# B Potential energy in a Mass spring system

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1. Dec 8, 2016

Why is the rate of change of potential energy always same the rate of change of kinetic energy in a mass spring system?
Additiinally, How do we determine the rate of change of potential energy in such case?

2. Dec 8, 2016

### ShayanJ

The total energy in a mass spring system is conserved and so any decrease in one form of energy should result in an increase in another because otherwise the total energy would change.
To determine the rate of change of potential energy:
Step1) Consider the most general trajectory for a harmonic oscillator:$x(t)=A\cos \omega t+B\sin\omega t$
Step2) Put $x(t)$ in $V(t)=\frac 1 2 kx(t)^2=\frac 1 2 m\omega^2 x(t)^2$.
Step3) Differentiate $V(t)$ w.r.t. time.
Step4) Differentiate $K(t)$ w.r.t. time.(you calculated it here!)
Step5) Compare the results from Step3 and Step4.

3. Dec 8, 2016

### TJGilb

Think about conservation of energy with regards to a mass-spring system: $\frac 1 2 m {\frac {dx} {dt}}^2+mg(h-x)+\frac 1 2 kx^2=constant$ where h is the max height. Imagine we drop it from when the displacement in the spring is zero. Putting this in terms of time we can get

$\frac 1 2 m(A \omega sin(\omega t))^2+mg(h-Acos(\omega t))+\frac 1 2 k(Acos(\omega t))^2=constant$

Taking the rate of change with respect to time we get

$mA^2 \omega^3 sin(\omega t)cos(\omega t)+mgA \omega sin(\omega t)-kA^2 cos(\omega t)sin(\omega t)=0$

Putting the rate of change of the potential energies on one side and the kinetic energy on the other, we get

$mA^2 \omega^3 sin(\omega t)cos(\omega t)=kA^2 cos(\omega t)sin(\omega t)-mgA \omega sin(\omega t)$

which simplifies to

$mA\omega^3 cos(\omega t) = kAcos(\omega t)-mg\omega$.

As you can see, the rate of change is driven by $cos(\omega t)$ for both.

This was quick and dirty, so if I made any errors somewhere, someone feel free to correct me.