2003

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PY2003 Oscillations

A course on oscillations. Apparently pendula are the only thing we understand (but that's okay because everything is a pendulum, approximately).

Lecturer: Prof. John McGilp

Website: Link

As taught in: 05-06, 07-08

Contents

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Simple harmonic motion

Harmonic motion is motion produced by a restoring force which is (linearly) proportional to displacement. Simple harmonic motion involves no damping or external forcing terms and is described by the following equation.

m \ddot{x} + k x = 0
\ddot{x} + \omega_0^2 x = 0

where \omega_0^2 = \frac{k}{m}

A solution to this equation is x = A \cos({\omega_0 t + \phi}) where A is the amplitude and \phi is the phase.

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Phase

We say there is a phase lead of \phi when \phi \leq \pi
We say there is a phase lag of 2\pi - \phi when \phi \geq \pi
Antiphase |\phi| = \pi
Quatrature |\phi| = \frac{\pi}{2}

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Energy

Oscillatory motion involves the transfer of energy between an inertial component capable of storing kinetic energy (T) and an elastic component capable of storing potential energy (U).

E = T + U = \frac{m\dot{x}^2}{2} + \frac{k x^2}{2} = \frac{k A^2}{2}
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Damped oscillator

There is normally a drag force which opposes motion and which we can approximate as \propto \dot{x}

m \ddot{x} + b \dot{x} + k x = 0 \cdots b > 0
\ddot{x} + \gamma \dot{x} + \omega_0^2 x = 0 \cdots \gamma = \frac{b}{m} = Width

Trial solution:

x = C e^{p t} \cdots C,p \in \mathbb{C}
\Rightarrow p^2 + \gamma p + \omega_0^2 = 0
\Rightarrow p = - \frac{\gamma}{2} \pm \omega_0 \sqrt{(\frac{\gamma}{2 \omega_0})^2 - 1}

3 cases:

  1. Light damping \gamma < 2 \omega_0
  2. Critical damping \gamma = 2 \omega_0
  3. Heavy damping \gamma > 2 \omega_0
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Light damping

p = - \frac{\gamma}{2} \pm i \omega_f \cdots \omega_f = \omega_0 \sqrt{1 - {(\frac{\gamma}{2 \omega_0})}^2}
\Rightarrow x = C e^{(- \frac{\gamma}{2} \pm i \omega_f) t}
\Rightarrow x = A_0 e^{(- \frac{\gamma}{2} t)} \cos(\omega_f t + \phi) \cdots taking the real parts only

Therefore the body oscillates with a frequency \omega_f while the amplitude A = A_0 e^{(- \frac{\gamma}{2} t)} decays exponentially.

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Critical damping

p_1 = p_2 = - \frac{\gamma}{2} = -\omega_0 \ldots \gamma = 2 \omega_0
x = e^{-\omega_0 t}

Another linearly independent solution is

x = t e^{-\omega_0 t}

Therefore the general solution is,

x = C e^{-\omega_0 t} + D t e^{-\omega_0 t}
x = (C + D t) e^{-\omega_0 t}

Exponential decay (aperiodic motion ie. not periodic).
However a slight change in \gamma will lead to either light or heavy damping.

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Heavy damping

p = - \frac{\gamma}{2} \pm \sqrt{(\frac{\gamma}{2})^2 - \omega_0^2}
p_1 = - \frac{\gamma}{2} - \sqrt{(\frac{\gamma}{2})^2 - \omega_0^2} < 0
p_2 = - \frac{\gamma}{2} + \sqrt{(\frac{\gamma}{2})^2 - \omega_0^2} < 0
x = C_1 e^{p_1 t} + C_2 e^{p_2 t}

After a short time , the term e^{p_1 t} decays away and x \approx e^{p_2 t}, which is exponential decay with no oscillatory behavior (aperiodic motion).

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Energy

b causes system to lose energy. For the lightly damped case A = A_0 e^{(- \frac{\gamma}{2} t)} therefore,

E = \frac{1}{2} m A^2 = \frac{1}{2} m A_0^2 e^{- \gamma t}
\Rightarrow \frac{dE}{dt} = \frac{1}{2} m A_0^2 (- \gamma) e^{- \gamma t} = - \gamma E
\Rightarrow E = E_0 e^{-\gamma t}

Therefore the system will lose energy exponentially with a decay constant \tau = 1/\gamma.

The fractional loss in energy per cycle is

-\frac{1}{E} \frac{dE}{dt} T = \gamma T \approx \gamma \frac{2\pi}{\omega_0} = \frac{2\pi}{Q}

where Q is the quality factor defined as Q = \frac{\omega_0}{\gamma }.

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Forced Damped Oscillations

Approximate driving force as a cosine

m \ddot{x} + b \dot{x} + k x = F_0 \cos(\omega t)
\ddot{x} + \gamma \dot{x} + \omega_0^2 x = \frac{F_0}{m} \cos(\omega t)

Initial motion will have both frequencies \omega and \omega_0. Eventually it will settle to \omega as \omega_0 is damped out.

Steady state solution:

x = A \cos(\omega t - \delta)

or

x = A e^{i (\omega t - \delta)}
\Rightarrow (-\omega^2 + i \omega \gamma + \omega_0^2 ) A e^{i (\omega t - \delta)} = \frac{F_0}{m}e^{i \omega t}
\Rightarrow (\omega_0^2 -\omega^2) A + i \omega \gamma A = \frac{F_0}{m}e^{i \delta}
\Rightarrow A(\omega) = \frac{F_0/m}{\sqrt{{(\omega_0^2 -\omega^2)}^2+ (\omega \gamma)^2}}

and

\tan \delta (\omega) = \frac{\gamma \omega}{\omega_0^2 -\omega^2}
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Resonance

For the damped forced oscillator A_{max} occurs when

\frac{dA}{d\omega} = 0 \Rightarrow \omega_{max}^2 = \omega_0^2 -\frac{\gamma^2}{2}

Hence resonance will occur when \omega \approx \omega_0

At low frequencies, \frac{\omega}{\omega_0} \ll 1, the system is in phase with the driving force \delta \approx 0.

k x \approx F_0 \cos(\omega t) \cdots x \gg \dot{x} \gg \ddot{x}
\Rightarrow A \approx \frac{F_0}{m}

At resonance, \frac{\omega}{\omega_0} \approx 1, the system is in quadrature with driving force \delta \approx \pi/2.

A_{max} \approx A(\omega = \omega_0) = Q\frac{F_0}{k}

At high frequencies, \frac{\omega}{\omega_0} \gg 1, the system is in anti-phase with driving force \delta \approx \pi.

m \ddot{x} \approx F_0 \cos(\omega t) \cdots \ddot{x} \gg \dot{x} \gg x

and

A \approx 0
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Quality factor

Qis a measure of the sharpness and size of resonance.

Q = \frac{\omega_0}{\gamma} = \frac{\omega}{\Delta\omega_{1/2}}

Therefore Q large \Rightarrow smaller \Delta\omega_{1/2}\Rightarrow sharper resonance

Q = \frac{A_{\omega_{max}}}{F_0/k} = \frac{A_{\omega_{max}}}{A_{\omega=0}}

Therefore Q large \Rightarrow greater resonance

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Response function

The response function allows us to simplify the notation.

\Re (\omega) = \frac{(\gamma \omega)^2}{(\omega_0^2 -\omega^2)^2 + (\gamma \omega)^2}
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Lorentzian

Assume \omega \approx \omega then,

(\omega_0^2 -\omega^2)^2 \approx 4 \omega_0^2 (\omega_0 -\omega)^2

and the response function becomes,

L(\omega) = \frac{(\gamma / 2)^2}{(\omega_0 -\omega)^2 + (\gamma / 2)^2}

which is called the Lorentzian and is symmetric about \omega/\omega_0 \approx 1

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Dispersive and absorptive amplitudes

x = A \cos (\omega t - \delta)
\Rightarrow x = A \cos \delta \cos \omega t + A \sin \delta \sin \omega t
\Rightarrow x = A_{dis} \cos \omega t + A_{abs} \sin \omega t

A_{dis} is the dispersive / elastic amplitude (in phase with driving force).
A_{abs} is the absorptive amplitude (in quatrature with driving force).

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Power absorption

Power absorption occurs due to the resistance term b

P(t) = F(t) v(t)

Therefore, the average power,

<P> = \frac{1}{T}\int_t^{t+T} F(t) v(t) \,dt
<P> = \frac{1}{T}\int_t^{t+T} - F_0 \cos (\omega t) \, A \omega \sin(\omega t - \delta) \,dt
<P> = \frac{F_0^2}{2m\gamma}\Re (w)

\Delta \omega_{1/2} = FWHM = Full Width at Half Maximum = \gamma

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Driven LCR series circuit

L \ddot{q} + \dot{q}R + \frac{q}{C} = V_0 \cos \omega t
\ddot{q} + \dot{q}\frac{R}{L} + \frac{q}{LC} = \frac{V_0}{L} \cos \omega t

which is the equation for a driven damped oscillator with,

R \equiv b
L \equiv m
V_0 \equiv F_0
\frac{1}{C} \equiv k

For tuning circuits we want the circuit to have a high selectivity ie. a high quality factor.

Q = \frac{\omega_0}{\gamma} = \frac{1}{R} \sqrt{\frac{L}{C}}

Therefore we make R low, L high and use C (variable capacitor) to tune the circuit.

The frequency at which resonance occurs is,

\omega_0 \approx \omega
\Rightarrow \sqrt{\frac{k}{m}} \approx 2 \pi f
\Rightarrow f \approx \frac{1}{2 \pi} \frac{1}{\sqrt{LC}}
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Non-sinosoidal driving force

For periodic driving forces we apply Fourier analysis. As an example we will take the sawtooth function as our driving force.

F(t) = t for -\pi \leq t \leq \pi and defined elsewhere by F(t + 2\pi) = F(t)

Fourier analysis gives,

F(t) = \sum_{n=1, n\,odd}^{\infty} \frac{4}{\pi n^2}\cos nt

and therefore our EOM is,

\ddot{x} + \gamma \dot{x} + \omega_0^2 x = \sum_{n=1, n\,odd}^{\infty} \frac{4}{\pi n^2}\cos nt

Solve for x_n and then use principle of superposition to obtain x = x_1 + x_2 + \cdots

\ddot{x}_n + \gamma \dot{x}_n + \omega_0^2 x_n = \frac{4}{\pi n^2}\cos nt

which has solution,

x_n = A_n \cos nt + B_n \sin nt

if,

-A_n n^2 \cos nt - B_n n^2 \sin nt + \gamma n A_n \sin nt + \gamma n B_n \cos nt + \omega_0^2 A_n \cos nt + \omega_0^2 B_n \sin nt = \frac{4}{\pi n^2}\cos nt
\Rightarrow -A_n n^2 + \gamma n B_n + \omega_0^2 A_n \cos nt = \frac{4}{\pi n^2}\cos nt and -B_n n^2 + \gamma n A_n + \omega_0^2 B_n = 0

Therefore,

A_n = \frac{4}{\pi n^2} \frac{\omega_0^2 - n^2}{(\omega_0^2 - n^2)^2 + (\gamma n)^2}
B_n = \frac{\gamma n A_n}{\omega_0^2 - n^2} = \frac{4 \gamma}{\pi n} \frac{1}{(\omega_0^2 - n^2)^2 + (\gamma n)^2}

and

x = \sum_{n=0, n\,odd}^{\infty} A_n \cos nt + B_n \sin nt

Note: The frequency which is closest to \omega_0 will have the greatest amplitude.

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Two coupled oscillators

The general motion of two coupled oscillators is a superposition of 2 independent simple harmonic motions occurring simultaneously. These independent motions are called modes or normal modes. There are 3 methods of finding modes:

  • physical reasoning
  • find normal co-ordinates directly from equations of motion
  • general approach
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General approach

The equations of motion for two coupled oscillators may be written as follows:

m_A \ddot{x}_A = - k_A x_A - k_{AB} (x_A - x_B)
m_B \ddot{x}_B = - k_B x_B - k_{AB} (x_B - x_A)

or,

\ddot{x}_A = - \Bigg( \frac{k_A}{m_A} + \frac{k_{AB}}{m_A}\Bigg) x_A - \Bigg( - \frac{k_{AB}}{m_A} \Bigg) x_B \Rightarrow \ddot{x}_A = - a_{11} x_A - a_{12} x_B
\ddot{x}_B = - \Bigg( - \frac{k_{AB}}{m_B}\Bigg) x_A - \Bigg( \frac{k_B}{m_B} + \frac{k_{AB}}{m_B}\Bigg) x_B \Rightarrow \ddot{x}_B = - a_{21} x_A - a_{22} x_B

To solve, we assume the system oscillates in a mode, implying that both x_A, x_B have the same frequency \omega and are in phase:

x_A = A \cos (\omega t + \phi) and x_B = B \cos (\omega t + \phi)
\Rightarrow \ddot{x}_A = - \omega^2 x_A and \ddot{x}_B = - \omega^2 x_B

Filling this in to our equations gives:

(a_{11} - \omega^2) x_A + a_{12}x_B = 0
a_{21} x_A + (a_{22} - \omega^2) x_B = 0

(Notice how the \omega^2 term acts as an eigenvalue for the matrix of coefficients.) Now, for this system to have non-trivial solution we want its determinant to be zero. Taking the determinant gives us an equation for \omega^2:

(\omega^2)^2 - (a_{11} + a_{22}) \omega^2 + (a_{11} a_{22} - a_{12}a_{21}) = 0

We can solve this, giving us two solutions \omega_1, \omega_2 and so:

x_A = A_1 \cos (\omega_1 t + \phi_1) + A_2 \cos (\omega_2 t + \phi_2)
x_B = B_1 \cos (\omega_1 t + \phi_1) + B_2 \cos (\omega_2 t + \phi_2)

This approach can be generalised to N coupled oscillators.

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N Coupled oscillators

A system of N coupled oscillators each with 1 degree of freedom, has N normal modes (one normal mode for each DOF) and the motion of the system is a superposition of these modes.

There are 3 methods for finding the modes:

  1. Physical reasoning: Find normal modes by inspection.
  2. Find N coupled EOM, uncouple them by a change of variable and solve.
  3. Find N coupled EOM and solve them using matrices and determinants.
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Solving using matrices and determinants

\ddot{x}_1 = a_{11} x_1 + a_{12} x_2 + \cdots + a_{1n}x_n
\ddot{x}_2 = a_{21} x_1 + a_{22} x_2 + \cdots + a_{2n}x_n
 \vdots            \vdots            \vdots                            \vdots
\ddot{x}_n = a_{n1} x_1 + a_{n2} x_2 + \cdots + a_{nn}x_n

In matrix notation,

\ddot{X} = A X

Trial solution,

X = X_0 e^{i \omega t}
\Rightarrow -\omega^2 X_0 e^{i \omega t} = A X_0 e^{i \omega t}
\Rightarrow (A -\omega^2 I) X_0 = 0

which has non-trivial solution if

det(A -\omega^2 I) = 0

Solve for \omega to get the eigenfrequencies.
For each eigenfrequency, sub \omega back into (A -\omega^2 I) X_0 = 0 to get the eigenvectors X_0.
The solution is a linear combination of these eigenmodes X_0 e^{i \omega t}

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Many coupled oscillators

Consider N beads of mass m on an elastic string.
Total length L = (N+1)a
Tension at equilibrium T_0 = k(a - a_0).


EOM of pth bead (in y direction)

m \ddot{y}_p = T_p \sin \alpha_p - T_{p-1} \sin \alpha_{p-1}

For small displacements y \ll a

T_p \approx T_{p-1} \approx T_0 and L_p \approx L_{p-1} \approx a
\sin \alpha_p = \frac{y_{p+1}-y_p}{L_p} \approx \frac{y_{p+1}-y_p}{a}
\sin \alpha_{p-1} = \frac{y_p-y_{p-1}}{L_{p-1}} \approx \frac{y_p-y_{p-1}}{a}

Therefore,

m \ddot{y}_p = \frac{T_0}{a}(y_{p+1} - 2 y_p + y_{p-1})
\ddot{y}_p = \omega_0^2 (y_{p+1} - 2 y_p + y_{p-1}) \ldots \omega_0 = \sqrt{\frac{T_0}{m a}}

which gives N coupled equations of motion and can be solved by solving the N \times N determinant.

However a simpler method is to solve for the normal modes. In a normal mode each particle has the same frequency \omega and phase \phi,

y_p = A_p \cos (\omega t + \phi)
\Rightarrow -\omega^2 A_p = \omega_0^2 (A_{p+1}-2A_p + A_{p-1})
\Rightarrow \frac{A_{p+1}- A_{p-1}}{A_p} = 2 - \frac{\omega^2}{\omega_0^2} =const

which has solution

A_p = C \sin p\theta or A_p = D \cos p\theta

because

\frac{A_{p+1}- A_{p-1}}{A_p} = \frac{C \sin (p+1)\theta + C sin(p-1)\theta}{C \sin p\theta} = 2 \cos \theta =const

Therefore,

A_p = C \sin p\theta + D \cos p\theta

To find \theta apply the boundary conditions

p=0 \Rightarrow y_p=0 \Rightarrow A_p=0 \Rightarrow D=0
p=N+1 \Rightarrow y_p=0 \Rightarrow A_p=0 \Rightarrow \theta(N+1)=n\pi\ \Rightarrow \theta = \frac{n}{N+1}\pi \ldots n=0,1,2,\ldots

Therefore,

A_{pn} = C \sin (\frac{n p \pi}{N+1}) \ldots n=0,1,2,\ldots

Also,

2 - \frac{\omega^2}{\omega_0^2} = 2 \cos \theta = 2 \cos (\frac{n \pi}{N+1})
\Rightarrow \omega^2 = 2 \omega_0^2 (1 - \cos (\frac{n \pi}{N+1}))
\Rightarrow \omega^2 = 2 \omega_0^2 (2 \sin^2 (\frac{n \pi}{2(N+1)}))
\Rightarrow \omega_n = 2 \omega_0 \sin (\frac{n \pi}{2(N+1)})

Therefore the displacement of the pth bead is given by the sum of all normal modes y_{pn} where,

y_{pn} = A_{pn} \cos(\omega_n t +\phi_n)\ldots n=0,1,2,\ldots
\omega_n = 2 \omega_0 \sin (\frac{n \pi}{2(N+1)})
A_{pn} = C_n \sin (\frac{p n \pi}{N+1})
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Transition to continuous systems

N \rightarrow \infty, a \rightarrow 0
L = (N+1)a =const
\rho = \frac{m}{a}=mass per unit length (const)


Taylor expand y_{p+1} and y_{p-1} about y_p

y_{p+1} = y_p + \frac{\partial y}{\partial x} a + \frac{1}{2}\frac{\partial^2y}{\partial x^2} a^2 + \cdots
y_{p-1} = y_p + \frac{\partial y}{\partial x}(-a) + \frac{1}{2}\frac{\partial^2y}{\partial x^2} a^2 + \cdots

Ignoring higher order terms we have,

\ddot{y}_p = \frac{T_0}{ma}(y_{p+1} - 2 y_p + y_{p-1})\approx \frac{T_0}{ma}\frac{\partial^2y}{\partial x^2} a^2

as a \rightarrow 0 this approximation becomes exact

\frac{\partial^2y}{\partial t^2} = \frac{T_0}{\rho}\frac{\partial^2y}{\partial x^2} \cdots \rho = \frac{m}{a}

which is the wave equation

\frac{\partial^2y}{\partial t^2} = v^2 \frac{\partial^2y}{\partial x^2}

with

v^2 = \frac{T_0}{\rho}
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Normal modes in a continuous medium

A_{pn} = C_n \sin (\frac{p n \pi}{N+1})
A_n(x) = C_n \sin (\frac{n \pi x}{(N+1)a}) \ldots x = pa
A_n(x) = C_n \sin (\frac{n \pi x}{L})\ldots L = (N+1)a


\omega_n = 2 \omega_0 \sin (\frac{n \pi}{2(N+1)})
\omega_n = 2 \omega_0 \sin (\frac{n a \pi}{2L}) \ldots L = (N+1)a
\omega_n = 2 \omega_0 \sin (\frac{a \pi}{\lambda_n})\ldots \lambda_n = \frac{2 L}{n}
\omega_n = 2 \sqrt{\frac{T_0}{ma}} \sin (\frac{a \pi}{\lambda_n})\ldots \lambda_n = \frac{2 L}{n}

For a \ll \lambda_n

\omega_n = 2 \sqrt{\frac{T_0}{ma}} \frac{a \pi}{\lambda_n}
\omega_n = \frac{2 \pi}{\lambda_n} \sqrt{\frac{T_0}{\rho}}
\omega_n = \frac{2 \pi}{\lambda_n} v
\omega_n = k_n v

This is a dispersion relation \omega (k)

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Anharmonic behaviour

Non-linear restoring force

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References

  • Lecture Notes of Prof. John McGlip 2005/2006
  • Vibrations and Waves by Main (SLEN 531.1 L8)
  • Vibrations and Waves by French (SLEN 531.32 L12)
  • The Physics of Vibrations and Waves by Pain (SLEN 531.1 L61)
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