Answer:
[tex]\mathbf{y(t) =0.408e^{-6.20t }-0.408e^{-25.80t}}}[/tex]
Explanation:
Given that:
The weight of ball = 4 pounds
The spring stretch x = 1/15 feet
Using the relation of weight on an object:
W = mg
m = W/g
m = 4 / 32
m = 1/8
Now, from Hooke's law:
F = kx
4 =k(1/5)
k = 20 lb/ft
However, since the air resistance is 4 times the velocity;
Then, we can say:
C = 4
Now, for the damped vibration in the spring-mass system, we have:
[tex]m\dfrac{d^2 y}{dx^2}+ c\dfrac{dy}{dt}+ky = 0[/tex]
[tex](\dfrac{1}{8})\dfrac{d^2 y}{dx^2}+ 4\dfrac{dy}{dt}+20y = 0[/tex]
[tex]\dfrac{d^2 y}{dx^2}+ 32\dfrac{dy}{dt}+160y = 0[/tex]
Solving the differential equation:
m² + 32m + 160 = 0
Solving the equation:
m = -25.80 or m = -6.20
So, the general solution for the equation is:
[tex]y (t)= c_1 e^{-6.20t}+c_2e^{-25.80t}[/tex]
[tex]y '(t)=-6.20 c_1 e^{-6.20t}-25.80c_2e^{-25.80t}[/tex]
y(0) = 0 ; y'(0) = 8
[tex]y (0)= c_1 e^{-6.20(0)}+c_2e^{-25.80(0)}[/tex]
[tex]c_1 +c_2 = 0 ---(1)[/tex]
At y'(0) = 8
[tex]y '(0)=-6.20 c_1 e^{-6.20(0)}-25.80c_2e^{-25.80(0)} \\ \\ 8=-6.20 c_1 e^{0}-25.80c_2e^{0} \\ \\ 8=-6.20 c_1 -25.80c_2--- (2)[/tex]
From (1), let [tex]c_1 = -c_2[/tex], then replace the value of c_1 into equation (2)
[tex]8=-6.20 (-c_2)-25.80c_2[/tex]
[tex]8=6.20c_2-25.80c_2[/tex]
[tex]8=-19.60c_2[/tex]
[tex]c_2=\dfrac{ 8}{-19.60}[/tex]
[tex]c_2 = -0.408[/tex]
From [tex]c_1 = -c_2[/tex]
[tex]c_1 = -(-0.408)[/tex]
[tex]c_1 = 0.408[/tex]
∴
The required solution in terms of t is:
[tex]\mathbf{y(t) =0.408e^{-6.20t }-0.408e^{-25.80t}}}[/tex]
