A brick of mass 4 kg hangs from the end of a spring. When the brick is at rest, the spring is stretched by 3 cm. The spring is then stretched an additional 2 cm and released. Assume there is no air resistance. Note that the acceleration due to gravity, g, is g=980 cm/s^2. Set up a differential equation with initial conditions describing the motion and solve it for the displacement s(t) of the mass from its equilibrium position (with the spring stretched 3 cm).

To set up the differential equation, we can use Hooke's Law for the spring force and Newton's second law for the dynamics of the mass.

Hooke's Law states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. It can be represented as:

F = -kx

where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position (positive when the spring is compressed and negative when stretched).

Newton's Second Law states that the force acting on an object is equal to the mass of the object times its acceleration:

F = ma

where m is the mass of the object, and a is the acceleration.

Since the spring force is the only force acting on the mass horizontally (excluding gravity which acts vertically), we can equate the spring force to the mass times acceleration.

ma = -kx

By rearranging, we get:

a = -(k/m)x

We know that acceleration a is the second derivative of displacement with respect to time, so:

d²s/dt² = -(k/m)s

Now we need to find k, the spring constant. The weight of the brick, which is mg, is balanced by the spring force when the brick is at rest, and the spring is stretched by 3 cm (0.03 m):

mg = kx

Substituting g = 980 cm/s² and x = 0.03 m, we get:

(4 kg)(980 cm/s²) = k(3 cm)

k = (4 kg)(980 cm/s²) / (0.03 m) = (4 kg)(980 cm/s²) / (0.03 m * 100 cm/m) = (3920 kg*cm/s²) / (3 cm) ≈ 1306.67 kg/s²

The differential equation is then:

d²s/dt² = -(1306.67 kg/s² / 4 kg)s

Simplifying:

d²s/dt² + 326.67 s⁄s² = 0

This is a second-order linear homogeneous differential equation with constant coefficients. The general solution to this type of differential equation is:

s(t) = A*cos(ωt) + B*sin(ωt)

where A and B are constants determined by initial conditions, and ω is the angular frequency of oscillation, ω = √(k/m) in this case.

Substituting k/m, we have:

ω = √(1306.67 kg/s² / 4 kg) = √(326.67 s⁻²) ≈ 18.073 rad/s

Now we need to apply the initial conditions to determine A and B. At time t=0, the spring is stretched an additional 2 cm (0.02 m) from its stretched equilibrium position. The initial displacement s(0) is thus 0.02 m, and the initial velocity v(0) is 0 since the brick is released from rest. Therefore:

s(0) = A*cos(0) + B*sin(0) = A = 0.02 m v(0) = -A*ω*sin(0) + B*ω*cos(0) = B*ω = 0 ⇒ B = 0

The solution is:

s(t) = 0.02m * cos(18.073 rad/s * t)

Extra: The differential equation we've solved describes simple harmonic motion, a type of periodic motion where the restoring force is directly proportional to the displacement (in this case, provided by Hooke's law). The solution demonstrates that the displacement oscillates with time in a cosine function.

The angular frequency (ω) is an important parameter in simple harmonic motion, as it determines how quickly the mass oscillates. The period of oscillation, which is the time it takes to complete a full cycle, can be calculated as T = 2π/ω. The frequency of oscillation, f, which is how many cycles occur in one second, is the reciprocal of the period: f = 1/T.

It's important to note that in the absence of any damping forces, such as air resistance or internal friction within the spring, the motion would continue indefinitely at the same amplitude and frequency. In the real world, such forces can't be ignored, and they would cause the amplitude to decrease over time, which is not captured in our basic model.