A 2.000-kg particle rests on a frictionless horizontal surface and is attached to the free end of a spring. The particle is pulled horizontally so that it stretches the spring 0.05 m and is then released from rest at t = 0. Determine the displacement, x, of the particle from the equilibrium position at t = 0.500 s.
givens:
spring force constant = 213.33
period= 0.608
angular frequency = 10.31
To determine the displacement, x, of the particle from the equilibrium position at t = 0.500 s, we can use the formula for the motion of a mass-spring system:
x(t) = A * cos(ωt + φ)
where:
- x(t) is the displacement of the particle from the equilibrium position at time t
- A is the amplitude of the motion (maximum displacement)
- ω is the angular frequency of the system
- φ is the phase constant
Since the particle is released from rest at t = 0, we can set φ = 0.
We are given the angular frequency (ω = 10.31), but we need to find the amplitude A. To find A, we can use the information about the spring.
The force exerted by the spring on the particle is given by Hooke's Law:
F = -kx
where:
- F is the force exerted by the spring
- k is the spring force constant
- x is the displacement from the equilibrium position
At the maximum displacement (A), the force exerted by the spring is equal to the weight of the particle (mg), since there is no vertical motion. Therefore, we have:
kA = mg
Substituting the values:
213.33A = 2.000 * 9.8
Solving for A:
A = 2.000 * 9.8 / 213.33
Now that we know A, we can find the displacement at t = 0.500 s by plugging the values into the equation for x(t):
x(t) = A * cos(ωt + φ)
where t = 0.500 s:
x(0.500) = A * cos(10.31 * 0.500 + 0)
Evaluating the expression:
x(0.500) = A * cos(5.155)
Therefore, to find the displacement, x, at t = 0.500 s, substitute the calculated value of A into the equation:
x(0.500) = (2.000 * 9.8 / 213.33) * cos(5.155)