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Sagot :
Correct question is;
A 4-foot spring measures 8 feet long after a mass weighing 8 pounds is attached to it. The medium through which the mass moves offers a damping force numerically equal to √2 times the instantaneous velocity. Find the equation of motion if the mass is initially released from the equilibrium position with a downward velocity of 7 ft/s. (Use g = 32 ft/s²)
Answer:
x(t) = 7te^(-2t√2)
Explanation:
We are given;
Weight; W = 8 lbs
mass; m = W/g
g = 32 ft/s²
Thus;
m = 8/32
m = ¼ slugs
From Newton's second law we can write the equation as;
m(d²x/dt²) = -kx - β(dx/dt)
Rearranging this, we have;
(d²x/dt²) + (β/m)(dx/dt) + (k/m)x = 0
Where;
β is damping constant = √2
k is spring constant = W/s
Where s = 8ft - 4ft = 4ft
k = 8/4
k = 2
Thus,we now have;
(d²x/dt²) + (√2/(¼))(dx/dt) + (2/(¼))x = 0
>> (d²x/dt²) + (4√2)dx/dt + 8x = 0
The auxiliary equation of this is;
m² + (4√2)m + 8 = 0
Using quadratic formula, we have;
m1 = m2 = -2√2
The general solution will be gotten from;
x_t = c1•e^(mt) + c2•t•e^(mt)
Plugging in the relevant values gives;
x_t = c1•e^(mt) + c2•t•e^(mt)
At initial condition of t = 0, x_t = 0 and thus; c1 = 0
Also at initial condition of t = 0, x'(0) = 7 and thus;
Since c1 = 0, then c2 = 7
Thus,equation of motion is;
x(t) = 7te^(-2t√2)
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