Difference between revisions of "Forest UCM Osc SHM"

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=Simple Harmonic Motion=
  
 
==Equation of motion==
 
==Equation of motion==
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then the analogous equation becomes
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then the analogous (known as the auxilary ) equation becomes
  
 
:<math> \left ( mO^2 + aO + k \right ) x = 0  </math>
 
:<math> \left ( mO^2 + aO + k \right ) x = 0  </math>
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::<math>  =  Ae^{i \omega t}  +  Be^{-i\omega t}  </math>
 
::<math>  =  Ae^{i \omega t}  +  Be^{-i\omega t}  </math>
  
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==Oscillator properties==
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:<math> x =  Ae^{i \omega t}  +  Be^{-i\omega t}  </math>
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Using Euler's formula for complex variables
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:<math>e^{\pm ix} = \cos(x) \pm  i \sin(x)</math>
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:<math>\Rightarrow x = A\left ( \cos(\omega t) + i \sin(\omega t)\right ) + B\left ( \cos(\omega t) - i \sin(\omega t)\right )</math>
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::: <math>= (A + B) \cos (\omega t) + i(A-B) \sin(\omega t)</math>
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::: <math>= (A + B) \cos (\omega t) + i(A-B) \sin(\omega t)</math>
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::: <math>= A^{\prime}\cos (\omega t) + B^{\prime} \sin(\omega t)</math>
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let
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:<math>C = \sqrt{\left(A^{\prime}\right)^2+ \left ( B^{\prime}\right)^2}</math>
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:<math>\cos\delta = \frac{A^{\prime}}{C}</math>
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:<math>\sin\delta = \frac{B^{\prime}}{C}</math>
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then
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:<math>x = A^{\prime}\cos (\omega t) + B^{\prime} \sin(\omega t)</math>
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::<math> = C \left [ \frac{A^{\prime}}{C}\cos (\omega t) +  \frac{B^{\prime}}{C} \sin(\omega t) \right ]</math>
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::<math> = C \left [\cos\delta\cos (\omega t) +  \sin\delta \sin(\omega t) \right ]</math>
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::<math> = C \cos (\omega t - \delta) </math>
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:<math>\cos(A-B) = \cos(A)\cos(B)+\sin(B)\sin(A)</math>
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:<math>x=A \cos(\omega t - \delta)</math>
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:<math>v = \dot x = \omega A \sin(\omega t - \delta)</math>
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==Kinetic (T) and potential (U) Energy ==
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:<math>U= \frac{1}{2} k x^2 =\frac{1}{2} m \omega^2 A^2 \cos^2(\omega t - \delta)</math>
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:<math>T= \frac{1}{2} m v^2 =\frac{1}{2} m \omega^2 A^2 \sin^2(\omega t - \delta)</math>
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:<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>
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:::<math>= \frac{1}{2} m \omega^2 A^2 =</math>constant
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:<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>
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:<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>
  
 
[[Forest_UCM_Osc#Simple_Harmonic_Motion_.28SHM.29]]
 
[[Forest_UCM_Osc#Simple_Harmonic_Motion_.28SHM.29]]

Latest revision as of 12:43, 5 October 2014


Simple Harmonic Motion

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 (known as the auxilary ) 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]

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]

Kinetic (T) and potential (U) Energy

[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} m \omega^2 A^2 \cos^2(\omega t - \delta)[/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=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=E[/math]

Forest_UCM_Osc#Simple_Harmonic_Motion_.28SHM.29