Expand knowledge base

1 files changed, 292 insertions, 0 deletions

diff --git a/content/know/concept/harmonic-oscillator/index.pdc b/content/know/concept/harmonic-oscillator/index.pdc
new file mode 100644
index 0000000..5d97a24
--- /dev/null
+++ b/content/know/concept/harmonic-oscillator/index.pdc
@@ -0,0 +1,292 @@
+---
+title: "Harmonic oscillator"
+firstLetter: "H"
+publishDate: 2021-06-02
+categories:
+- Physics
+- Mathematics
+
+date: 2021-03-09T20:35:14+01:00
+draft: false
+markup: pandoc
+---
+
+# Harmonic oscillator
+
+A **harmonic oscillator** obeys
+the simple 1D version of [Hooke's law](/know/concept/hookes-law/):
+to displace the system away from its equilibrium,
+the needed force $F_d(x)$ scales linearly with the displacement $x(t)$:
+
+$$\begin{aligned}
+    F_d(x) = k x
+\end{aligned}$$
+
+Where $k$ is a system-specific proportionality constant,
+called the **spring constant**,
+since a spring is a good example of a harmonic oscillator,
+at least for small displacements.
+Hooke's law is also often stated for
+the restoring force $F_r(x)$ instead:
+
+$$\begin{aligned}
+    F_r(x) = - k x
+\end{aligned}$$
+
+Let a mass $m$ be attached to the end of the spring.
+After displacing it, we let it go $F_d = 0$,
+so Newton's second law for the restoring force $F_r$ demands that:
+
+$$\begin{aligned}
+    F_r = m x''
+\end{aligned}$$
+
+But $F_r = - k x$,
+meaning $m x'' = - k x$,
+leading to the following equation for $x(t)$:
+
+$$\begin{aligned}
+    \boxed{
+        x'' + \omega_0^2 x = 0
+    }
+\end{aligned}$$
+
+Where $\omega_0 \equiv \sqrt{k / m}$ is the **natural frequency** of the system.
+This differential equation has the following general solution:
+
+$$\begin{aligned}
+    \boxed{
+        x(t)
+        = C_1 \sin\!(\omega_0 t) + C_2 \cos\!(\omega_0 t)
+    }
+\end{aligned}$$
+
+Where $C_1$ and $C_2$ are constants determined by the initial conditions.
+For example, for $x(0) = 1$ and $x'(0) = 0$, the solution becomes:
+
+$$\begin{aligned}
+    x(t) = \cos\!(\omega_0 t)
+\end{aligned}$$
+
+When using Lagrangian mechanics or Hamiltonian mechanics,
+we need to know the potential energy $V(x)$
+added to the system by a displacement to $x$.
+This equals the work done by the displacement,
+and is therefore given by:
+
+$$\begin{aligned}
+    V(x) = \int_0^x F_d(x) \:dx = \frac{1}{2} k x^2 = \frac{1}{2} m \omega_0^2 x^2
+\end{aligned}$$
+
+
+## Damped oscillation
+
+If there is a **friction force** $F_f$ affecting the system,
+then the oscillation amplitude will decrease,
+or it might not oscillate at all.
+We define $F_f$ using a **viscous damping coefficient** $c$:
+
+$$\begin{aligned}
+    F_f = - c x'
+\end{aligned}$$
+
+Both $F_r$ and $F_f$ are acting on the system,
+so Newton's second law states that:
+
+$$\begin{aligned}
+    m x'' = - c x' - k x
+\end{aligned}$$
+
+This can be rewritten in the following conventional form
+by defining the **damping coefficient** $\zeta \equiv c / (2 \sqrt{m k})$,
+which determines the expected behaviour of the system:
+
+$$\begin{aligned}
+    \boxed{
+        x'' + 2 \zeta \omega_0 x' + \omega_0^2 x = 0
+    }
+\end{aligned}$$
+
+The general solution is found from the roots $u$ of the auxiliary quadratic equation:
+
+$$\begin{aligned}
+    u^2 + 2 \zeta \omega_0 u + \omega_0^2 = 0
+\end{aligned}$$
+
+The discriminant $D = 4 \zeta^2 \omega_0^2 - 4 \omega_0^2$
+tells us that the behaviour changes substantially
+depending on the damping coefficient $\zeta$,
+with three possibilities: $\zeta < 1$ or $\zeta = 1$ or $\zeta > 1$.
+
+If $\zeta < 1$, there is **underdamping**:
+the system oscillates with exponentially decaying
+amplitude and reduced frequency $\omega_1 \equiv \omega_0 \sqrt{1 - \zeta^2}$.
+The general solution is:
+
+$$\begin{aligned}
+    \boxed{
+        x(t)
+        = \big( C_1 \sin\!(\omega_1 t) + C_2 \cos\!(\omega_1 t) \big) \exp\!(- \zeta \omega_0 t)
+    }
+\end{aligned}$$
+
+If $\zeta = 1$, there is **critical damping**:
+the system returns to its equilibrium point in minimum time.
+The general solution is given by:
+
+$$\begin{aligned}
+    \boxed{
+        x(t)
+        = \big( C_1 + C_2 t \big) \exp\!(- \omega_0 t)
+    }
+\end{aligned}$$
+
+If $\zeta > 1$, there is **overdamping**:
+the system returns to equilibrium slowly.
+The general solution is as follows,
+where $\omega_1 \equiv \omega_0 \sqrt{\zeta^2 - 1}$:
+
+$$\begin{aligned}
+    \boxed{
+        x(t)
+        = \big( C_1 \exp\!(\omega_1 t) + C_2 \exp\!(- \omega_1 t) \big) \exp\!(- \zeta \omega_0 t)
+    }
+\end{aligned}$$
+
+
+## Forced oscillation
+
+In the differential equations given above,
+the right-hand side has always been zero,
+meaning that the oscillator is not affected by any external forces.
+What if we put a function there?
+
+$$\begin{aligned}
+    x'' + 2 \zeta \omega_0 x' + \omega_0^2 x = f(t)
+\end{aligned}$$
+
+Obviously, there exist infinitely many $f(t)$ to choose from,
+and each needs a separate analysis.
+However, there is one type of $f(t)$ that deserves special mention,
+namely sinusoids:
+
+$$\begin{aligned}
+    \boxed{
+        x'' + 2 \zeta \omega_0 x' + \omega_0^2 x = \frac{F}{m} \cos\!(\omega t + \chi)
+    }
+\end{aligned}$$
+
+Where $F$ is a constant force, $\chi$ is an arbitrary phase,
+and the frequency $\omega$ is not necessarily $\omega_0$.
+We solve this case for $x(t)$ in detail.
+Consider the complex version of the equation:
+
+$$\begin{aligned}
+    X'' + 2 \zeta \omega_0 X' + \omega_0^2 X = \frac{F}{m} \exp\!\big(i (\omega t + \chi)\big)
+\end{aligned}$$
+
+Then $x(t) = \Re\{X(t)\}$.
+Inserting the ansatz $X(t) = C \exp\!(i \omega t)$,
+for some constant $C$:
+
+$$\begin{aligned}
+    - C \omega^2 + C 2 i \zeta \omega_0 \omega + C \omega_0^2 = \frac{F}{m} \exp\!(i \chi)
+\end{aligned}$$
+
+Where $\exp\!(i \omega t)$ has already been divided out.
+We isolate this equation for $C$:
+
+$$\begin{aligned}
+    C
+    = \frac{F}{m \big((\omega_0^2 - \omega^2) + 2 i \zeta \omega_0 \omega\big)} \exp\!(i \chi)
+    = \frac{F \big((\omega_0^2 - \omega^2) - 2 i \zeta \omega_0 \omega\big)}
+    {m \big((\omega_0^2 - \omega^2)^2 + 4 \zeta^2 \omega_0^2 \omega^2\big)}
+    \exp\!(i \chi)
+\end{aligned}$$
+
+We would like to rewrite this in polar form $C = r \exp\!(i \theta)$,
+which turns out to be as follows:
+
+$$\begin{aligned}
+    C
+    &= \frac{F}{m \sqrt{(\omega_0^2 - \omega^2)^2 + 4 \zeta^2 \omega_0^2 \omega^2}}
+    \exp\!\bigg(i \chi - i \arctan\!\Big(\frac{2 \zeta \omega_0 \omega}{\omega_0^2 - \omega^2}\Big)\bigg)
+\end{aligned}$$
+
+For brevity, let us define the **impedance** $Z$
+and the **phase shift** $\phi$
+in the following way:
+
+$$\begin{aligned}
+    Z
+    \equiv \sqrt{(\omega_0^2 - \omega^2)^2 / \omega^2 + 4 \zeta^2 \omega_0^2}
+    \qquad \quad
+    \phi
+    \equiv \arctan\!\Big(\frac{2 \zeta \omega_0 \omega}{\omega_0^2 - \omega^2}\Big)
+\end{aligned}$$
+
+Returning to the original ansatz $X(t) = C \exp\!(i \omega t)$,
+we take its real part to find $x(t)$:
+
+$$\begin{aligned}
+    \boxed{
+        x(t)
+        = \frac{F}{m \omega Z} \sin\!(\omega t + \chi - \phi)
+    }
+\end{aligned}$$
+
+Two things are noteworthy here.
+Firstly, $f(t)$ and $x(t)$ are out of phase by $\phi$; there is some lag.
+This is caused by damping, because if $\zeta = 0$, it disappears $\phi = 0$.
+
+Secondly, the amplitude of $x(t)$ depends on $\omega$ and $\omega_0$.
+This brings us to **resonance**,
+where the amplitude can become extremely large.
+Actually, resonance has two subtly different definitions,
+depending on which one of $\omega$ and $\omega_0$ is a free parameter,
+and which one is fixed.
+
+If the natural $\omega_0$ is fixed and the driving $\omega$ is variable,
+we find for which $\omega$ resonance occurs by minimizing the amplitude denominator $\omega Z$.
+We thus find:
+
+$$\begin{aligned}
+    0
+    = \dv{(\omega Z)}{\omega}
+    = \frac{- 4 \omega_0^2 \omega + 4 \omega^3 + 8 \zeta^2 \omega_0^2 \omega}{2 \sqrt{(\omega_0^2 - \omega^2)^2 + 4 \zeta^2 \omega_0^2 \omega^2}}
+    \quad \implies \quad
+    \boxed{
+        \omega = \omega_0 \sqrt{1 - 2 \zeta^2}
+    }
+\end{aligned}$$
+
+Meaning the resonant $\omega$ is lower than $\omega_0$,
+and resonance can only occur if $\zeta < 1 / \sqrt{2}$.
+
+However, if the driving $\omega$ is fixed and the natural is $\omega_0$ is variable,
+the problem is bit more subtle:
+the damping coefficient $\zeta = c / (2 m \omega_0)$
+depends on $\omega_0$.
+This leads us to:
+
+$$\begin{aligned}
+    0
+    = \dv{(\omega Z)}{\omega_0}
+    = \frac{4 \omega_0^3 - 4 \omega^2 \omega_0}{2 \sqrt{(\omega_0^2 - \omega^2)^2 + c^2 \omega^2 / m^2}}
+    \quad \implies \quad
+    \boxed{
+        \omega_0 = \omega
+    }
+\end{aligned}$$
+
+Surprisingly, the damping does not affect $\omega_0$, if $\omega$ is given.
+However, in both cases, the damping *does* matter for the eventual amplitude:
+$c \to 0$ leads to $x \to \infty$,
+and resonance disappears or becomes negligible for $c \to \infty$.
+
+
+
+## References
+1.  M.L. Boas,
+    *Mathematical methods in the physical sciences*, 2nd edition,
+    Wiley.