Answer:

The equation is x(t) = 6sin(2t)

x(t) = 6sin(2t)

Step-by-step 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)