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.

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) =

=

The damping force

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

Then the force due to the spring is :

where;

x = the mass distance from the equilibrium point.

Total Force F =

⇒ 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

⇒   = 0

⇒

The auxiliary equation is can be written as:

⇒ r =

r =

r =  OR  

Now,the general equation of motion is :

Initial Conditions are:

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

x(0) = -1

⇒ -1 =

x' (0) =0 ⇒ 0 = -3(-1) + (4)

⇒ =

Thus:

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step-by-step explanation: