Solany6527
04.02.2021 •
Physics
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Ответ:
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)