Final answer:
The motion for a 0.350-kg mass on a spring with a constant of 265 N/m and an amplitude of 28.0 cm is described by the equation x(t) = 0.28 m * cos(27.4 rad/s * t). The spring's length is longest at half-period intervals and shortest at full-period intervals, with a calculated period of approximately 0.229 seconds.
Step-by-step explanation:
The motion of a mass attached to a spring that oscillates vertically can be described using the principles of simple harmonic motion (SHM). For a 0.350-kg mass and a spring constant of 265 N/m with an amplitude of 28.0 cm, the equation of motion as a function of time is given by:
x(t) = A cos(ωt + φ)
Where x(t) is the displacement from equilibrium at time t, A is the amplitude, ω (omega) is the angular frequency, and φ (phi) is the phase constant.
Given that the mass passes through the equilibrium position with positive velocity at t=0, the phase constant φ is 0. The angular frequency ω can be calculated using the formula:
ω = √(k/m)
Substituting the given values:
ω = √(265 N/m / 0.350 kg)
≈ 27.4 rad/s
Thus, the equation of motion is:
x(t) = 0.28 m * cos(27.4 rad/s * t)
For (b), the length of the spring will be longest when the mass is at the maximum displacement from the equilibrium point, which happens at t = T/2, T/2 + T, etc., where T is the period of the motion.
The length will be shortest when the mass is at the equilibrium position which occurs at t = 0, T, 2T, etc.
The period T can be calculated as:
T = 2 * π / ω
Using the computed ω, we find:
T ≈ 2 * π / 27.4 rad/s ≈ 0.229 s
Hence, the spring is longest at approximately t = 0.115 s, 0.344 s, etc., and shortest at t = 0 s, 0.229 s, etc.