Question

A mass weighing 9 lb stretches a spring 1.5 in. The mass is displaced 5 inch in the downward direction from its equilibrium position and released with no initial velocity. Assuming that there is no damping, and that the mass is acted on by an external force of 5 cos(6t) lb, solve the initial value problem describing the motion of the mass. For this problem, please remember to use English units: ft, lb, sec.

Answer 1

Answer:

$9u" + 2304u = 160cos(6t)$

$u(0) = \frac{5}{12}ft$ and $u'(0) =0$

Step-by-step explanation:

Given

$w =9lb$ --- weight

$L = 1.5in$ --- stretch

$u(0) =5in$ --- initial position

$u'(0) =0$ --- initial velocity

$\gamma =0\ lbs/ft$ --- no damping

$F(t) = 5cos(6t)\ lb$ -- External force

Required

Solve the initial value problem

The initial value problem is calculated using:

$mu" + ku = F(t)$

First, convert all lengths to ft

$L = 1.5in$

$L = \frac{1.5}{12}ft$

$L = \frac{1}{8}ft$

$u(0) =5in$

$u(0) = \frac{5}{12}ft$

Weight is calculated as:

$w = mg$ and $w = kL$

Where

$g = 32fts^{-1}$

Make m the subject in $w = mg$

$m = \frac{w}{g}$

$m = \frac{9}{32}$

Make k the subject in $w = kL$

$k = \frac{w}{L}$

$k = \frac{9}{1/8}$

$k = 72$

Recall that: $mu" + ku = F(t)$

$\frac{9}{32}u" + 72u = 5cos(6t)$

Multiply through by 32

$9u" + 2304u = 160cos(6t)$

$u(0) = \frac{5}{12}ft$ and $u'(0) =0$