A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest from a point 4 inches above the equilibrium position. Find the equation of motion. (Use g = 32 ft/s² for the acceleration due to gravity.) x(t) = cos(4√61) Need Help? Read It ft Watch It
To find the equation of motion, we can use Hooke's law and Newton's second law:
Hooke's law: F = -kx
where F is the force applied to the spring, x is the displacement of the spring from its equilibrium position, and k is the spring constant.
Newton's second law: F = ma
where F is the net force acting on the mass, m is the mass, and a is the acceleration of the mass.
Since the mass is released from rest, its initial velocity is zero. The only force acting on the mass is the force due to the spring, so we can write:
F = -kx = ma
Solving for a, we get:
a = -(k/m) x
where k/m is the spring constant divided by the mass.
To find the spring constant, we can use Hooke's law:
F = -kx
where F is the weight of the mass (in pounds), and x is the displacement of the spring (in inches). When the mass is attached to the spring, the spring is stretched 4 inches from its equilibrium position. So we have:
24 lb = k * 4 in
Solving for k, we get:
k = 6 lb/in
Substituting k/m and k into the equation for a, we get:
a = -(6/24) x = -(1/4) x
This is a second-order differential equation with constant coefficients, and its solution is of the form:
x(t) = A cos(wt) + B sin(wt)
where A and B are constants determined by the initial conditions, and w is the natural frequency of the system
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