The characteristics of a horizontal and undamped spring-block system are k=400N/m and M=0.3kg.The position of the block is determined by its abscissa along an x'ox axis. When the spring has its natural length then x=0.
a) the spring is stretched by shifting the block by 10 cm from its equilibrium position and then releasing it at to=0.
determine the time equation of motion.
b) the spring is compressed by shifting the block by 10cm from its equilibrium position and then releasing it at to=0.
determine the time equation of motion.
To determine the time equation of motion for both cases (spring stretched and spring compressed), we can use the equation of motion for a spring-block system. The equation is represented as:
M * d^2x/dt^2 + k * x = 0
where M is the mass of the block, k is the spring constant, x is the displacement of the block from the equilibrium position, t is time, and d/dt represents the derivative with respect to time.
Let's solve each case separately:
a) Spring stretched:
Given: k = 400 N/m, M = 0.3 kg, x = 0.1 m (10 cm)
We can rearrange the equation as follows to solve for the time equation of motion:
d^2x/dt^2 + (k/M) * x = 0
Substituting the given values, we get:
d^2x/dt^2 + (400 N/m) / (0.3 kg) * x = 0
d^2x/dt^2 + (400/0.3) * x = 0
d^2x/dt^2 + (4000/3) * x = 0
This equation represents a second-order linear homogeneous ordinary differential equation. The general solution is of the form:
x(t) = A * cos(ωt) + B * sin(ωt)
where A and B are constants, and ω is the angular frequency.
To find the values of A, B, and ω, we need to use the initial conditions. At t = 0, x = 0.1 m, and dx/dt = 0 (initially at rest). Thus, we have two conditions:
x(0) = 0.1 = A * cos(0) + B * sin(0) = A
dx/dt(0) = 0 = -A * ω * sin(0) + B * ω * cos(0) = B * ω
From the first condition, A = 0.1.
From the second condition, B = 0 (since ω ≠ 0).
Therefore, the time equation of motion for the spring stretched case is:
x(t) = 0.1 * cos(ωt)
b) Spring compressed:
Given: k = 400 N/m, M = 0.3 kg, x = -0.1 m (-10 cm)
Similarly, we can rearrange the equation to solve for the time equation of motion:
d^2x/dt^2 + (k/M) * x = 0
Substituting the values, we get:
d^2x/dt^2 + (400 N/m) / (0.3 kg) * x = 0
d^2x/dt^2 + (400/0.3) * x = 0
d^2x/dt^2 + (4000/3) * x = 0
Following the same process as in case a, we find that the time equation of motion for the spring compressed case is also:
x(t) = 0.1 * cos(ωt)
Note: In both cases, the displacement x is multiplied by 0.1 because the given displacements are 10 cm.
To find the angular frequency ω, we can use the relation:
ω = sqrt(k/M) = sqrt(400/0.3) = sqrt(4000/3)
Therefore, the final time equation of motion for both cases is:
x(t) = 0.1 * cos(sqrt(4000/3) * t)