Final answer:
To find the position of the mass at t = 0.150 s, we used the equation for simple harmonic motion, considering the mass started at the amplitude of its motion. After calculating the angular frequency and applying the cosine function to the given time, we determined that the position is approximately -21.5 cm, which corresponds to option 4.
Step-by-step explanation:
To determine the position of a mass attached to a spring undergoing simple harmonic motion (SHM) at a specific time, we can use the equation of motion for SHM:
x(t) = A cos(ωt + φ)
Where:
- x(t) is the displacement at time t
- A is the amplitude
- ω is the angular frequency, ω = √(k/m)
- φ is the phase constant
Since the mass is at its amplitude x = A at t = 0s, we can set φ to 0 for our case (cosine starts at its maximum value at φ = 0). Thus, the equation simplifies to:
x(t) = A cos(ωt)
Given that A = 30.0 cm, m = 2.00 kg, and k = 500.0 N/m, we can calculate ω as follows:
ω = √(k/m) = √(500.0 N/m / 2.00 kg) = 15.8 s-1
Now, we can find the displacement at t = 0.150 s:
x(0.150 s) = 30.0 cm × cos(15.8 s-1 × 0.150 s)
Calculate the cosine part which gives approximately -0.707, and the displacement x(0.150 s) ≈ 30.0 cm × (-0.707) ≈ -21.2 cm. The negative sign indicates that the mass is on the opposite side of the equilibrium position compared to where it started.
Therefore, the correct answer is option 4) -21.5 cm.