Question

A spring contains a weight of 32 lbs. It is stretched to a distance of 1.6 ft and is
currently at rest. The damping force is 9 ˙x and an external force of 2 cos 4t
applied. Set up the differential equation, including the initial conditions, but
you do not need to solve.

Answer 1

Weight is given as 32lbs

stretched to a distance,x=1.6 ft

damping force= 9x

external force=2cos4t

First we need to make equation for each of the forces. So let us list the total number of forces is acting

F₁=Force due to gravity. That is the weight of the body and it is given as 32 lbs. Weight ,mg=32lbs,so m=32/g=1

F₂=Restoring force. Due to the weight of the body the spring will get elongated. And the elongation is given as x+l. and according to the Hookes law this force has a magnitude of,k(x+l). Here the mass is at rest. So when the mass is at rest, in its equilibrium position the restoring force F₂ is equal in magnitude and opposite in direction to the force of gravity. So,F₂=mg

-mg=-k(x+l)

Since the body is at the equilibrium position,x=0

=>mg=kl

so F₂=-kx-mg

mg=32lbs

F₃=restoring force of the medium. It is alsi called as the damping force.

F₃=a*dx/dt=9x

F₄ is the external force, and it is given as 2cos 4t

Apply Newtons law, F=ma

F=F₁+F₂+F₃+F₄

md²x/dt²=mg-kx-mg-a*dx/dt+F(t)

mx''+ax'+kx=F(t)

k=f/x=32/1.6=20

x''+9x+20x=2cos4t