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