Answer to Question #276484 in Calculus for Fernando

Question

A spring is such that it would be stretched 8 in by a 20 lb weight. Let the weight be attached to

a spring and pulled down 5 in below the equilibrium point. If the weight is started with an upward

velocity of 4 flt/sec, describe the motion. No damping force or impressed force is present.

Answer 1

spring constant $=k=\frac{20}{8}=2.5$

$\implies k=2.5lb/in$

$m\ddot{x}+c\dot{x}+kx=0$

$\implies \ddot{x}+0+\frac{k}{m}x=0$

$\implies \ddot{x}+0+\frac{2.5}{0.625}x=0$

$\implies\ddot{x}+4x=0$

$\implies (x^2+4)x=0$

$\implies m=±2i$

$\therefore x(t) =c_1cos(2t)+c_2sin(2t)$

$\dot{x}(t) =-2c_1sin(2t)+2c_2cos(2t)$

$x(0)=-5in$

$\dot{x}(0)=r(0)=4ft/y$

$\therefore x(0)=c_1+0 \\c_1=-5$

$\dot{x}(0)=\dot{r}(0)=2c_2$

$\implies 4=2c_2\\c_2=2$

$\therefore required\ general \ \dot{y}$

$x(t)=-5cos(2t)+2sin(2t)$