Forest UCM Osc HookesLaw - Revision history

Foretony: /* Equation of Motion from Cons of Energy */

2021-10-04T14:29:34Z

Equation of Motion from Cons of Energy

Foretony: /* Oscillator properties */

2014-10-04T14:29:32Z

Oscillator properties

Foretony: /* 2-D Oscillators */

2014-10-04T14:29:07Z

2-D Oscillators

Foretony: /* Kinetic (T) and potential (U) Energy */

2014-10-04T14:28:23Z

Kinetic (T) and potential (U) Energy

Foretony: /* Equation of motion */

2014-10-04T14:26:25Z

Equation of motion

Foretony: /* Kinetic (T) and potential (U) Energy */

2014-10-04T14:20:48Z

Kinetic (T) and potential (U) Energy

Foretony: /* Kinetic (T) and potential (U) Energy */

2014-10-04T14:19:40Z

Kinetic (T) and potential (U) Energy

Foretony: /* Kinetic (T) and potential (U) Energy */

2014-10-04T13:01:31Z

Kinetic (T) and potential (U) Energy

Foretony: /* Kinetic (T) and potential (U) Energy */

2014-10-04T13:00:49Z

Kinetic (T) and potential (U) Energy

Foretony: /* Kinetic (T) and potential (U) Energy */

2014-10-04T12:50:31Z

Kinetic (T) and potential (U) Energy

@@ Line 53: / Line 53: @@
 if we make the following definitions
-: <math> \left . \frac{\partial^2 U}{\partial x^2} \right |_{x=x_0} (x-x_0)^2 \; = k  </math>
+: <math> \left . \frac{\partial^2 U}{\partial x^2} \right |_{x=x_0}  \; = k   </math>
-: <math>U(x) =  \frac{1}{6} \left . \frac{\partial^4 U}{\partial x^4} \right |_{x=x_0} (x-x_0)^4 \; = \epsilon </math>
+: <math>U(x) =  \frac{1}{6} \left . \frac{\partial^4 U}{\partial x^4} \right |_{x=x_0}  \; = \epsilon </math>
 and that  the equilibrium point is located at the orgin

← Older revision		Revision as of 14:29, 4 October 2014
Line 102:		Line 102:
	2.) The work done is independent of path ( <math>\vec \nabla \times \vec F = 0</math> in 1-D and 3-D)		2.) The work done is independent of path ( <math>\vec \nabla \times \vec F = 0</math> in 1-D and 3-D)

−
−	~~==Oscillator properties==~~
−
−	~~:<math> x = Ae^{i \omega t} + Be^{-i\omega t} </math>~~
−
−	~~Using Euler's formula for complex variables~~
−
−	~~:<math>e^{\pm ix} = \cos(x) \pm i \sin(x)</math>~~
−
−	~~:<math>\Rightarrow x = A\left ( \cos(\omega t) + i \sin(\omega t)\right ) + B\left ( \cos(\omega t) - i \sin(\omega t)\right )</math>~~
−	~~::: <math>= (A + B) \cos (\omega t) + i(A-B) \sin(\omega t)</math>~~
−	~~::: <math>= (A + B) \cos (\omega t) + i(A-B) \sin(\omega t)</math>~~
−	~~::: <math>= A^{\prime}\cos (\omega t) + B^{\prime} \sin(\omega t)</math>~~
−
−	~~let~~
−
−	~~:<math>C = \sqrt{\left(A^{\prime}\right)^2+ \left ( B^{\prime}\right)^2}</math>~~
−
−	~~:<math>\cos\delta = \frac{A^{\prime}}{C}</math>~~
−	~~:<math>\sin\delta = \frac{B^{\prime}}{C}</math>~~
−
−	~~then~~
−
−	~~:<math>x = A^{\prime}\cos (\omega t) + B^{\prime} \sin(\omega t)</math>~~
−	~~::<math> = C \left [ \frac{A^{\prime}}{C}\cos (\omega t) + \frac{B^{\prime}}{C} \sin(\omega t) \right ]</math>~~
−	~~::<math> = C \left [\cos\delta\cos (\omega t) + \sin\delta \sin(\omega t) \right ]</math>~~
−	~~::<math> = C \cos (\omega t - \delta) </math>~~
−
−	~~:<math>\cos(A-B) = \cos(A)\cos(B)+\sin(B)\sin(A)</math>~~
−
−	~~:<math>x=A \cos(\omega t - \delta)</math>~~
−	~~:<math>v = \dot x = \omega A \sin(\omega t - \delta)</math>~~


	[[Forest_UCM_Osc#Hooke.27s_Law]]		[[Forest_UCM_Osc#Hooke.27s_Law]]

@@ Line 134: / Line 134: @@
 :<math>x=A \cos(\omega t - \delta)</math>
 :<math>v = \dot x = \omega A \sin(\omega t - \delta)</math>
 [[Forest_UCM_Osc#Hooke.27s_Law]]

@@ Line 134: / Line 134: @@
 :<math>x=A \cos(\omega t - \delta)</math>
 :<math>v = \dot x = \omega A \sin(\omega t - \delta)</math>
 = 2-D Oscillators=

← Older revision		Revision as of 14:26, 4 October 2014
Line 101:		Line 101:

	2.) The work done is independent of path ( <math>\vec \nabla \times \vec F = 0</math> in 1-D and 3-D)		2.) The work done is independent of path ( <math>\vec \nabla \times \vec F = 0</math> in 1-D and 3-D)
−
−
−	~~==Equation of motion==~~
−
−	~~In solving the differential equation~~
−
−	~~:<math> m\ddot x =-kx </math>energy is constant with time~~
−
−
−	~~and observe that the above differential equation is a special case of the more general differential equation~~
−
−	~~:<math> m\ddot x + c \dot x =-kx </math>energy is constant with time~~
−
−	~~one could rewrite the above as~~
−
−	~~:<math> \left ( \frac{d}{dt}\frac{d}{dt} +\frac{d}{dt} + k \right ) x = 0 </math>~~
−
−	~~One could cast the above differential equation into an analogous quadratic equation if you~~
−
−	~~let~~
−
−	~~:<math>O = \frac{d}{dt}</math>~~
−
−
−	~~then the analogous equation becomes~~
−
−	~~:<math> \left ( mO^2 + aO + k \right ) x = 0 </math>~~
−
−	~~where m, a, and k are constants~~
−
−	~~factoring this quadratic you would have~~
−
−	~~:<math> \left ( O - \gamma \right )\left ( O - \beta \right ) x = 0 </math>~~
−
−	~~where a non-trivial solution would exist if one of the terms in the parentheses were zero~~
−
−	~~this basically reduces our 2nd order differential equation down to two first order differential equations~~
−
−	~~:<math> \left ( \frac{d}{dt} - \gamma \right ) \left ( \frac{d}{dt} + \beta \right ) x = 0 </math>~~
−
−	~~one of the solutions would be~~
−
−	~~:<math> \frac{d}{dt} x - \gamma x= 0 </math>~~
−	~~:<math> \frac{dx}{x} = \gamma dt </math>~~
−	~~:<math> x = Ae^{\gamma t} </math>~~
−
−
−	~~:<math> x = Ae^{\gamma t} + Be^{\beta t} </math>~~
−
−
−	~~For the special case where there isn't a first derivative term (a=0)~~
−
−	~~You simply have~~
−
−	~~:<math> \left ( m\frac{d}{dt}\frac{d}{dt} + k \right ) x = 0 </math>~~
−
−	or
−	~~:<math> \left ( mO^2 + k \right ) x = 0 </math>~~
−
−	~~:<math> O= \pm \sqrt{-\frac{k}{m}} = \pm i \sqrt{-\frac{k}{m}}</math>~~
−	~~:<math> \gamma \equiv +i\sqrt{\frac{k}{m}} = i \omega</math>~~
−	~~:<math> \beta = -i\sqrt{\frac{k}{m}} = -i \omega</math>~~
−
−	~~then you have~~
−
−	~~:<math> x = Ae^{\gamma t} + Be^{\beta t} </math>~~
−	~~::<math> = Ae^{i \omega t} + Be^{-i\omega t} </math>~~

@@ Line 208: / Line 208: @@
 :<math>E = T + U= \frac{1}{2} m \omega^2 A^2 \sin^2(\omega t - \delta) +\frac{1}{2} m \omega^2 A^2 \cos^2(\omega t - \delta)</math>
-:::<math>= \frac{1}{2} m \omega^2 A^2 </math>
+:::<math>= \frac{1}{2} m \omega^2 A^2 =</math>constant
-:<math>U_{\mbox{max}}= \frac{1}{2} k x^2_{\mbox{max}} =\frac{1}{2} m \omega^2 A^2 \cos^2(\omega t - \delta)_{\mbox{max}}= \frac{1}{2} m \omega^2 A^2=U_{\mbox{max}}</math>
+:<math>U_{\mbox{max}}= \frac{1}{2} k x^2_{\mbox{max}} =\frac{1}{2} m \omega^2 A^2 \cos^2(\omega t - \delta)_{\mbox{max}}= \frac{1}{2} m \omega^2 A^2=E</math>
-:<math>T_{\mbox{max}}= \frac{1}{2} m v^2_{\mbox{max}} =\frac{1}{2} m \omega^2 A^2 \sin^2(\omega t - \delta)_{\mbox{max}}= \frac{1}{2} m \omega^2 A^2=U_{\mbox{max}}</math>
+:<math>T_{\mbox{max}}= \frac{1}{2} m v^2_{\mbox{max}} =\frac{1}{2} m \omega^2 A^2 \sin^2(\omega t - \delta)_{\mbox{max}}= \frac{1}{2} m \omega^2 A^2=E</math>
 = 2-D Oscillators=

@@ Line 204: / Line 204: @@
 ==Kinetic (T) and potential (U) Energy ==
-:<math>U= \frac{1}{2} k x^2 =\frac{1}{2} m \omega^2 A^2 \sin^2(\omega t - \delta)</math>
+:<math>U= \frac{1}{2} k x^2 =\frac{1}{2} m \omega^2 A^2 \cos^2(\omega t - \delta)</math>
 :<math>T= \frac{1}{2} m v^2 =\frac{1}{2} m \omega^2 A^2 \sin^2(\omega t - \delta)</math>
-:<math>E = T + U= \frac{1}{2} m \omega^2 A^2 \sin^2(\omega t - \delta) + \frac{1}{2} k x^2 =\frac{1}{2} m \omega^2 A^2 \sin^2(\omega t - \delta)</math>
+:<math>E = T + U= \frac{1}{2} m \omega^2 A^2 \sin^2(\omega t - \delta) +\frac{1}{2} m \omega^2 A^2 \cos^2(\omega t - \delta)</math>
-:<math>U_{\mbox{max}}= \frac{1}{2} k x^2_{\mbox{max}} =\frac{1}{2} m \omega^2 A^2 \sin^2(\omega t - \delta)_{\mbox{max}}= \frac{1}{2} m \omega^2 A^2=U_{\mbox{max}}</math>
+:<math>U_{\mbox{max}}= \frac{1}{2} k x^2_{\mbox{max}} =\frac{1}{2} m \omega^2 A^2 \cos^2(\omega t - \delta)_{\mbox{max}}= \frac{1}{2} m \omega^2 A^2=U_{\mbox{max}}</math>
 :<math>T_{\mbox{max}}= \frac{1}{2} m v^2_{\mbox{max}} =\frac{1}{2} m \omega^2 A^2 \sin^2(\omega t - \delta)_{\mbox{max}}= \frac{1}{2} m \omega^2 A^2=U_{\mbox{max}}</math>

← Older revision		Revision as of 12:50, 4 October 2014
Line 204:		Line 204:
	==Kinetic (T) and potential (U) Energy ==		==Kinetic (T) and potential (U) Energy ==

−	:<math>U= \frac{1}{2} k x^2 =\frac{1}{2} m \omega^2 A^2 \sin^2(\omega t - \delta)}</math>	+	:<math>U= \frac{1}{2} k x^2 =\frac{1}{2} m \omega^2 A^2 \sin^2(\omega t - \delta)</math>