In simple harmonic motion, the displacement of an object from its equilibrium position can be described by the equation:
x = A * cos(2πt/T)
Where: x is the displacement from the equilibrium position, A is the amplitude of the motion, t is the time, and T is the period of the motion.
In this case, we are given that the amplitude (A) is 2.0 cm and the period (T) is 1.0 s. We need to find the distance moved from the center of the oscillations at a specific time of 0.4 s.
Plugging the given values into the equation, we have:
x = 2.0 cm * cos(2π * 0.4 / 1.0)
Calculating the cosine value:
cos(2π * 0.4 / 1.0) ≈ cos(2π * 0.4) ≈ cos(0.8π)
Using the value of cosine, we find:
x ≈ 2.0 cm * cos(0.8π)
Evaluating the cosine value:
cos(0.8π) ≈ -0.5878
Multiplying by the amplitude, we get:
x ≈ 2.0 cm * (-0.5878)
x ≈ -1.1756 cm
Therefore, the distance moved from the center of the oscillations at 0.4 s is approximately -1.1756 cm (negative because it indicates a displacement to the left of the center).