Oscillations
Oscillations
Introduction
Oscillations appear everywhere in nature: in astrophysics, mechanics, quantum mechanics, optics, and, of course, in music. All these oscillations have one thing in common: they possess a physical quantity that changes periodically as a function of time. We call such a system an oscillating system. A physical system in stable equilibrium can absorb energy, which is then stored in two different forms. The system oscillates as energy continuously transforms back and forth between these two forms over time.
Harmonic Oscillations
Whenever a system is displaced from a stable equilibrium, a restoring force acts to return it to that position, often resulting in harmonic oscillation.
A restoring force has the form: \[\begin{equation} F(x) = -kx \end{equation}\] This formula is commonly known as Hooke’s law, and it represents the restoring force of a spring, where \(k\) is a constant that depends on the spring and \(x\) is the distance from the equilibrium position.
When we talk about Hooke’s law, we are mainly referring to springs and elastic forces, but this linear relationship applies to many other restoring forces for small displacements.
The minus sign in the formula indicates that the force is a restoring force, meaning that it opposes the motion.
Let us now apply Newton’s second law to a point mass \(m\) on which an elastic force acts:
\[\begin{equation} F(x) = -kx = ma \iff -kx = m\ddot{x} \end{equation}\]
We thus obtain the equation of motion for a harmonic oscillator:
\[\begin{equation} \label{diff_eq_harmonic oscillator} \ddot{x} + \omega_0^2x = 0 \end{equation}\] where \(\omega_0\) is defined as: \[\begin{equation} \omega_0 := \sqrt{\frac{k}{m}} \end{equation}\] and is called the natural frequency of the harmonic oscillator.
Note that while \(\omega_0\) is called the natural frequency, dimensionally it represents an angular frequency (or angular velocity).
The equation of motion of a simple harmonic oscillator is a second-order homogeneous linear differential equation; to solve it, we consider the characteristic equation:
\[\begin{equation} \lambda^2 + \omega_0^2 = 0 \implies \lambda_{1,2} = \pm i\omega_0 \end{equation}\]
It follows:
\[\begin{equation} x(t) = C_1e^{i\omega_0 t} + C_2e^{-i\omega_0 t} \end{equation}\]
Since the exponents are imaginary, we can express this solution in terms of sine and cosine:
\[\begin{equation} x(t) = A\sin(\omega_0 t) + B\cos(\omega_0 t) \end{equation}\]
With \(A, B \in \mathbb{R}\). Using trigonometric identities, this can be rewritten as:
\[\begin{equation} x(t) = A\cos(\omega_0 t + \delta) \end{equation}\]
The period of oscillation
The period \(T\) is defined as the time required to complete one full oscillation.
We know that a sinusoidal function completes a full cycle every \(2\pi\) radians, therefore:
\[\begin{equation} T = \frac{2\pi}{\omega_0} \end{equation}\]
In the same way, we can define the frequency as the number of oscillations per second:
\[\begin{equation} \nu = \frac{1}{T} = \frac{\omega_0}{2\pi} \end{equation}\]
The unit of measurement of frequency is the Hertz (\(\text{Hz} = \text{s}^{-1}\)).
Amplitude, velocity, and acceleration
Since the sine (or cosine) function always oscillates between \(-1\) and \(1\), the coefficient \(A\) that multiplies the wave function represents the maximum amplitude of oscillation.
\[\begin{equation} x(t) = A\sin(\omega_0 t + \delta) \implies -A \le x(t) \le A \end{equation}\]
The amplitude represents the maximum displacement from the origin.
By differentiating with respect to time, we obtain the oscillation velocity:
\[\begin{equation} v(t) = \dv{x}{t} = A\omega_0 \cos(\omega_0 t + \delta) \implies -A\omega_0 \le v(t) \le A\omega_0 \end{equation}\]
Similarly, we obtain the acceleration:
\[\begin{equation} a(t) = \dv{v}{t} = -A\omega_0^2 \sin(\omega_0 t + \delta ) \implies -A\omega_0^2 \le a(t) \le A\omega_0^2 \end{equation}\]
Energy balance of the harmonic oscillator
To calculate the energy of the harmonic oscillator, we consider the differential equation of the harmonic oscillator [diff_eq_harmonic oscillator], and we set \(\omega := \omega_0\):
\[\begin{equation} \ddot{x} + \omega^2x = 0 \qquad \big| \cdot m\dot{x} \end{equation}\]
\[\begin{equation} \implies m\dot{x}\dv{\dot{x}}{t} + m\omega^2x \dv{x}{t} = 0 \end{equation}\]
\[\begin{equation} \implies m\dv{}{t} \left ( \frac{\dot{x}^2}{2} \right ) + m\omega^2 \dv{}{t} \left ( \frac{x^2}{2} \right ) = 0 \end{equation}\]
\[\begin{equation} \implies \dv{}{t} \left ( \frac{1}{2}m\dot{x}^2 + \frac{1}{2} m\omega^2x^2 \right ) = 0 \end{equation}\]
\[\begin{equation} \underbrace{\frac{1}{2}m\dot{x}^2}_{E_{kin}} + \underbrace{\frac{1}{2}m\omega^2 x^2}_{E_{pot}} = \text{const.} := E_{tot} \end{equation}\]
Where:
\[\begin{equation} x(t) = A\sin(\omega t + \delta) \end{equation}\] \[\begin{equation} \dot{x}(t) = A \omega \cos(\omega t + \delta) \end{equation}\]
We then get:
\[\begin{equation} E_{tot} = \frac{1}{2}m\omega^2A^2\cos^2(\omega t + \delta) + \frac{1}{2}m\omega^2A^2\sin^2(\omega t + \delta) \end{equation}\]
The total energy of the harmonic oscillator is therefore constant with respect to time and is equal to:
\[\begin{equation} E_{tot} = \frac{1}{2}m\omega^2A^2 \end{equation}\]
It can be shown that the time-averaged kinetic and potential energies are equal:
\[\begin{equation} \ev{E_{kin}} = \ev{E_{pot}} = \frac{1}{2}E_{tot} \end{equation}\]
Damped oscillations
The damped harmonic oscillator
Consider a cart attached to a spring on a horizontal plane; by displacing the cart from its resting position, an elastic force \(F = -kx\) will act on it. The equation of motion is:
\[\begin{equation} m\ddot{x} = -kx \implies \ddot{x} +\frac{k}{m}x = 0 \end{equation}\]
Now we also introduce a frictional damping force proportional to the velocity, of the form \(F_a = -\varkappa \dot{x}\). We then obtain:
\[\begin{equation} m\ddot{x} = -kx - \varkappa \dot{x} \implies \ddot{x} + \frac{\varkappa}{m}\dot{x} + \frac{k}{m}x = 0 \end{equation}\]
We define the following quantities:
\[\begin{equation} \omega_0 = \sqrt{\frac{k}{m}} \quad \text{and} \quad \rho := \frac{\varkappa}{2m} \end{equation}\]
We therefore obtain the equation of the damped harmonic oscillator:
\[\begin{equation} \ddot{x}+2\rho \dot{x} + \omega_0^2 x = 0 \end{equation}\]
This is again a second-order linear differential equation. The characteristic polynomial is:
\[\begin{equation} \lambda^2 + 2\rho\lambda + \omega_0^2 = 0 \end{equation}\]
The discriminant of this equation is given by:
\[\begin{equation} \Delta = 4(\rho^2 - \omega_0^2) \end{equation}\]
We distinguish the following three cases:
1. Underdamping (\(\rho < \omega_0\))
In this case \(\Delta < 0\), therefore the general solution to the differential equation is:
\[\begin{equation} x(t) = \left( C_1 \cos(\omega t) + C_2\sin(\omega t) \right) e^{-\rho t} \end{equation}\]
Where \(C_1\) and \(C_2\) are real coefficients and \(\omega = \sqrt{\omega_0^2 - \rho^2}\).
The frequency of the damped harmonic oscillator is lower than the natural frequency of the undamped oscillation.
The equation of the underdamped harmonic oscillator can also be written in the form:
\[\begin{equation} x(t) = Ae^{-\rho t}\sin(\omega t + \delta) \end{equation}\]
2. Critical damping (\(\rho = \omega_0\))
With \(\Delta = 0\), the solution to the differential equation yields:
\[\begin{equation} x(t) = (C_1 + C_2t )e^{-\rho t} \end{equation}\]
3. Overdamping (\(\rho > \omega_0\))
In this case the discriminant is \(\Delta > 0\). The solution to the differential equation is therefore:
\[\begin{equation} x(t) = C_1e^{\lambda_1 t} + C_2e^{\lambda_2 t} \end{equation}\]
Where \(\lambda_{1,2} = -\rho \pm \sqrt{\rho^2 - \omega_0^2} < 0\) are both real coefficients.
In all three cases we have: \[\begin{equation} \lim_{t \rightarrow \infty}x(t) = 0 \end{equation}\] So a damped harmonic oscillator (without external driving forces) will inevitably come to a stop.
Energy balance of the damped oscillation
To compute the energy, we multiply the differential equation for the damped oscillator by the momentum \(m \dot{x}\):
\[\begin{equation} \ddot{x} + 2 \rho \dot{x} + \omega_0^2x = 0 \qquad \big| \cdot m\dot{x} \end{equation}\] \[\begin{equation} \implies \dv{}{t} \left( \frac{1}{2}m\dot{x}^2 + \frac{1}{2}m\omega_0^2x^2 \right) = -2\rho m \dot{x}^2 = F_R \cdot \dot{x} = P_R \end{equation}\]
where \(F_R\) is the friction force and \(P_R\) is the friction power.
Then the rate of change of the total energy is equal to the dissipated power:
\[\begin{equation} \dv{}{t} \{E_{kin} + E_{pot} \} = P_R \end{equation}\]
We consider now the case of underdamping, so that we can make the following approximation:
\[\begin{equation} \rho \ll \frac{1}{T} \end{equation}\]
This means that during one period, the amplitude of the oscillation decreases very slightly, and the exponential function can be approximated as \(1\) over this short time interval.
We then have the following:
\[\begin{equation} \ev{E_{kin}}_T = \frac{1}{2}E_{tot} = \frac{1}{2}m\ev{\dot{x}^2}_T \end{equation}\]
The friction power averaged over one period is equal to:
\[\begin{equation} P_R = -2\rho m \ev{\dot{x}^2}_T \implies P_R = -2\rho E_{tot} \end{equation}\]
\[\begin{equation} \dv{}{t}E_{tot} = -2\rho E_{tot} \implies E_{tot}(t) = \underbrace{E_{tot}(0)}_{:= E_0} \cdot e^{-2\rho t} \end{equation}\]
This means that the energy decreases exponentially with the decay constant \(\tau = \frac{1}{2\rho}\):
\[\begin{equation} E_{tot} (t) = E_0e^{-t/\tau} \end{equation}\]
The relative decrease in energy during one period is often described by the quality factor (Q-factor):
\[\begin{equation} Q := 2\pi \frac{E_{tot}(t)}{E_{tot}(t) - E_{tot}(t+T)} = 2\pi \frac{E_0e^{-t/\tau}}{E_0e^{-t/\tau}(1-e^{-T/\tau})} \end{equation}\]
Since \(\tau \gg T\), we can use a first-order Taylor approximation as follows:
\[\begin{equation} (1-e^{-T/\tau}) \approx 1-\left(1-\frac{T}{\tau}\right) = \frac{T}{\tau} \end{equation}\]
So the Q-factor becomes:
\[\begin{equation} Q \approx 2\pi \frac{\tau}{T} \end{equation}\]
Forced Oscillations and Resonance
Steady-state and transient solutions
Up to this point, we have examined systems that are displaced from equilibrium and then allowed to oscillate freely. We now extend this analysis to encompass oscillatory motion driven by an externally applied periodic force (or torque).
Let
\[\begin{equation} F_{ext}(t) = F_0 \cos (\Omega t) \quad F_0, \Omega \in \mathbb{R} \end{equation}\]
with \(\Omega\) being the angular frequency of the external periodic force. The differential equation of the forced oscillation is:
\[\begin{equation} \ddot{x} + 2\rho\dot{x} + \omega_0^2 x = \frac{F_0}{m}\cos (\Omega t) \end{equation}\]
The solution to a second-order non-homogeneous differential equation is the sum of the general solution to the homogeneous equation and a particular solution:
\[\begin{equation} x(t) = x_h(t) + x_p(t) \end{equation}\]
We can take the solution of the damped oscillator as the homogeneous solution:
\[\begin{equation} x_h(t) = A_0e^{-\rho t}\cos(\omega t + \delta) \end{equation}\]
Note that \(\lim_{t\rightarrow \infty} x_h(t) = 0\). So for \(t \gg \frac{1}{\rho}\) we have \(x(t) \rightarrow x_p(t)\). The term \(x_h\) is called the transient solution, while \(x_p\) represents the steady-state solution.
We guess a particular solution of the form:
\[\begin{equation} x_p(t) = x_0e^{i\Omega t} \end{equation}\]
Resonance
If we substitute the particular solution \(x_p\) into the equation of motion, we get:
\[\begin{equation} x_0 [-\Omega^2 + 2i \rho \Omega + \omega_0^2]e^{i\Omega t} = \frac{F_0}{m}e^{i\Omega t} \end{equation}\]
Recall that:
\[\begin{equation} F_{ext}(t) = \text{Re}\{F_0e^{i\Omega t}\} = F_0 \cos(\Omega t) \end{equation}\]
Then the complex amplitude is:
\[\begin{equation} x_0 = \frac{F_0/m}{(\omega_0^2 -\Omega^2) + 2i\rho\Omega} \end{equation}\]
Moreover, if we define:
\[\begin{equation} x_0 := \abs{x_0}e^{i\delta_0} \quad \text{and} \quad a_0 = \frac{F_0}{m} \end{equation}\]
Then the magnitude of the amplitude \(\abs{x_0}\) and the phase \(\delta_0\) of the forced oscillator are given by:
\[\begin{equation} \frac{\abs{x_0}}{a_0} = \frac{1}{\sqrt{(\omega_0^2 -\Omega^2)^2 + 4\rho^2\Omega^2}} := V(\rho, \Omega) \end{equation}\]
\[\begin{equation} \delta_0 = \arctan \left\{ \frac{-2\rho\Omega}{\omega_0^2-\Omega^2} \right\} \end{equation}\]
\(V(\rho,\Omega)\) is called the Amplitude Response.
The phase \(\delta_0\) describes the phase shift of the oscillation relative to the driving force.
Both \(V\) and \(\delta_0\) depend on the damping coefficient \(\rho\) and the driving frequency \(\Omega\).
These relationships can be elucidated by introducing two dimensionless variables, denoted by \(\eta\) and \(\xi\): \[\begin{equation} \eta \coloneqq \frac{\Omega}{\omega_0} \quad \text{and} \quad \xi \coloneqq \frac{\rho}{\omega_0} \end{equation}\] \[\begin{equation} \implies V(\eta, \xi ) = \frac{1}{\omega_0^2}\frac{1}{\sqrt{(1-\eta^2)^2+4\xi^2\eta^2}} \end{equation}\] \[\begin{equation} \delta_0 = \arctan\left\{\frac{-2\xi\eta}{1-\eta^2}\right\} \end{equation}\]
The peak of the resonance curve occurs where the derivative is zero:
\[\begin{equation} \left. \frac{\partial V}{\partial \eta} \right|_{\eta = \eta_{res}} = 0 \end{equation}\]
\[\begin{equation} \implies \eta_{res} = \sqrt{1-2\xi^2} \end{equation}\]
\[\begin{equation} \implies \Omega_{res} = \omega_0\sqrt{1-2\xi^2} \end{equation}\]
At this maximum point, we reach the resonance amplitude. Note that the resonance frequency \(\Omega_{res}\) is slightly smaller than the natural frequency of the oscillator.
Damping limits the maximum amplitude of oscillation, preventing catastrophic mechanical failures, such as the destruction of bridges under resonant wind loads.
Amplitude and Phase of Resonance
The resonance amplitude \(\abs{x_0}\) is:
proportional to the amplitude of the external driving force,
inversely proportional to both the detuning \(\abs{\omega_0 - \Omega}\) and the damping coefficient \(\rho\).
We can analyze the response in the following three limiting cases:
1. Low-frequency limit: \(\Omega \ll \omega_0\) and \(\eta \rightarrow 0\):
\[\begin{equation} \implies \abs{x_0} \approx \frac{a_0}{\omega_0^2} \end{equation}\]
\[\begin{equation} \delta_0 \approx 0 \end{equation}\]
In this regime, damping plays a negligible role; the (small) amplitude is determined primarily by the restoring force.
The oscillator moves in phase with the driving force.
2. Resonance condition: \(\Omega \approx \omega_0\) and \(\eta \rightarrow 1\):
\[\begin{equation} \implies \abs{x_0} \approx \frac{a_0}{2\rho} \end{equation}\] \[\begin{equation} \delta_0 = -\frac{\pi}{2} \end{equation}\] \[\begin{equation} \text{Re}(x) = \abs{x_0}\sin(\Omega t) \end{equation}\]
Energy is continuously and efficiently transferred to the oscillator.
3. High-frequency limit: \(\Omega \gg \omega_0\) and \(\eta \gg 1\): \[\begin{equation} \implies \abs{x_0} \approx \frac{a_0}{\Omega^2} \end{equation}\] \[\begin{equation} \delta_0 \approx -\pi \end{equation}\]
The (small) amplitude is now determined only by the inertia (mass \(m\)), and tends to zero asymptotically for \(\Omega \rightarrow \infty\).
The oscillator moves in antiphase (perfectly opposed) to the driving force.
Energy balance of the forced oscillation
We use the same approach as before to calculate the energy of the forced oscillation:
\[\begin{equation} m\ddot{x} + 2m\rho \dot{x} + m\omega_0^2x = F_{ext} \qquad \big| \cdot \dot{x} \end{equation}\] \[\begin{equation} \implies \dv{}{t} \underbrace{\left( \frac{1}{2}m \dot{x}^2 + \frac{1}{2}m\omega_0^2x^2 \right)}_{E_{osc}} = -2m\rho\dot{x}^2 + F_{ext}\dot{x} \end{equation}\]
In the steady state, the total energy of the oscillator remains constant on average over a cycle:
\[\begin{equation} \dv{}{t} \langle E_{osc} \rangle = 0 \implies \langle F_{ext}\dot{x} \rangle = \langle 2m \rho \dot{x}^2 \rangle \end{equation}\]
The power supplied by the external force is entirely dissipated by the damping force.
The mean power supplied during one period \(T_\Omega\) is given by:
\[\begin{equation} \ev{P}_{T_\Omega} = \frac{1}{T_\Omega} \int_0^{T_\Omega} F_{ext}\dot{x} \, dt \end{equation}\]
Since \(\dot{x}(t) = -\abs{x_0}\Omega \sin(\Omega t + \delta_0)\), it follows:
\[\begin{equation} \ev{P}_{T_\Omega} = -\frac{1}{2}F_0\Omega\abs{x_0}\sin(\delta_0) \end{equation}\]
Now recall that \(\abs{x_0}\sin(\delta_0) = \text{Im}(x_0)\), so we get:
\[\begin{equation} \ev{P}_{T_\Omega} = -\frac{1}{2}F_0\Omega \left \{ \frac{F_0}{m} \frac{-2\rho\Omega}{(\omega_0^2 - \Omega^2)^2+4 \rho^2 \Omega^2}\right \} \end{equation}\]
After some algebraic manipulations, we arrive at this important result:
\[\begin{equation} \ev{P}_{T_\Omega} = \frac{ma_0^2}{4\rho} \frac{1}{1 + \left (\frac{\omega_0^2 - \Omega^2}{2\rho\Omega}\right)^2} \end{equation}\]
We notice that the maximum power transfer (power resonance) occurs exactly when \(\Omega = \omega_0\).
We can use the dimensionless variables \(\eta, \xi\) defined earlier to rewrite this result:
\[\begin{equation} \ev{P}_{T_\Omega}(\eta, \xi) = \frac{ma_0^2}{4\omega_0}\frac{1}{\xi}\frac{1}{1+ \left (\frac{1-\eta^2}{2\xi\eta}\right)^2} \end{equation}\]