PX155 - D5 - energy and SHM

work done against spring to stretch from $x^{'} = 0$ to $x^{'} = x$ :

W = \int_{0}^{x} k x^{'} d x^{'} = \frac{1}{2} k x^{2}

if mass on a spring is executing $S H M$ with $x = A \cos ω t$ :

U = \frac{1}{2} k A^{2} \cos^{2} ω t

T = \frac{1}{2} m {\dot{x}}^{2} = \frac{1}{2} m (- A ω \sin ω t)^{2}

T = \frac{1}{2} k A^{2} \sin^{2} (ω t)

we have, $$total ; E=T+U = \frac{1}{2}kA^{2}(cos^{2}\omega t+sin^{2}\omega t)$$

∴ E = \frac{1}{2} k A^{2}

- ie: constant total energy

on average
- $< T >= \frac{1}{4} k A^{2}$
- $< U >= \frac{1}{4} k A^{2}$
  - equipartition of energy