Where I assumed the initial conditions are that the initial speed is zero
and the initial location of the mass is the relaxed position of the spring.
and then proceeded to solve via Newton's 2nd law
x'' = - (k/m)*x + (c/m)*t
Consequently the PE in the spring is
0.5*k*x(T)^2
and is different from what Leigh wrote.
Disclaimer: I worked this out quickly, and make no warrantees regarding its
validity. Any use of this solution in a life threatening situation is
strictly unadvised
Joel
______________________
Finally, let me illustrate the application of these versions in a
simple example. Consider a particle of mass m to which is affixed
a massless Hooke spring of spring constant k. Starting at time t=0
I pull on the spring with a force F=Ct, where C is a constant. The
system is initally at rest.
After a time T:
(a) How much elastic potential energy will be stored in the system?
(b) What will the kinetic energy of the system be?
(c) How much work will I have done on the system?
Answers:
(a) F(T) = CT = k x where x is the spring's elongation.