Respuesta :
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!!!
Answer:
x(t) = 7te^(-6.83t)
Explanation:
m = mass attached
k = spring's constant
β = positive damping constant = 2
From W=mg, we can find m
m = W/g = 8/32 = 0.25 slugs
From Hooke's law, we can find k.
W = kΔs; k = W/Δs
k = 8/(8-4) = 8/4 = 2 lb/ft
Applying Newton's second law to the system, we have a Differential eqaution (DE) as;
m(d²x/dt²) = - kx - β(dx/dt)
Divide through by m;
(d²x/dt²) = -kx/m - β(dx/dt)
Rearranging, we have;
(d²x/dt²) + kx/m + (β/m)(dx/dt) = 0
x(t) is the displacement from the equilibrium position
Plugging in the relevant values to obtain ;
(d²x/dt²) + 2x/0.25 + (2/0.25)(dx/dt) = 0
(d²x/dt²) + 8x+ 8(dx/dt) = 0
The auxiliary equation of this is;
m² + 8m + 8 = 0
The roots are m = -1.17 and m= -6.83
So, the general solution is;
x(t) = A•e^(-1.17t) + B•t•e^(-6.83t)
From the initial condition,
x(0) = 0 ft and x'(0) = 7 ft /s
Thus,
At x(0) = 0; 0 = Ae^(0) + 0
So,A = 0
Thus, x(t) = B•t•e^(-6.83t)
Now, x'(t) = Be^(-6.83t)
At x'(0) = 7 ft /s
7 = Be^(0)
B = 7
Thus, the equation of motion is;
x(t) = 7te^(-6.83t)
