Answer:
x = -5 sin (2t)
Explanation:
k is the spring stiffness. The unstretched length of the spring is L.
When the mass is added, the spring stretches to an equilibrium position of L+s, where mg = ks. When the mass is displaced a distance x (where x is positive if the displacement is down and negative if it's up), the spring is stretched a total distance s + x.
There are two forces on the mass: weight and force from the spring. Sum of the forces in the downward direction:
∑F = ma
mg − k(s + x) = ma
mg − ks − kx = ma
Since mg = ks:
-kx = ma
Acceleration is second derivative of position, so:
-kx = m d²x/dt²
Let's find k:
F = kx
400 = 2k
k = 200
We know that m = 50. Substituting:
-200x = 50 d²x/dt²
-4x = d²x/dt²
d²x/dt² + 4x = 0
This is a linear second order differential equation of the form:
x" + ω² x = 0
The solution to this is:
x = A cos (ωt) + B sin (ωt)
Here, ω² = 4, so ω = 2.
x = A cos (2t) + B sin (2t)
We're given initial conditions that x(0) = 0 and x'(0) = -10 (remember that down is positive and up is negative).
Finding x'(t):
x' = -2A sin (2t) + 2B cos (2t)
Plugging in the initial conditions:
0 = A
-10 = 2B
Therefore:
x = -5 sin (2t)