The position of a particle is given by the expression x = 2.00 cos (6.00πt + π), where x is in meters and t is in seconds.
(a) Determine the frequency.
______Hz
(b) Determine period of the motion.
______s
(c) Determine the amplitude of the motion.
______m
(d) Determine the phase constant.
______rad
(e) Determine the position of the particle at t = 0.270 s.
______m
when t = 0, argument of cos = pi
when does it equal pi + 2 pi?
when 6 pi t = 2 pi
or T = (1/3) second
or f = 1/T = 3 Hz
amplitude is 2 of course
phase = -pi or however your book defines phase
2 cos (6*pi*.27 + pi)
2 cos (8.23 radians)
2 cos (471.6 degrees) because I am too lazy to convert calculator to radians
= -.717 meters
To determine the frequency, period, amplitude, phase constant, and position of the particle at a specific time, we can use the given expression for the position of the particle.
(a) The frequency of the motion can be determined using the formula:
frequency = (6.00π) / (2π)
Simplifying, we get:
frequency = 6.00 / 2 = 3.00 Hz
Therefore, the frequency is 3.00 Hz.
(b) The period of the motion can be determined using the formula:
period = 1 / frequency
Substituting the value of frequency, we get:
period = 1 / 3.00 = 0.333 s
Therefore, the period is 0.333 s.
(c) The amplitude of the motion is given by the coefficient of the cosine function in the expression. In this case:
amplitude = 2.00 m
Therefore, the amplitude is 2.00 m.
(d) The phase constant is determined by the argument inside the cosine function. In this case:
phase constant = π rad
Therefore, the phase constant is π rad.
(e) To find the position of the particle at t = 0.270 s, we substitute this value into the expression for x:
x = 2.00 cos (6.00π(0.270) + π)
Evaluating the expression, we get:
x = 2.00 cos (1.62π + π)
Simplifying further, we have:
x = 2.00 cos (2.62π)
Since the cosine function repeats every 2π radians, we can write:
x = 2.00 cos (0.62π)
Evaluating this expression, we get:
x = 2.00 cos (0.62 * 3.14)
x = 2.00 cos (1.9468)
x ≈ 2.00 (-0.2236)
x ≈ -0.4472 m
Therefore, the position of the particle at t = 0.270 s is approximately -0.4472 m.
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.