Answer:
![Y=(\dfrac{3}{16}+t \dfrac{3}{8})e^{-2t}-\dfrac{3}{16}cos 4t](https://tex.z-dn.net/?f=Y%3D%28%5Cdfrac%7B3%7D%7B16%7D%2Bt%20%5Cdfrac%7B3%7D%7B8%7D%29e%5E%7B-2t%7D-%5Cdfrac%7B3%7D%7B16%7Dcos%204t)
Explanation:
Given that m= 1 slug and given that spring stretches by 2 feet so we can find the spring constant K
mg=k x
1 x 32= k x 2
K=16
And also give that damping force is 8 times the velocity so damping constant C=8.
We know that equation for spring mass system
my''+Cy'+Ky=F
Now by putting the values
1 y"+8 y'+ 16y=6 cos 4 t ----(1)
The general solution of equation Y=CF+IP
Lets assume that at steady state the equation of y will be
y(IP)=A cos 4t+ B sin 4t
To find the constant A and B we have to compare this equation with equation 1.
Now find y' and y" (by differentiate with respect to t)
y'= -4A sin 4t+4B cos 4t
y"=-16A cos 4t-16B sin 4t
Now put the values of y" , y' and y in equation 1
1 (-16A cos 4t-16B sin 4t)+8( -4A sin 4t+4B cos 4t)+16(A cos 4t+ B sin 4t)=6sin4 t
So by comparing the coefficient both sides
-16A+32B+16A=0 So B=0
-16 B-32 A+16B=6 So A=-3/16
y=-3/16 cos 4t
Now to find the CF of differential equation 1
y"+8 y'+ 16y=6 cos 4 t
Homogeneous version of above equation
![m^2+8m+16=0](https://tex.z-dn.net/?f=m%5E2%2B8m%2B16%3D0)
So ![CF =(C_1+tC_2)e^{-2t}](https://tex.z-dn.net/?f=CF%20%3D%28C_1%2BtC_2%29e%5E%7B-2t%7D)
So the general equation
![Y=(C_1+tC_2)e^{-2t}-3/16 cos 4t](https://tex.z-dn.net/?f=Y%3D%28C_1%2BtC_2%29e%5E%7B-2t%7D-3%2F16%20cos%204t)
Given that t=0 Y=0 So
![C_1=\dfrac{3}{16}](https://tex.z-dn.net/?f=C_1%3D%5Cdfrac%7B3%7D%7B16%7D)
t=0 Y'=0 So
![C_2 =\dfrac{3}{8}](https://tex.z-dn.net/?f=C_2%20%3D%5Cdfrac%7B3%7D%7B8%7D)
![Y=(\dfrac{3}{16}+t \dfrac{3}{8})e^{-2t}-\dfrac{3}{16}cos 4t](https://tex.z-dn.net/?f=Y%3D%28%5Cdfrac%7B3%7D%7B16%7D%2Bt%20%5Cdfrac%7B3%7D%7B8%7D%29e%5E%7B-2t%7D-%5Cdfrac%7B3%7D%7B16%7Dcos%204t)
The above equation is the general equation for motion.