# Regarding Energy and Work done in a Spring Mass System

1. Nov 10, 2014

### S S Gautam

1. The problem statement, all variables and given/known data
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.

2. Relevant 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.

3. The attempt at a solution

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.

2. Nov 15, 2014

### Staff: Admin

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

3. Nov 18, 2014

### rude man

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

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