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My solution to the first part of part (a) is:spring.
x(T) = -(c*m^2/k^2)SIN(SQRT(k/m)*T) + (c*m/k)*T
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
its
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
validity. Any use of this solution in a life threatening situation is
strictly unadvised