Answer:
The equation is x(t) = 6sin(2t)
x(t) = 6sin(2t)
Explanation:
Let m represent the mass attached
k represent the string constant
β represent the positive damping constant
Applying Newtons second law on the system;
m d²x/dt² = -kx - β dx/dt + f(t)
Where the displacement from equilibrium point and f(t) is represented by x
The equation can be rewritten as
1/m f(t) = d²x/dt²+ 1/m(kx + β dx/dt)
Given that m = 1 slug
k = 5lb/ft
β = 2
f(t) = 24cos2t + 6sin2t
By substituton
1/m f(t) = d²x/dt²+ 1/m(kx + β dx/dt)
Becomes
1/1 * 24cos2t + 6sin2t = d²x/dt² + 1/1(5x + 2 dx/dt)
= 24cos2t + 6sin2t = d²x/dt² + 5x + 2dx/dt
Writing the above solution in an homogeneous equation;
d²x/dt² + 5x + 2dx/dt becomes
x" + 2x' + 5x = 24cos2t + 6sin2t ----- (1)
The auxiliary equation is written as
m² + 2m + 5 = 0
Solving the above quadratic equation;
m = -1 - 2i or -1 + 2i where i is a complex number
The complementary solution is written as;
x(t) = e^(-t)(c1 cos2t + c2 sin2t)
x(t) = e^(-t)c1 cos2t + e^(-t)c1 sin2t
Represent e^(-t)c1 with A and e^(-t)c2 with B
x(t) = e^(-t)c1 cos2t + e^(-t)c1 sin2t becomes
x(t) = Acos2t + Bsin2t
Differentiate
x'(t) = 2Bcos2t - 2Asin2t
Differentiate
x"(t) = -4Acos2t - 4Bsin2t
Go back to (1)
x" + 2x' + 5x = 24cos2t + 6sin2t becomes
(A + 4B) cos2t + (-4A + B)sin2t = 24cos2t + 6sin2t
By comparison
A + 4B = 24 ---- (1)
-4A + B = 6 ----- (2)
Solving (1) and (2) simultaneously,
From (2)
B = 6 + 4A
Substitute 6 + 4A for B in (1)
A + 4(6 + 4A) = 24
A + 24 + 16A = 24
17A = 0
A = 0
Substitute 0 for A in (B = 6 +4A)
B = 6 + 4A
B = 6 + 4(0)
B = 6 + 0
B = 6.
So, the equation is x(t) = (0) cos2t + 6sin(2t)
x(t) = 6sin(2t)