Forest UCM Osc Damped

1-D Damped Oscillations

An example of a damped oscillator is the case of a mass attached to a spring with one end fixed. The block slide on top of a table and feels a frictional force from the table.

Newton's 2nd Law

As in the case of air resistance, it was assume that there was a frictional force proportional to the velocity of the oscillation body. The same assumption is made for a block oscillating via a spring with friction between the block and the table top.

[math] \sum \vec{F}_{ext} = -k\vec r - b \vec \dot v = m \vec \ddot r[/math]

[math] \sum F_{ext} = -kx - b \dot x = m \ddot x[/math]: in 1-D

or

[math] m \ddot x + kx + b \dot x = 0[/math]

or

[math] \ddot x + \frac{k}{m}x + \frac{b}{m} \dot x = 0[/math]

let

[math]\frac{k}{m} = \omega^2_0 =[/math] undamped oscillation frequency

[math]\frac{b}{m} \equiv 2 \beta =[/math] damping constant

then

[math] \ddot x + 2 \beta \dot x + \omega^2_0x = 0[/math]

Solve for the Equation of Motion

As see in section Forest_UCM_Osc_SHM#Equation_of_motion, you can determine solutions to the above by writing the analogous auxilary equation:

[math] \left ( O^2 + 2 \beta O + \omega^2_0 \right ) x = 0 \;\;\;\;\;\; O \equiv \frac{d}{dt}[/math]

Setting the term in parentheses to zero and using the quadratic formula

[math]O = \frac{- 2\beta \pm \sqrt{(2\beta)^2 -4\omega^2_0}}{2} = - \beta \pm \sqrt{\beta^2 -\omega^2_0}[/math]

[math] \left ( O + \beta + \sqrt{\beta^2 -\omega^2_0} \right ) \left ( O + \beta - \sqrt{\beta^2 -\omega^2_0}\right ) x = 0 [/math]

You have change the second order differential equation into two first order differential equations

[math] \left ( \frac{d}{dt} + \beta + \sqrt{\beta^2 -\omega^2_0} \right ) x = 0 [/math]

[math] \Rightarrow \frac{dx}{x} = \left ( -\beta - \sqrt{\beta^2 -\omega^2_0} \right ) dt [/math]

[math] x= e^{\left (- \beta - \sqrt{\beta^2 -\omega^2_0} \right )t} [/math]

[math] \left (\frac{d}{dt} + \beta - \sqrt{\beta^2 -\omega^2_0}\right ) x = 0 [/math]

[math] \Rightarrow \frac{dx}{x} = \left ( - \beta + \sqrt{\beta^2 -\omega^2_0} \right ) dt [/math]

[math] x= e^{\left ( - \beta + \sqrt{\beta^2 -\omega^2_0} \right )t} [/math]

constructing a complete solution from the two solutions (orthogonal functions) above.

[math] x= \left ( C_1 e^{ \sqrt{\beta^2 -\omega^2_0} t} + C_2 e^{ -\sqrt{\beta^2 -\omega^2_0} t} \right) e^{- \beta t} [/math]

Undamped oscillator

If [math]\beta[/math] = 0

Then

[math] x= \left ( C_1 e^{ \sqrt{\beta^2 -\omega^2_0} t} + C_2 e^{ -\sqrt{\beta^2 -\omega^2_0} t} \right) e^{- \beta t} [/math]

[math] = \left ( C_1 e^{ i\omega_0 t} + C_2 e^{ -i\omega_0 t} \right) [/math] the SHM solution derived before at Forest_UCM_Osc_SHM#Equation_of_motion

Under damped Oscillator

[math]\beta \lt \omega_0[/math]

In this case the term

[math]\sqrt{\beta^2 -\omega^2_0} =\sqrt{(-1)(\omega^2_0- \beta^2 } = i \sqrt{\omega^2_0- \beta^2 } \equiv i \omega_1[/math]

[math] x= \left ( C_1 e^{ \sqrt{\beta^2 -\omega^2_0} t} + C_2 e^{ -\sqrt{\beta^2 -\omega^2_0} t} \right) e^{- \beta t} [/math]

[math] = \left ( C_1 e^{i\omega_1 t} + C_2 e^{ -i\omega_1t} \right) e^{- \beta t} [/math]

[math] = Ae^{- \beta t} \cos(\omega_1 t -\delta) [/math]

There are two terms above, the first term is an exponential decay and the second is the usual harmonic oscilator.

They combine to produce oscillations whoase amplitudes decay with time.

Over damped Oscillator

[math]\beta \gt \omega_0[/math]

[math] x= \left ( C_1 e^{ \sqrt{\beta^2 -\omega^2_0} t} + C_2 e^{ -\sqrt{\beta^2 -\omega^2_0} t} \right) e^{- \beta t} [/math]

[math] = \left ( C_1 e^{ \left (-\beta +\sqrt{\beta^2 -\omega^2_0} \right ) t} + C_2 e^{ \left (-\beta-\sqrt{\beta^2 -\omega^2_0}\right )t} \right) [/math]

For the overdamped case you have two exponentials with negative exponents

[math]-\beta + \sqrt{\beta^2 -\omega^2_0} \lt 0 [/math]

Critically damped Oscillator

[math]\beta = \omega_0[/math]

[math] x= \left ( C_1 e^{ \sqrt{\beta^2 -\omega^2_0} t} + C_2 e^{ -\sqrt{\beta^2 -\omega^2_0} t} \right) e^{- \beta t} [/math]

[math] = A e^{- \beta t} [/math]

In this case the method of using the auxilary equation has yielded two identical solutions. A second solution should exist.

Problem 5.21 and 5.24

[math] x= C_1 e^{- \beta t} +C_1 t e^{- \beta t} [/math]

Physics Example; The LRC circuit

An electrical circuit composed of an inductor [math](L)[/math] , a resister [math](R)[/math], and a capacitor [math](C)[/math] is known as an LRC circuit.

The resistor in the circuit serves to simply resists the movement of charge through the circuit ( analogous to friction; causing energy to leave the system in teh form of heat) such that the voltage drop accross the resistor is given by

[math]V= R \dot q[/math]

The capacitor in the ciruit plays role of storing charge ( like a spring stores energy) such that it become harder to store charge on the capacitor as you increase the amount of charge you try to store in it. The voltage drop across a capacitor is given by

[math]V = \frac{C}{q}[/math]

An Inductor trys to resist the change in the rate that the current moves around the system (like inertia; produce by inducing a reverse EMF). The voltage drop across an inductor is given by

[math]V = L \ddot q[/math]

If you apply conservation of energy (potential) by traversing the circuit in a complete loop (Kirchoff's 2nd law) then you can write the conservation law as

[math]L \ddot q + R \dot q + \frac{1}{C} q = 0[/math]

The above differential equation has the same form as the damped oscillator differential equation where

[math]2\beta = \frac{R}{L}[/math] and [math] \omega_0^2 = \frac{1}{LC}[/math]

Forest_UCM_Osc#Damped_Oscillations