values into the formula:
T = 2π√(0.25/9.8)
Calculating this further, we get:
T ≈ 2π√0.0255 ≈ 2π * 0.159936 ≈ 1.005 seconds
Hence, the period of small-amplitude oscillations of the sign is approximately 1.005 seconds.