1.3 Energy in the simple harmonic oscillator

Energy in sho

Recall that the total energy of a system is: $$E=KE+PE=K+U$$ We also know that the kinetic energy is

$$K=\frac{1}{2}m{\text{v}}^2$$ But what is $U$ ? For a conservative Force( $\oint \vec{F}d\vec{x}=0$ ) - eg. gravity, electrical... (no friction) we know that the work done by anexternal force is stored as $U$ . For the case of a mass on a spring, the external force is opposite the springForce (That is it has the opposite sign from the spring force).: $$F_{ext}=kx$$ (i.e. This is the force you use to pull the mass and stretch the spring beforeletting go and making it oscillate.)

Thus $$U={\int }_0^{x}kxdx=\frac{1}{2}k{x}^2$$ This gives: $${\displaystyle \begin{array}{c}E=\frac{1}{2}m{\text{v}}^2+\frac{1}{2}k{x}^2\\ =\frac{1}{2}m{\left(\frac{dx}{dt}\right)}^2+\frac{1}{2}k{x}^2\end{array}}$$ It is important to realize that any system that is represented by either of these two equations below represents oscillating system $$m\frac{d^{2}x}{d{t}^2}+kx=0$$ $$\frac{1}{2}m{\left(\frac{dx}{dt}\right)}^2+\frac{1}{2}k{x}^2=E$$

To calculate the energy in the system it is helpful to take advantage of the fact that we can calculate the energy at any point in x. For example in thecase of the simple harmonic oscillator we have that: $$x=A{e}^{i(\omega t+\alpha )}$$ We can choose $t$ such that $$x=A$$ Now remember that when I write $$x=A{e}^{i(\omega t+\alpha )}$$ I "really" (pun intended) mean $$x=Re\left[A{e}^{i(\omega t+\alpha )}\right]$$ Likewise then $$\dot{x}=Re\left[i\omega A{e}^{i(\omega t+\alpha )}\right]$$

At the point in time where $x=A$ this gives us $$\dot{x}=Re\left[i\omega A\right]=0$$ Thus at that point in time we have $\dot{x}=0$ . We can now substitute that and $x=A$ into $$E=\frac{1}{2}m{\left(\frac{dx}{dt}\right)}^2+\frac{1}{2}k{x}^2$$ we obtain $$E=\frac{1}{2}k{A}^2$$ This is an important point. The energy in the oscillator is proportional to the amplitude squared!