To calculate the period of oscillations for the triangular plate, you can use the equation:
T = 2π√(I/ (Mgd))
where T is the period of oscillation, I is the moment of inertia, M is the mass of the plate, g is the acceleration due to gravity, and d is the distance from the pivot axis to the center of mass.
In this case, the given values are:
M = 0.700 kg (mass)
ICM = 5.60 × 10^(-3) kg m^2 (moment of inertia about the axis through the center of mass)
d = 0.200 m (distance from the pivot axis to the center of mass)
θ = 15° (initial angle of release)
First, we need to calculate the moment of inertia about the pivot axis. The parallel axis theorem states that I = ICM + Md^2.
Given:
ICM = 5.60 × 10^(-3) kg m^2
M = 0.700 kg
d = 0.200 m
Using the parallel axis theorem, we can calculate:
I = ICM + Md^2
I = 5.60 × 10^(-3) + 0.700 × (0.200)^2
I = 5.60 × 10^(-3) + 0.700 × 0.040
I = 5.60 × 10^(-3) + 0.028
I = 5.60 × 10^(-3) + 0.028
I = 5.628 × 10^(-3) kg m^2
Now, we can calculate the period using the equation:
T = 2π√(I / (Mgd))
Given:
I = 5.628 × 10^(-3) kg m^2
M = 0.700 kg
g = 9.8 m/s^2
d = 0.200 m
T = 2π√(5.628 × 10^(-3) / (0.700 × 9.8 × 0.200))
Simplifying:
T = 2π√(5.628 × 10^(-3) / 1.372)
T = 2π√(0.004107 / 1.372)
T = 2π√0.002992
T = 2π × 0.054768
T ≈ 0.3446 s (to 4 significant figures)
Therefore, the period of oscillation is approximately 0.3446 seconds.