Response:
y= 240/901 cos 2t+ 8/901 sin 2t
Clarification:
To determine mass m=weight/g
m=8/32=0.25
To calculate the spring constant
Kx=mg (with c=6 inches and mg=8 pounds)
K(0.5)=8 (6 inches converts to 0.5 feet)
K=16 lb/ft
The governing equation for the spring-mass system is
my''+Cy'+Ky=F
Inserting the known values yields
0.25 y"+0.25 y'+16 y=4 cos 20 t ----(1) (given C=0.25 lb.s/ft)
Assuming the steady state equation for y is
y=A cos 2t+ B sin 2t
To determine constants A and B, we must equate this with equation 1.
Next, we find y' and y" by differentiating with respect to t.
y'= -2A sin 2t+2B cos 2t
y"=-4A cos 2t-4B sin 2t
Now, substitute the values of y", y' and y into equation 1
0.25 (-4A cos 2t-4B sin 2t)+0.25(-2A sin 2t+2B cos 2t)+16(A cos 2t+ B sin 2t)=4 cos 20 t
By comparing coefficients on both sides
30 A+ B=8
A-30 B=0
From this, we find
A=240/901 and B=8/901
Thus, the steady state response
y= 240/901 cos 2t+ 8/901 sin 2t
