Final answer:
The force constant of the spring is 206.67 N/m. The angular frequency is 10.00 rad/s, the frequency is 1.59 Hz, and the period is 0.63 s. The total energy of the system is 0.0933 J. The amplitude of the motion is 0.0300 m. The maximum velocity and maximum acceleration of the particle are 0.300 m/s and 0.300 m/s^2, respectively. The displacement of the particle from the equilibrium position at t = 0.500 s cannot be determined without the phase constant. The velocity and acceleration of the particle at t = 0.500 s can be determined using the derivative of the position equation.
Step-by-step explanation:
(a) To find the force constant of the spring, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. Hooke's Law is given by the equation F = k * x, where F is the force, k is the force constant, and x is the displacement. From the given information, we can set up the equation as follows:
6.20 N = k * 0.0300 m
Solving for k, we find:
k = 206.67 N/m
(b) The angular frequency (ω) of the motion can be found using the equation ω = √(k/m), where k is the force constant and m is the mass of the particle. The frequency (f) is then given by f = ω/(2π), and the period (T) is given by T = 1/f. From the given information, we can calculate:
ω = √(206.67 N/m / 0.490 kg) = 10.00 rad/s
f = 10.00 rad/s / (2π) = 1.59 Hz
T = 1/f = 1 s / 1.59 Hz = 0.63 s
(c) The total energy of the system can be calculated using the equation E = (1/2)kx^2, where k is the force constant and x is the amplitude of the motion. From the given information, we can substitute the values:
E = (1/2)(206.67 N/m)(0.0300 m)^2 = 0.0933 J
(d) The amplitude of the motion is equal to the maximum displacement from the equilibrium position. In this case, the amplitude is given as 0.0300 m.
(e) The maximum velocity (vmax) of the particle can be found using the equation vmax = ωA, where ω is the angular frequency and A is the amplitude of the motion. The maximum acceleration (amax) can be found using the equation amax = ω^2A. From the given information, we can calculate:
vmax = (10.00 rad/s)(0.0300 m) = 0.300 m/s
amax = (10.00 rad/s)^2(0.0300 m) = 0.300 m/s^2
(f) To determine the displacement (x) of the particle from the equilibrium position at t = 0.500 s, we can use the equation x = Acos(ωt + φ), where A is the amplitude, ω is the angular frequency, t is the time, and φ is the phase constant. From the given information, we can substitute the values:
x = (0.0300 m)cos((10.00 rad/s)(0.500 s) + φ)
Since the phase constant (φ) is not given, we cannot determine the exact displacement at t = 0.500 s without additional information.
(g) To determine the velocity and acceleration of the particle when t = 0.500 s, we can take the derivative of the position equation with respect to time. The velocity (v) is given by v = -Aωsin(ωt + φ), and the acceleration (a) is given by a = -Aω^2cos(ωt + φ). From the given information, we can substitute the values:
v = -(0.0300 m)(10.00 rad/s)sin((10.00 rad/s)(0.500 s) + φ)
a = -(0.0300 m)(10.00 rad/s)^2cos((10.00 rad/s)(0.500 s) + φ)