Answer:
Equation of motion
x(t) = 1.24 [e⁻¹•¹⁷ᵗ - e⁻⁶•⁸³ᵗ]
Explanation:
If x = displacement from equilibrium position,
m = mass attached to the spring
k = spring constant
β = damping constant = 2
The force balance for the system is as follows
ma + βv + kx = 0
a = (d²x/dt²)
v = (dx/dt)
m(d²x/dt²) + β(dx/dt) + kx = 0
W = mg
m = (8/32) = 0.25 lbs²/ft
W = k (extension)
8 = k (8-4)
k = (8/4) = 2 lb/ft
β = 2
0.25(d²x/dt²) + 2(dx/dt) + 2x = 0
(d²x/dt²) + 8(dx/dt) + 8x = 0
x" + 8x' + 8x = 0
This is a second order differential equation.
Solving this ODE, we obtain the complimentary solution to be
x = A e⁻¹•¹⁷ᵗ + B e⁻⁶•⁸³ᵗ
where A and B are constants.
The initial conditions given, the mass is initially at the equilibrium position, at t = 0, x = 0 and the initial velocity of the body = 7 ft/s
x(0) = 0 ft, x'(0) = 7 ft /s
x(t) = A e⁻¹•¹⁷ᵗ + B e⁻⁶•⁸³ᵗ
x(0) = A + B = 0
x'(t) = -1.17A e⁻¹•¹⁷ᵗ - 6.83B e⁻⁶•⁸³ᵗ
x'(0) = -1.17A - 6.83B = 7
A + B = 0 (eqn 1)
-1.17A - 6.83B = 7 (eqn 2)
From eqn 1, A = - B; substituting this into eqn 2,
-1.17(-B) - 6.83B = 7
1.17B - 6.83B = 7
-5.66B = 7
B = -(7/5.66) = -1.24
A = - B = - (-1.24) = 1.24
x(t) = A e⁻¹•¹⁷ᵗ + B e⁻⁶•⁸³ᵗ
Becomes
x(t) = 1.24 e⁻¹•¹⁷ᵗ - 1.24 e⁻⁶•⁸³ᵗ
x(t) = 1.24 [e⁻¹•¹⁷ᵗ - e⁻⁶•⁸³ᵗ]
Hope this Helps!!!