Question

A spring is such that it would be stretched 15.36 in by a 40 lb weight. Let the weight be attached

to a spring and pulled down 6.5 in below the equilibrium point. If the weight is started with an

upward velocity of 7 flt/sec, describe the motion. No damping force but an impressed force of

F(t) = 10lb is present.

Answer 1

in equilibrium

$k\delta=mg\\\implies k=\frac{mg}{g}$

$k=\frac{386.0885}{15.36}\\k=1005.4388lb/in$

Equation of motion regarding given system

$m\frac{d^2y(t)}{dt^2}+ky(t)=f(t)$

$D^2+\frac{k}{m}y_t=\frac{f(t)}{m}$

$|D^2+w^2|y(t)=\frac{sin5t}{0}$

$D^2+w^2=0\\D=±w$

$w=\sqrt{\frac{k}{m}}$

$w=\sqrt{\frac{10005.4388}{40}}$

$D=±5i$

solution

$y(t)=Asin(5.0136\theta)+Bcos(5.0136\theta)$

$y_p(t)=\frac{1}{40(D^2+w^2)}sin5\theta$

$y_p(t)=\frac{sin5\theta}{(40\times0.13597)}\\=0.1839sin(5t)$

$y(t)=y_p(t)+y_p(t)=Asin(5.0136\theta)+Bcos(5.0136\theta)+0.1839sin(5t)$

$y(0)=5=0+Bx+0\\\implies B=5in$

$y'(t)=5.0136Asin(5.0136\theta)-5.0136Bcos(5.0136\theta)+0.1839\times 5sin(5t)$

$y'(0)=5.0136A+0.9195=48$

$5.0136A=48-0.9195$

$A=9.39in$

$y(t)=9.398sin(5.0136\theta)+5cos(5.0136\theta)+0.1839sin(5t)$