A 2.24 kg mass on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 4.45 N/m. The mass is displaced 3.06 m to the right from its equilibrium position and then released, which initiates simple harmonic motion. How many times does the mass oscillate in 24.2 s?

1 answer

First, find the equations that are pertinent to the problem.

F(spring) = -k*x
where F(spring) is the force on a spring, k is the spring constant, and x is the displacement.

w = (k/m)^1/2
where w is the angular acceleration, m is the mass of the spring

T = 2*PI/w
where T is the period of oscillation

w = (4.45/2.24)^0.5

T = 2*PI/w

The number of times the mass will oscillate in 24.2 s is 24.2/T