Final answer:
The frequency of oscillation of the two attached 200.0-gram masses on the spring is approximately 1.29 Hz when they oscillate together.
Step-by-step explanation:
When a 200.0-gram mass is attached to a spring, the spring constant k can be calculated using Hooke's Law
F = kx,
where F is the force and x is the displacement (stretch of the spring). Since the weight of the mass (mg) provides the force that stretches the spring and g = 9.8 m/s2, we find k as follows:
k = mg / x = (0.2 kg * 9.8 m/s2) / 0.075 m = 26.1333 N/m.
With a second 200.0-gram mass attached, the total mass becomes 0.4 kg.
The frequency of oscillation can be determined using the formula
f = 1/(2π)*√(k/m):
f = 1/(2π)*√(26.1333 N/m / 0.4 kg) ≈ 1.29 Hz.
Therefore, the frequency of their oscillation is approximately 1.29 Hz when both masses oscillate together.