Question

A 6lb weight is attached to a spring suspended from a ceiling. The weight stretches the spring 4 inches. The external force is f(t)=27sin(4t)-3cos(4t). The medium offers a resistance of 3dy/dt (ft/sec). Find the equation of the motion.

Answer 1

Answer:

displacement x = - 0.046sin4t +0.006cos4t

Step-by-step explanation:

The model of the equation of motion is a forced motion equation and to determine the displacement of the weight as a function of time; we have:

the weight balances of the  elastic force in the spring  to be expressed by the relation:

mg = kx

where;

x=4 in (i.e  1/3 ft )

mass m = 6lb

let make k the subject; then:

k = mg/x = 6×32/(1/3) = 576

assuming  x to be the displacement form equilibrium;

Then;

$F = 27sin 4t-3cos4t +k(x+1/3) - mg -3v$

(since F(t)=27sin 4t-3cos4t somehow faces downwards, mg=downwards and k(x+1/3)= upwards and medium resistance 3v =  upwards)

SO;

$d2x/dt2 = 27sin 4t-3cos4t +kx - 3dx/dt$

$d2x/dt2 +3dx/dt - 576x = 27sin 4t-3cos4t$

Assuming : $x = asin4t + bcos4t$

$dx/dt = 4acos4t - 4bsin4t$

$d2x/dt2 = -16asin4t - 14bcos4t$

replacing  these values in the above equation

$= -16asin4t - 14bcos4t + 12acos4t - 12bsin4t -576asin4t-576bcos4t = 27sin 4t-3cos4t$

$= sin4t (-592a-12b) + cos4t(12a -590b) = 27sin 4t-3cos4t$

equating sin and cos terms

a = - 0.046 ;  b = 0.006

displacement x = - 0.046sin4t +0.006cos4t