A triangular plate with a non-uniform areal density has a mass M=0.700 kg. It is suspended by a pivot at P and can oscillate as indicated below. Its center of mass is a distance d=0.200 m from the pivot axis and its moment of inertia about an axis through the CM and parallel to the pivot axis is ICM=5.60×10-3 kg m2.

The plate is released from an angle θ=15°. Calculate the period the oscillations.

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.