Question

A force of 5 pounds stretches a spring 1 foot. A mass weighing 6.4 pounds is attached to the spring, and the system is then immersed in a medium that offers a damping force numerically equal to 1.2 times the instantaneous velocity. (a) Find the equation of motion if the mass is initially released from rest from a point 1 foot above the equilibrium position.

Answer 1

Answer:

$x(t) = e ^{-3t} [-cos (4t) - \frac{3}{4}sin (4t)]$

Step-by-step explanation:

A force of 5 pounds stretches a spring 1 foot.

This implies that, the string constant (K) = 5 lbs? ft

Mass (m) = $\frac{6.4}{32}$

= $\frac{1}{5}$

The damping force $F_d = -1.2 v$

The negativity of the damping force is due to the fact that it opposes the motion.

Then the force due to the spring is :

$F_5 = -kx$    

where;

x = the mass distance from the equilibrium point.

Total Force F = $F_d + F_s$

⇒ F = -1.2v - 5.0 x

⇒ mx = -1.2v - 5.0 x

where v = x' and a = x"

Hence, the equation becomes mx" =  -1.2x' - 5.0 x

⇒  $\frac{1}{5} x"+1.2x'+5v$ = 0

⇒ $x" +6x'+25x =0$

The auxiliary equation is can be written as:

$r^2 +6x+25=0$

⇒ r = $\frac{-6 \pm\sqrt{36-100} }{2}$

r = $-3\pm4i$

r = $-3+4i$  OR   $-3-4i$

Now,the general equation of motion is :

$x(t)=e^{-3t}(C_1cos(4t)+C_2sin(4t))$

$x'(t)=-3e^{-3t}(C_1cos(4t)+C_2sin(4t)) + e^{-3t}(-4C_1sin(4t)+4C_2 cos(4t))$

Initial Conditions are:

x(0) = -1 and x' (0) = 0

x(0) = -1

⇒ -1 = $C_1$

x' (0) =0 ⇒ 0 = -3(-1) + (4$C_2$)

⇒ $C_2$ = $-\frac{3}{4}$

Thus:

$x(t) = e ^{-3t} [-cos (4t) - \frac{3}{4}sin (4t)]$