Regarding Energy and Work done in a Spring Mass System

[S S Gautam]

Homework Statement

I am trying to solve a spring mass system (mass m and constant spring stiffness k) where instead of applied force a displacement is applied as u=sin(ωt) where ω is the frequency and t is the time. I need to find the difference of the total work done and the energy (KE and PE) of the system. The spring constant is k and the mass of the spring is m. The spring is tied at one end and at the other end mass is attached to which the displacement u is applied.

Homework Equations

The equation of motion is m*a(t) + k *u(t) = F(u) where F(u) is the unknown force which depends on the displacement applied u.

The total work done during a time interval Δt is given by ∫[F(u)*du] with limits from t to t + Δt.

The total change in the KE (=ΔKE) and PE (=ΔPE) is (I think) given by ΔKE = 0.5*m*(V_t+Δt - V_t)^2 and ΔPE as 0.5*k*(u_t+Δt-u_t)^2.

The Attempt at a Solution

[/B]
Now, since there is no dissipation the total work done, W, from t to t + Δt should be equal to ΔKE+ΔPE i.e. change in KE and PE from t to t + Δt . I am calculating the work done as integral from t to t + Δt of F(u)*du where F(u) = m*a + k*u where a = double derivative of u. The change in KE and PE is computed from the formula ΔKE = 0.5*m*(V_t+Δt - V_t)^2 and ΔPE as 0.5*k*(u_t+Δt-u_t)^2. However, I do not get W = ΔKE+ΔPE. I have also tried getting the ΔKE and ΔPE using the integration from t to t+Δt. Can somebody suggest what is wrong that I am doing.

Your advice, suggestions is highly appreciated.

Thank you.

 

[rude man]

I need to find the difference of the total work done and the energy (KE and PE) of the system.

What does this mean? Difference between what and what? Please quote the problem in its original.