To solve the problems, we can interpret the given equation as a simple harmonic motion equation of the form x = Acos(ωt + φ), where A is the amplitude, ω is the angular frequency, t is the time, and φ is the phase constant.
(a) To determine the frequency, we can use the relation between angular frequency and frequency, which is given by ω = 2πƒ, where ω is the angular frequency and ƒ is the frequency. In this case, ω = 6.00π rad/s. Solving for ƒ, we have:
ƒ = ω / (2π) = (6.00π rad/s) / (2π) = 3.00 Hz
So the frequency is 3.00 Hz.
(b) The period of motion, T, is the reciprocal of the frequency, which is given by T = 1/ƒ. Substituting the frequency we found in part (a), we have:
T = 1 / 3.00 Hz = 0.333 s
So the period of the motion is 0.333 s.
(c) The amplitude, A, is the maximum displacement from the equilibrium position. In this case, it is given as 2.00 meters, so the amplitude is 2.00 m.
(d) The phase constant, φ, is the value of the phase angle when t = 0. To find it, we can compare the given equation to the general form x = Acos(ωt + φ). In this case, we have:
x = 2.00 cos(6.00πt + π)
Comparing the equation to the general form, we can see that φ = π. So the phase constant is π radians.
(e) To determine the position of the particle at t = 0.270 s, we can substitute the value of t into the given equation:
x = 2.00 cos(6.00π(0.270) + π)
x = 2.00 cos(1.62π + π)
Using the trigonometric identity cos(a + b) = cos(a)cos(b) - sin(a)sin(b), we can simplify further:
x = 2.00 [cos(1.62π)cos(π) - sin(1.62π)sin(π)]
x = 2.00 [-cos(0.38π)]
Using the trigonometric identity cos(π/2 - θ) = sin(θ), we can simplify further:
x = 2.00 sin(0.38π)
x = 2.00 sin(0.38 * 180°)
Using the numerical value of sin(0.38 * 180°) ≈ 0.596, we can calculate:
x ≈ 2.00 * 0.596
x ≈ 1.19 m
So the position of the particle at t = 0.270 s is approximately 1.19 m.