Answer to Question #276485 in Calculus for Fernando

Question

A spring with constant 1.5lb/ft, lies on a long smooth (frictionless) table. An 8 lb weight is

attached to the spring and is at rest at equilibrium position. A 6 lb force is applied to the support

along the line of action of the spring for 5 secs and is removed. Discuss the motion.

Answer 1

Solution;

By Newton's law of motion;

$ma+kx=F$

Let ;

$x=Asin(wt)$

$x''(t)=-Aw^2sin(wt)$

But;

$w=\sqrt{\frac{k}{m}}=\sqrt{\frac{1.5}{8}}=0.4330$