Forest UCM Osc SHM

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 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