SHM Chapter 18 ke,pe,te vs x

(b)    Variation with Displacement of Energy in Simple Harmonic Motion

Consider a spring of spring constant k connected to a mass m as shown. When the mass is displaced from its equilibrium position by a distance x0 and released, it will oscillate backwards and forwards.



At the time of release, the energy of the system will consist totally of the spring’s potential energy (PE). As the spring pulls the mass toward the equilibrium position, the potential energy is transformed into kinetic energy (KE) until at the equilibrium position the kinetic energy will reach a maximum. After the mass crosses the equilibrium position, the spring is compressed and the PE is converted to KE. Maximum PE occurs at maximum compression of spring.


The constant exchange between potential energy PE and kinetic energy KE is essential in producing oscillations. The total energy TE, which is the summation of kinetic and potential energy, will always add up to a constant value as shown.





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



recalling Energy formula
$$KE=\frac{1}{2}m{v}^2$$
$$PE=\frac{1}{2}k{x}^2$$
therefore, total mechanical energy TE is both KE and PE, can be written as
$$TE=KE+PE=\frac{1}{2}m{v}^2+\frac{1}{2}k{x}^2$$
since  ${\displaystyle v=\pm \omega \sqrt{(x_0^2-x^2)}}$

$$KE=\frac{1}{2}m(\omega \sqrt{(x_0^2-x^2)}{)}^2$$
$$KE=\frac{1}{2}m {\omega }^2(x_0^2-x^2)$$
therefore,

$$$$$$$$$$TE=KE+PE=\frac{1}{2}m{\omega }^2(x_0^2-x^2)+\frac{1}{2}k{x}^2$$
by substitution, when x = -x₀,

$$TE=KE+PE=\frac{1}{2}m{\omega }^2(x_0^2-(-x_0{)}^2)+\frac{1}{2}k(-x_0{)}^2=\frac{1}{2}k(x_0{)}^2$$
hence or otherwise the table below shows the various formula and their values

general energy formula	SHM energy formula	when x = -x0	when x = 0	when x = x0
$$KE=\frac{1}{2}m{v}^2$$	$$KE=\frac{1}{2}m {\omega }^2(x_0^2-x^2)$$	0	$$\frac{1}{2}m{\omega }^2{x}_0^2$$	0
$$PE=\frac{1}{2}k{x}^2$$	$$PE=\frac{1}{2}k{x}^2$$	$$\frac{1}{2}m{\omega }^2{x}_0^2$$	0	$$\frac{1}{2}m{\omega }^2{x}_0^2$$
$$TE=KE+PE$$	$$TE= \frac{1}{2}m{\omega }^2(x_0^2-x^2)+\frac{1}{2}k{x}^2$$	$$\frac{1}{2}m{\omega }^2{x}_0^2$$	$$\frac{1}{2}m{\omega }^2{x}_0^2$$	$$\frac{1}{2}m{\omega }^2{x}_0^2$$

If no energy is lost during an oscillation, then total energy remains constant.

Model:

http://dl.dropboxusercontent.com/u/44365627/lookangEJSworkspace/export/ejss_model_SHM18/SHM18_Simulation.xhtml