Final answer:
Using the spring constant calculated from the initial 3.9 kg mass that extends the spring by 0.4 m, the period of oscillation for a new 3.0 kg mass is approximately 1.09 seconds when this mass is attached to the same spring.
Step-by-step explanation:
To calculate the period of a mass-spring system oscillating in simple harmonic motion (SHM), we can use the formula T = 2\u03c0\u221a(m/k), where T is the period, m is the mass, and k is the spring constant. Given that a 3.9 kg mass extends the spring by 0.4 m, we can find the spring constant (k) using Hooke's Law (F = kx), where F is the force and x is the extension. Since the force exerted by the mass due to gravity is F = mg, where g is the acceleration due to gravity (9.81 m/s2), we can solve for k.
The force exerted by the mass is F = 3.9 kg \u00d7 9.81 m/s2 = 38.259 N. Using this force to solve for k, we have k = F/x = 38.259 N / 0.4 m = 95.6475 N/m. Now, using the spring constant and the new mass (3.0 kg), we can calculate the new period of oscillation.
Plugging the values into the period formula, T = 2\u03c0\u221a(m/k) = 2\u03c0\u221a(3.0 kg / 95.6475 N/m), we get T ≈ 1.09 s. Thus, the period of the oscillator with a 3.0 kg mass is approximately 1.09 seconds.