Question

A spring is such that a 2 lb weight stretches it by 6 in. An impressed force of F(t) = ¼ sin 8t is

acting on the spring. If the 2-lb weight is released from a point 3 in below the equilibrium point,

describe the motion.

Answer 1

$my''+ky=F(t)$

we have:

$mg=2$

$m=2/g=2/32=1/16$ lb

$k=mg/l=2/0.5=4$

then:

$y''/16+4y=sin8t/4,y(0)=1/4,y'(0)=0$

$r^2/16+4=0$

$r=\pm8i$

$y_h=c_1cos8t+c_2sin8t$

$y_p=Atcos8t+Btsin8t$

$y'_p=Acos8t-8Atsin8t+Bsin8t+8Btcos8t=$

$=(A+8Bt)cos8t+(B-8At)sin8t$

$y''_p=8Btcos8t-8(A+8Bt)sin8t-8Asin8t+8(B-8At)cos8t$

$8Btcos8t-8(A+8Bt)sin8t-8Asin8t+8(B-8At)cos8t+$

$+64(Atcos8t+Btsin8t)=4sin8t$

$8B-64A+64A=0\implies B=0$

$-8A-8A=4$

$A=-1/4$

$y_p=-tcos8t/4$

$y=y_h+y_p=c_1cos8t+c_2sin8t-tcos8t/4$

$y(0)=c_1=1/4$

$y'=-2sin8t+8c_2cos8t-cos8t/4+2tsin8t$

$y'(0)=8c_2-1/4=0$

$c_2=1/32$

$y(t)=cos8t/4+sin8t/32-tcos8t/4$