To solve the problem, we need to use the law of conservation of energy which states that the initial energy of the system is equal to the final energy of the system.
The initial energy of the system is gravitational potential energy:
U = mgh
where m is the mass, g is the gravitational acceleration, and h is the height. Since the spring is stretched 0.25 m, the mass is raised by this amount, so h = 0.25 m. Substituting the values, we have:
U = (3.0 kg)(9.8 m/s2)(0.25 m) = 7.35 J
The final energy of the system is the sum of kinetic energy and elastic potential energy:
K + U_elastic = 1/2mv^2 + 1/2kx^2
where v is the velocity of oscillation, k is the spring constant, and x is the amplitude of oscillation. The amplitude is the total distance from the equilibrium position, so x = 0.25 m + 0.2 m = 0.45 m.
The elastic potential energy can be calculated using Hooke's law:
U_elastic = 1/2kx^2 = 1/2(410 N/m)(0.45 m)^2 = 41.9 J
Setting the initial and final energies equal to each other, we get:
U = K + U_elastic
7.35 J = 1/2mv^2 + 41.9 J
1/2mv^2 = 34.55 J
v^2 = 69.1 m^2/s^2
Taking the square root, we have:
v = 8.3 m/s
Therefore, the maximum velocity achieved by the mass as it undergoes harmonic oscillation is 8.3 m/s.