Vibrations

#OrdinaryDifferentialEquations #ClassicalMechanics

Subjects: Ordinary Differential Equations, Classical Mechanics
Links: Second Order Linear Differential Equations

Undamped Free Vibration

If we have a simple harmonic oscillator with the differential equation $$ m u'' -k u=0 $$we can define the natural frequency $\omega_0 {#2} = k/m$ . Then the solutions are of the form $$ u = A\cos \omega_0 t + B\sin\omega_0 t $$We can rewrite this as$$ u = R\cos(\omega_0 t -\delta) $$The way we can change between these representations by $$ \begin{align*} A = R\cos \delta& \qquad B = R\sin \delta \ R= \sqrt{A^2+B^2}& \qquad \tan\delta = B/A \end{align*} $$The period of the motion is

T = \frac{2 π}{ω_{0}}

$R$ is called the amplitude of the motion, and $δ$ is called the phase.

Damped Free Vibrations

If we include the damping effect, on the differential equation we get that, with $m, γ, k > 0$

m u^{″} + γ u^{'} + k u = 0

Solving the characteristic polynomial we get that the roots are $$ r_1, r_2 = \frac{\gamma}{2m}\left(-1\pm \sqrt{1-\frac{4km}{\gamma^2}}\right) $$

There are various cases,

$γ^{2} - 4 k m > 0$ , then $u = A e^{r_{1} t} + B e^{r_{2} t}$ , this is called overdamped.
$γ^{2} - 4 k m = 0$ , then $u = (A + B t) e^{- γ t / 2 m}$ , this is called critically damped.
$γ^{2} - 4 k m < 0$ , then $μ = \frac{\sqrt{4 k m - γ^{2}}}{2 m} > 0$ , and $u = e^{- γ t / 2 m} (A \cos μ t + B \sin μ t)$ , this is called underdamped.

In this last case, the motion isn’t periodic but it oscillates back and forth, $μ$ is called the quasi frecuency. By comparing $μ$ to the frequency $ω_{0}$ of an undamped motion we get that $$ \frac{\mu}{\omega_0} = \left(1-\frac{\gamma^2}{4km}\right)^{1/2} \approx 1-\frac{\gamma^2}{8km} $$

if $γ^{2} / 4 k m$ is small. By analogy, the quantity $T_{d} = 2 π / μ$ is called the quasi period. If we compare $T_{d}$ and $T$ we get that $$ \frac{T_d}{T} =\frac{\mu}{\omega_0} = \left(1-\frac{\gamma^2}{4km}\right)^{1/2} \approx 1-\frac{\gamma^2}{8km} $$

The time between successive maxima is $T_{d}$ , and it behaves more like a period.

The ratio of the displacements at two successive maxima is given by $\exp (γ T_{d} / 2 m)$ . No matter the maxima chosen. The natural logarithm of this ratio is called the logarithmic decremet and is denoted as $Δ$ . Thus $Δ = π γ / m μ$ .

Forced Vibrations

Forced Vibrations with Damping

The equation of a general spring mass system subject to an external force $F (t)$ is

m u^{″} + γ u^{'} + k u = F (t)

We will make the external force to be given by $F_{0} \cos ω t$ , where $F_{0}$ and $ω$ are positive constants. Then the solutions are of the form

u = c_{1} u_{1} + c_{2} u_{2} + A \cos ω t + B \sin ω t = u_{c} + U

Intuitive Way

A way I found to do it is to non-dimensionalize, to make the equation easier to understand. We divide with respect to $m$ , we get the equation

u^{″} + \frac{γ}{m} u^{'} + \frac{k}{m} u = \frac{F_{0}}{m} \cos (ω t)

We take advantage and define $ω_{0}$ to be the natural frequency, defined as $$ \omega_0^2 = \frac{k}{m} $$The equation now looks like $$ u'' + \frac{\gamma}{m}u' + \omega_0^2u =\frac{F_0}{m} \cos(\omega t) $$Finally we make a weird step, and take that $τ = ω_{0} t$ . Then the next step, is to look at how the differential equation changes given that

\frac{d u}{d t} = \frac{d u}{d τ} \frac{d τ}{d t}, \frac{d^{2} u}{d t^{2}} = \frac{d^{2} u}{d τ^{2}} {(\frac{d τ}{d t})}^{2} + \frac{d u}{d τ} \frac{d^{2} τ}{d t^{2}}

We get that is now:

ω_{0}^{2} \frac{d^{2} u}{d τ^{2}} + \frac{γ ω_{0}}{m} \frac{d u}{d τ} + ω_{0}^{2} u = \frac{F_{0}}{m} \cos (\frac{ω}{ω_{0}} τ)

We define $β = ω / ω_{0}$ , and $α = γ / m ω_{0} > 0$ , getting the equation and diving by $ω_{0}^{2}$

u^{″} + α u^{'} + u = \frac{F_{0}}{m ω_{0}^{2}} \cos (β τ)

Finally, we make one last subsitution $x = \frac{u}{F_{0} / m ω_{0}^{2}}$ , meaning we can divide the whole experesion by $F_{0} / m ω_{0}^{2}$ , then we get

x^{″} + α x^{'} + x = \cos (β τ)

This is much easier to analize. We can examine the homogeneous solutions. We get that if $u = e^{r t}$ , then

r = \frac{- α \pm \sqrt{α^{2} - 4}}{2}

We can examine it, but no matter the $α$ , we get that $ℜ (r) < 0$ , then the particular solutions must vanish. If $α \neq 2$ , then $x_{h} (t) = A e^{r_{1} t} + B e^{r_{2} t}$ , then since $ℜ (r_{1}), ℜ (r_{2}) < 0$ . then $x_{h} \to 0$ as $t \to \infty$ . If $α = 2$ , then $x_{h} (t) = e^{- t} (A + B t)$ , which also $x_{h} \to 0$ as $t \to \infty$ . This solutions are called transient solutions since vanish when $t \to \infty$ .

The solutions that don’t vanish are called the steady-state solutions. Then for the particular solution we get that if we define $z$ such that $ℜ (z) = x$ , then we can solve the differential equation with $z (t) = A e^{i β τ}$ , getting for the value of $A$

A = \frac{1}{1 - β^{2} + i α β}

Doing some algebra we get

x (τ) = \frac{1}{(1 - β^{2})^{2} + (α β)^{2}} [(1 - β^{2}) \cos (β τ) + α β \sin (β τ)]

We can make it a little less ugly, using the harmonic addition theorem, getting

x (τ) = \frac{1}{\sqrt{(1 - β^{2})^{2} + (α β)^{2}}} \sin (β τ + \arctan (\frac{1 - β^{2}}{α β}))

Resonance happens when $ω = ω_{0}$ , then $β = 1$ . If resonance happens then

x (τ) = \frac{\sin (τ)}{α}

If we try to translate back to $u$ , we get the following

Weird Way

Since $m, γ, k > 0$ , then $r_{1}$ and $r_{2}$ have that $ℜ (r_{1}), ℜ (r_{2}) < 0$ . Then we know that as $t \to \infty$ , then $u_{1}$ and $u_{2}$ tend to $0$ . Since $u_{c}$ dies out as $t$ increases, it is called the transient solution.

The remaining terms $U (t)$ do not die out as $t \to \infty$ , but persists indefinetely, as long as the force is applied. They represent a steady oscillation with the same frequency as the external force and are called the steady state solution or the forced response or the forced response.

It is convinient to write $U (t)$ to write it as a single trignometric expresion

U (t) = R \cos (ω t - δ)

We need a couple of useful terms before getting the expression, with $ω_{0}$ being the natural frequency of the system.

ω_{0}^{2} = \frac{k}{m} Δ = \sqrt{m^{2} (ω_{0}^{2} - ω^{2}} + γ^{2} ω^{2}

Then

R = \frac{F_{0}}{Δ} \cos δ = \frac{m (ω_{0}^{2} - ω^{2})}{Δ} \sin δ = \frac{γ ω}{Δ}

If we try to look how $R$ of the steady state oscillation depends on $ω$ we get that

\frac{R k}{F_{0}} = {[{(1 - \frac{ω^{2}}{ω_{0}^{2}})}^{2} + Γ \frac{ω^{2}}{ω_{0}^{2}}]}^{- 1 / 2}

where $Γ = γ^{2} / m k$ . If we try to maximize $R$ , we need to differentiate with respct to $ω$ and set it to $0$ . Then we get the expresion

ω_{max}^{2} = ω_{0}^{2} - \frac{γ^{2}}{2 m^{2}} = ω_{0}^{2} (1 - \frac{γ^{2}}{2 m k})

We can see that $ω_{max} < ω$ , and it is close when $γ$ is small. Then

R_{max} = \frac{F_{0}}{γ ω_{0} \sqrt{1 - γ^{2} / 4 m k}} \approx \frac{F_{0}}{γ ω_{0}} (1 + \frac{γ^{2}}{8 m k})

The last expression is true for small $γ$ . If $Γ > 2$ , then $ω_{max}$ is imaginary and in this case $R_{max}$ is given by $ω = 0$ . The critical damping is equivalent of $Γ = 4$ .

For small $γ$ , we have that $R_{max} \approx F_{0} / γ ω_{0}$ . Thus for lightly damped systems, the amplitud $R$ of the forced response when $ω$ is near $ω_{0}$ is quite large even for small relatively small external forces, and the smaller $γ$ , the more pronounced effect. This is called resonance.

Forced Vibrations without Damping

We will assume that $γ = 0$ , now we have the differential equation

m u^{″} + k u = F_{0} \cos ω t

Without Resonance

The form of the solution of depends on the value of $ω$ , with $ω_{0} = \sqrt{k / m}$ , in case of $ω \neq ω_{0}$ , then the solution is of the form

u = c_{1} \cos ω_{0} t + c_{2} \sin ω_{0} t + \frac{F_{0}}{m (ω_{0}^{2} - ω^{2})} \cos ω t

The constants $c_{1}$ and $c_{2}$ are constants determined by initial conditions. If look at the particular case where $u (0) = 0$ and $u^{'} (0) = 0$ , meaning all the energy of the system comes from the external force we get that

c_{1} = - \frac{F_{0}}{m (ω_{0}^{2} - ω^{2})} c_{2} = 0

Meaning we have

u = \frac{F_{0}}{m (ω_{0}^{2} - ω^{2})} (\cos ω t - \cos ω_{0} t)

using trigonometric identities we get that

u = (\frac{2 F_{0}}{m (ω_{0}^{2} - ω^{2})}) (\sin \frac{(ω_{0} - ω)}{2} t) (\sin \frac{(ω_{0} + ω)}{2} t)

If $| ω_{0} - ω |$ is small, then $ω_{0} + ω$ is much greater than $| ω_{0} - ω |$ . Consquently, $\sin (ω_{0} + ω) t / 2$ is rapidly oscilating compared to $\sin (ω_{0} - ω) t / 2$ . Thus the motion is rapidly oscilation with a frequncy of $(ω_{0} + ω) / 2$ , but slowly varying amplitud of

(\frac{2 F_{0}}{m | ω_{0}^{2} - ω^{2} |}) | \sin \frac{(ω_{0} - ω)}{2} t |

This type of motion exhibits what is called a beat. In electronics, the variation of amplitud with time is called amplitud modulation.

With Resonance

In the case that $ω = ω_{0}$ we get that the solution to the differential equation

m u^{″} + k u = F_{0} \cos ω t

is of the form

u = c_{1} \cos ω_{0} t + c_{0} \sin ω_{0} t + \frac{F_{0}}{2 m ω_{0}} t \sin ω_{0} t

and will become unbounded.