Answer to Question #286768 in Calculus for Monu

Question

A spring is such that a 16 lb weight stretches it by 1.5 in. The weight is pulled down to a point

4 in below the equilibrium point and given an initial downward velocity of 4 ft/sec. An impressed

force of F(t) = 2 cos 74t is acting on the spring. Describe the motion.

Answer 1

We have the parameters

$\displaystyle\\w=16lb\\L=1.5m=\frac{125}{1000}ft\\\gamma=0\frac{s•lb}{ft}\\F(t)=2cos(74t)\\u(0)=4 \hspace{0.1cm}in=0.167ft\\u\prime(0)=4\frac{ft}{sec}$

Some further computation gives

$m=\frac{w}{g}\approx\frac{16lb}{32\frac{m}{s^{2}}}=\frac{1}{2}\frac{lb•s^{2}}{m}$

$k=\frac{mg}{L}=\frac{16lb}{\frac{125}{1000}ft}=\frac{16000}{125}\frac{lb}{ft}=128\frac{lb}{ft}$

It follows that the initial value problem modeling the motion of the mass is

$\displaystyle\\\frac{1}{2}u^{\prime\prime}+128u=2cos(74t); u(0)=0.167$

$\displaystyle\\u^{\prime}(0)=4$