SHM Chapter 17 ke,pe,te vs t

a)    Variation with time of energy in simple harmonic motion

http://dl.dropboxusercontent.com/u/44365627/lookangEJSworkspace/export/ejss_model_SHM17/SHM17_Simulation.xhtml

If the variation with time of displacement is as shown, then the energies should be drawn as shown.


recalling Energy formula
$$KE=\frac{1}{2}m{v}^2$$
$$PE=\frac{1}{2}k{x}^2$$ in terms of time t,
x = x₀ sin(ωt)
differentiating with t gives
v = v₀ cos (ωt)

$$KE=\frac{1}{2}m{v}^2=\frac{1}{2}m(x_0\omega cos(\omega t){)}^2=\frac{1}{2}m{\mathrm{(x}}_0^2{\omega }^2\mathrm{)c}o{s}^2(\omega t))$$similarly
$$PE=\frac{1}{2}k{x}^2=\frac{1}{2}k(x_{0}sin(\omega t){)}^2=\frac{1}{2}(m{\omega }^2)x_0^{2}si{n}^2(\omega t))$$ 
therefore total energy is a constant value in the absence of energy loss due to drag (resistance)
$$TE=KE+PE=\frac{1}{2}m{\omega }^2{x}_0^2(si{n}^2\omega t+co{s}^2\omega t)=\frac{1}{2}m{\omega }^2{x}_0^2$$





this is how the x vs t looks together of the energy vs t graphs



the table shows some of the common values

general energy formula	SHM energy formula	when $$t=0$$	when  $$t=\frac{T}{4}$$	when $$t=\frac{T}{2}$$	when $$t=\frac{3T}{4}$$	when $$t=T$$
$$KE=\frac{1}{2}m{v}^2$$	$$K$$	$$\frac{1}{2}m{\omega }^2{x}_0^2$$	$$$$0	$$\frac{1}{2}m{\omega }^2{x}_0^2$$	0	$$\frac{1}{2}m{\omega }^2{x}_0^2$$
$$PE=\frac{1}{2}k{x}^2$$	$$PE=\frac{1}{2}(m{\omega }^2)x_0^{2}si{n}^2(\omega t))$$	0	$$\frac{1}{2}m{\omega }^2{x}_0^2$$	0	$$\frac{1}{2}m{\omega }^2{x}_0^2$$	0
$$TE=KE+PE$$	$$TE=\frac{1}{2}m{\omega }^2{x}_0^2$$	$$\frac{1}{2}m{\omega }^2{x}_0^2$$	$$\frac{1}{2}m{\omega }^2{x}_0^2$$	$$\frac{1}{2}m{\omega }^2{x}_0^2$$	$$\frac{1}{2}m{\omega }^2{x}_0^2$$	$$\frac{1}{2}m{\omega }^2{x}_0^2$$

Model:

http://dl.dropboxusercontent.com/u/44365627/lookangEJSworkspace/export/ejss_model_SHM17/SHM17_Simulation.xhtml