# Engineering Acoustics/Simple Oscillation

## The Position Equation

This section shows how to form the equation describing the position of a
mass on a spring.

For a simple oscillator consisting of a mass *m* attached to one end of
a spring with a spring constant *s*, the restoring force, *f*, can be
expressed by the equation

```{=html}
<center>
```
$f = -sx\,$

```{=html}
</center>
```
where *x* is the displacement of the mass from its rest position.
Substituting the expression for *f* into the linear momentum equation,

```{=html}
<center>
```
$f = ma = m{d^2x \over dt^2}\,$

```{=html}
</center>
```
where *a* is the acceleration of the mass, we can get

```{=html}
<center>
```
$m\frac{d^2 x}{d t^2 }= -sx$

```{=html}
</center>
```
or,

```{=html}
<center>
```
$\frac{d^2 x}{d t^2} + \frac{s}{m}x = 0$

```{=html}
</center>
```
Note that the frequency of oscillation $\omega_0$ is given by

```{=html}
<center>
```
$\omega_0^2 = {s \over m}\,$

```{=html}
</center>
```
To solve the equation, we can assume

```{=html}
<center>
```
$x(t)=A e^{\lambda t} \,$

```{=html}
</center>
```
The force equation then becomes

```{=html}
<center>
```
$(\lambda^2+\omega_0^2)A e^{\lambda t} = 0,$

```{=html}
</center>
```
Giving the equation

```{=html}
<center>
```
$\lambda^2+\omega_0^2 = 0,$

```{=html}
</center>
```
Solving for $\lambda$

```{=html}
<center>
```
$\lambda = \pm j\omega_0\,$

```{=html}
</center>
```
This gives the equation of *x* to be

```{=html}
<center>
```
$x = C_1e^{j\omega_0 t}+C_2e^{-j\omega_0 t}\,$

```{=html}
</center>
```
Note that

```{=html}
<center>
```
$j = (-1)^{1/2}\,$

```{=html}
</center>
```
and that *C~1~* and *C~2~* are constants given by the initial conditions
of the system

If the position of the mass at *t* = 0 is denoted as *x~0~*, then

```{=html}
<center>
```
$C_1 + C_2 = x_0\,$

```{=html}
</center>
```
and if the velocity of the mass at *t* = 0 is denoted as *u~0~*, then

```{=html}
<center>
```
$-j(u_0/\omega_0) = C_1 - C_2\,$

```{=html}
</center>
```
Solving the two boundary condition equations gives

```{=html}
<center>
```
$C_1 = \frac{1}{2}( x_0 - j( u_0 / \omega_0 ))$

```{=html}
</center>
```
\

```{=html}
<center>
```
$C_2 = \frac{1}{2}( x_0 + j( u_0 / \omega_0 ))$

```{=html}
</center>
```
\
The position is then given by

```{=html}
<center>
```
**$x(t) = x_0 cos(\omega_0 t) + (u_0 /\omega_0 )sin(\omega_0 t)\,$**

```{=html}
</center>
```
\
This equation can also be found by assuming that *x* is of the form

```{=html}
<center>
```
$x(t)=A_1 cos(\omega_0 t) + A_2 sin(\omega_0 t)\,$

```{=html}
</center>
```
And by applying the same initial conditions,

```{=html}
<center>
```
$A_1 = x_0\,$

```{=html}
</center>
```
\

```{=html}
<center>
```
$A_2 = \frac{u_0}{\omega_0}\,$

```{=html}
</center>
```
\
This gives rise to the same position equation

```{=html}
<center>
```
$x(t) = x_0 cos(\omega_0 t) + (u_0 /\omega_0 )sin(\omega_0 t)\,$

```{=html}
</center>
```
Back to Main page

## Alternate Position Equation Forms

If *A~1~* and *A~2~* are of the form

```{=html}
<center>
```
$A_1 = A cos( \phi)\,$

```{=html}
</center>
```
```{=html}
<center>
```
$A_2 = A sin( \phi)\,$

```{=html}
</center>
```
\
Then the position equation can be written

```{=html}
<center>
```
**$x(t) = Acos( \omega_0 t - \phi )\,$**

```{=html}
</center>
```
\
By applying the initial conditions (*x(0)=x~0~, u(0)=u~0~*) it is found
that

```{=html}
<center>
```
$x_0 = A cos(\phi)\,$

```{=html}
</center>
```
\

```{=html}
<center>
```
$\frac{u_0}{\omega_0} = A sin(\phi)\,$

```{=html}
</center>
```
\
If these two equations are squared and summed, then it is found that

```{=html}
<center>
```
**$A = \sqrt{x_0^2 + (\frac{u_0}{\omega_0})^2}\,$**

```{=html}
</center>
```
\
And if the difference of the same two equations is found, the result is
that

```{=html}
<center>
```
**$\phi = tan^{-1}(\frac{u_0}{x_0 \omega_0})\,$**

```{=html}
</center>
```
The position equation can also be written as the Real part of the
imaginary position equation

```{=html}
<center>
```
$\mathbf{Re} [x(t)] = x(t) = A cos(\omega_0 t - \phi)\,$

```{=html}
</center>
```
\
Due to euler\'s rule (e^jφ^ = cosφ + jsinφ), **x**(t) is of the form

```{=html}
<center>
```
$x(t) = A e^{j(\omega_0 t - \phi)}\,$

```{=html}
</center>
```
{m\_{TOTAL}}} = \\sqrt{\\frac{2s}{M}} `</math>`{=html}

$\mathbf{f_0} = \frac{\omega_{0}}{2\pi} = \mathbf{\frac{1}{2\pi}\sqrt{\frac{2s}{M}}}$

2.  !Simple
    Oscillator-1.2.1.b{width="100"}

$\omega_{0} = \sqrt{\frac{s_{TOTAL}}{m_{TOTAL}}} = \sqrt{\frac{s}{2M}}$

$\mathbf{f_0} = \frac{\omega_{0}}{2\pi} = \mathbf{\frac{1}{2\pi}\sqrt{\frac{s}{2M}}}$

3.  !Simple
    Oscillator-1.2.1.c{width="100"}

!Simple
Oscillator-1.2.1.c-solution{width="300"}

$\mathbf{1.}\text{ }
s(x_1-x_2) = sx_2$

$\mathbf{2.}\text{ }
-s(x_1-x_2) = m \frac{d^2x}{dt^2}$

$\frac{d^2x_1}{dt^2} + \frac{s}{2m}x_1 = 0$

$\omega_0 = \sqrt{\frac{s}{2m}}$

$\mathbf{f_0 = \frac{1}{2\pi}\sqrt{\frac{s}{2m}}}$

4.  !Simple
    Oscillator-1.2.1.d{width="100"}

$\omega_0=\sqrt{\frac{2s}{m}}$

$\mathbf{f_0 = \frac{1}{2\pi}\sqrt{\frac{2s}{m}}}$

}}

Back to Main page


# Engineering Acoustics/Mechanical Resistance

## Mechanical Resistance

For most systems, a simple oscillator is not a very accurate model.
While a simple oscillator involves a continuous transfer of energy
between kinetic and potential form, with the sum of the two remaining
constant, real systems involve a loss, or dissipation, of some of this
energy, which is never recovered into kinetic nor potential energy. The
mechanisms that cause this dissipation are varied and depend on many
factors. Some of these mechanisms include drag on bodies moving through
the air, thermal losses, and friction, but there are many others. Often,
these mechanisms are either difficult or impossible to model, and most
are non-linear. However, a simple, linear model that attempts to account
for all of these losses in a system has been developed.

## Dashpots

The most common way of representing mechanical resistance in a damped
system is through the use of a dashpot. A dashpot acts like a shock
absorber in a car. It produces resistance to the system\'s motion that
is proportional to the system\'s velocity. The faster the motion of the
system, the more mechanical resistance is produced.

As seen in the graph above, a linear relationship is assumed between the
force of the dashpot and the velocity at which it is moving. The
constant that relates these two quantities is $R_M$, the mechanical
resistance of the dashpot. This relationship, known as the viscous
damping law, can be written as:

```{=html}
<center>
```
$F = R\cdot u$

```{=html}
</center>
```
Also note that the force produced by the dashpot is always in phase with
the velocity.

The power dissipated by the dashpot can be derived by looking at the
work done as the dashpot resists the motion of the system:

```{=html}
<center>
```
$P_D = \frac{1}{2}\Re\left[\hat{F}\cdot\hat{u^*}\right]= \frac{|\hat{F}|^{2}}{2R_{M}}$

```{=html}
</center>
```
## Modeling the Damped Oscillator

In order to incorporate the mechanical resistance (or damping) into the
forced oscillator model, a dashpot is placed next to the spring. It is
connected to the mass ($M_M$) on one end and attached to the ground on
the other end. A new equation describing the forces must be developed:

```{=html}
<center>
```
$F - S_Mx - R_Mu = M_Ma  \rightarrow  F = S_Mx + R_M\dot{x} + M_M\ddot{x}$

```{=html}
</center>
```
It\'s phasor form is given by the following:

```{=html}
<center>
```
$\hat{F}e^{j\omega t} = \hat{x}e^{j\omega t}\left[S_M + j\omega R_M + \left(-\omega ^2\right)M_M\right]$

```{=html}
</center>
```
## Mechanical Impedance for Damped Oscillator

Previously, the impedance for a simple oscillator was defined as
$\mathbf{\frac{F}{u}}$. Using the above equations, the impedance of a
damped oscillator can be calculated:

```{=html}
<center>
```
$\hat{Z_M} = \frac{\hat{F}}{\hat{u}} = R_M + j\left(\omega M_M - \frac{S_M}{\omega}\right) = |\hat{Z_M}|e^{j\Phi_Z}$

```{=html}
</center>
```
For very low frequencies, the spring term dominates because of the
$\frac{1}{\omega}$ relationship. Thus, the phase of the impedance
approaches $\frac{-\pi}{2}$ for very low frequencies. This phase causes
the velocity to \"lag\" the force for low frequencies. As the frequency
increases, the phase difference increases toward zero. At resonance, the
imaginary part of the impedance vanishes, and the phase is zero. The
impedance is purely resistive at this point. For very high frequencies,
the mass term dominates. Thus, the phase of the impedance approaches
$\frac{\pi}{2}$ and the velocity \"leads\" the force for high
frequencies.

\
Based on the previous equations for dissipated power, we can see that
the real part of the impedance is indeed $R_M$. The real part of the
impedance can also be defined as the cosine of the phase times its
magnitude. Thus, the following equations for the power can be obtained.

```{=html}
<center>
```
$W_R = \frac{1}{2}\Re\left[\hat{F}\hat{u^{*}}\right] = \frac{1}{2}R_M|\hat{u}|^2 = \frac{1}{2}\frac{|\hat{F}|^2}{|\hat{Z_M}|^2}R_M = \frac{1}{2}\frac{|\hat{F}|^2}{|\hat{Z_M}|}cos(\Phi_Z)$

```{=html}
</center>
```
Back to Main page


# Engineering Acoustics/Characterizing Damped Mechanical Systems

## Characterizing Damped Mechanical Systems

Characterizing the response of Damped Mechanical Oscillating system can
be easily quantified using two parameters. The system parameters are the
resonance frequency ($'''w  resonance'''$ and the damping of the system
$'''Q  (quality factor)or B  (Temporal Absorption''')$. In practice,
finding these parameters would allow for quantification of unknown
systems and allow you to derive other parameters within the system.

------------------------------------------------------------------------

Using the mechanical impedance in the following equation, notice that
the imaginary part will equal zero at resonance.

($Z_m = F/u = R_m + j(w*M_m - s/w)$)

Resonance case:($w*M_m = s/w$)

## Calculating the Mechanical Resistance

The decay time of the system is related to 1 / B where B is the Temporal
Absorption. B is related to the mechancial resistance and to the mass of
the system by the following equation.

$B = Rm / 2*Mm$

The mechanical resistance can be derived from the equation by knowing
the mass and the temporal absorption.

## Critical Damping

The system is said to be critically damped when:

$Rc = 2*M*sqrt(s/Mm) = 2*sqrt(s*Mm) = 2*Mm*wn$

A critically damped system is one in which an entire cycle is never
completed. The absorption coefficient in this type of system equals the
natural frequency. The system will begin to oscillate, however the
amplitude will decay exponentially to zero within the first oscillation.

## Damping Ratio

$Damping Ratio = Rm/Rc$

The damping ratio is a comparison of the mechanical resistance of a
system to the resistance value required for critical damping. Rc is the
value of Rm for which the absorption coefficient equals the natural
frequency (critical damping). A damping ratio equal to 1 therefore is
critically damped, because the mechanical resistance value Rm is equal
to the value required for critical damping Rc. A damping ratio greater
than 1 will be overdamped, and a ratio less than 1 will be underdamped.

## Quality Factor

The Quality Factor (Q) is way to quickly characterize the shape of the
peak in the response. It gives a quantitative representation of power
dissipation in an oscillation.

$Q = wresonance / (wu - wl)$

Wu and Wl are called the half power points. When looking at the response
of a system, the two places on either side of the peak where the point
equals half the power of the peak power defines Wu and Wl. The distance
in between the two is called the half-power bandwidth. So, the resonant
frequency divided by the half-power bandwidth gives you the quality
factor. Mathematically, it takes Q/pi oscillations for the vibration to
decay to a factor of 1/e of its original amplitude.

Back to Main page


# Engineering Acoustics/Electro-Mechanical Analogies

## Why Circuit Analogs?

Acoustic devices are often combinations of mechanical and electrical
elements. A common example of this would be a loudspeaker connected to a
power source. It is useful in engineering applications to model the
entire system with one method. This is the reason for using a circuit
analogy in a vibrating mechanical system. The same analytic method can
be applied to Electro-Acoustic
Analogies.

## How Electro-Mechanical Analogies Work

An electrical circuit is described in terms of its potential (voltage)
and flux (current). To construct a circuit analog of a mechanical system
we define flux and potential for the system. This leads to two separate
analog systems. The Impedance Analog denotes the force acting on an
element as the potential and the velocity of the element as the flux.
The Mobility Analog equates flux with the force and velocity with
potential.

                         Mechanical   Electrical Equivalent
  ---------------------- ------------ -----------------------
  **Impedance Analog**
  Potential:             Force        Voltage
  Flux:                  Velocity     Current
  **Mobility Analog**
  Potential:             Velocity     Voltage
  Flux:                  Force        Current

For many, the mobility analog is considered easier for a mechanical
system. It is more intuitive for force to flow as a current and for
objects oscillating the same frequency to be wired in parallel. However,
either method will yield equivalent results and can also be translated
using the dual (dot) method.

## The Basic Elements of an Oscillating Mechanical System

**The Mechanical Spring:**

:   ![](resistor11.jpg "fig:resistor11.jpg")

The ideal spring is considered to be operating within its elastic limit,
so the behavior can be modeled with Hooke\'s
Law.
It is also assumed to be massless and have no damping effects.

$$F=-cx, \$$

**The Mechanical Mass**

In a vibrating system, a mass element opposes acceleration. From
Newton\'s Second Law:

$$F=mx^{\prime\prime}=ma=m\frac{du}{dt}$$

$$F=K\int\,u dt$$

**The Mechanical Resistance**

:   ![](Dashpot.png "fig:Dashpot.png")

The dashpot is an ideal viscous damper which opposes velocity.

$$F=R u\displaystyle$$

**Ideal Generators**

The two ideal generators which can drive any system are an ideal
velocity and ideal force generator. The ideal velocity generator can be
denoted by a drawing of a crank or simply by declaring $u(t)=f(t)$, and
the ideal force generator can be drawn with an arrow or by declaring
$F(t)=f(t)$

**Simple Damped Mechanical Oscillators**

:   ![](Forced_Oscillator.PNG "fig:Forced_Oscillator.PNG")

In the following sections we will consider this simple mechanical system
as a mobility and impedance analog. It can be driven either by an ideal
force or an ideal velocity generator, and we will consider simple
harmonic motion. The m in the subscript denotes a mechanical system,
which is currently redundant, but can be useful when combining
mechanical and acoustic systems.

## The Impedance Analog

**The Mechanical Spring**

In a spring, force is related to the displacement from equilibrium. By
Hooke\'s Law,

$$F(t)=c_m \Delta x = c_m \int_{0}^{t} u( \tau )d \tau$$

The equivalent behaviour in a circuit is a capacitor:

$$V(t)=\frac{1}{C}\int_{0}^{t} \,i(\tau) d\tau$$

**The Mechanical Mass**

The force on a mass is related to the acceleration (change in velocity).
The behaviour, by Newton\'s Second Law, is:

$$F(t)=m_ma=m_m\frac{d}{dt}u(t)$$

The equivalent behaviour in a circuit is an inductor:

$$V(t)=L\frac{d}{dt}i(t)$$

**The Mechanical Resistance**

For a viscous damper, the force is directly related to the velocity

$$F=R_m u\displaystyle$$

The equivalent is a simple resistor of value $R_m\displaystyle$

$$V=R i\displaystyle$$

**Example:**

Thus the simple mechanical oscillator in the previous section becomes a
series RCL Circuit:

![](RLC_series_circuit_v1.svg "RLC_series_circuit_v1.svg")

The current through all three elements is equal (they are at the same
velocity) and that the sum of the potential drops across each element
will equal the potential at the generator (the driving force). The ideal
voltage generator depicted here would be equivalent to an ideal force
generator.

**IMPORTANT NOTE**: The velocity measured for the spring and dashpot is
the relative velocity ( velocity of one end minus the velocity of the
other end). The velocity of the mass, however, is the absolute velocity.

**Impedances:**

  Element               Impedance
  --------- ----------- -------------------------------------------------
  Spring    Capacitor   $Z_c = \frac{V_c}{I_c} = \frac{c_m}{j \omega }$
  Mass      Inductor    $Z_m = \frac{V_m}{I_m} = j \omega m_m$
  Dashpot   Resistor    $Z_d = \frac{V_m}{I_m} = R_m$

## The Mobility Analog

Like the Impedance Analog above, the equivalent elements can be found by
comparing their fundamental equations with the equations of circuit
elements. However, since circuit equations usually define voltage in
terms of current, in this case the analogy would be an expression of
velocity in terms of force, which is the opposite of convention.
However, this can be solved with simple algebraic manipulation.

**The Mechanical Spring**

$$F(t)= c_m \int u(t)d t$$

The equivalent behavior for this circuit is the behavior of an inductor.

$$\int V dt=\int L \frac{d}{dt} i(t) dt$$

$$i=\frac{1}{L}\int\,V dt$$

**The Mechanical Mass**

$$F=m_ma=m_m\frac{d}{dt}u(t)$$

Similar to the spring element, if we take the general equation for a
capacitor and differentiate,

$$\frac{d}{dt}V(t)=\frac{d}{dt}\frac{1}{C}\int \,i(t) dt$$

$$i(t)=C\frac{d}{dt}V(t)$$

**The Mechanical Resistance**

Since the relation between force and velocity is proportionate, the only
difference is that the mechanical resistance becomes inverted:

$$F=\frac{1}{r_m} u=R_m u$$

$$i=\frac{1}{R}V$$

**Example:**

The simple mechanical oscillator drawn above would become a parallel RLC
Circuit. The potential across each element is the same because they are
each operating at the same velocity. This is often the more intuitive of
the two analogy methods to use, because you can visualize force
\"flowing\" like a flux through your system. The ideal voltage generator
in this drawing would correspond to an ideal velocity generator.

![](RLC_parallel_circuit.png "RLC_parallel_circuit.png")

**IMPORTANT NOTE:** Since the measure of the velocity of a mass is
absolute, a capacitor in this analogy must always have one terminal
grounded. A capacitor with both terminals at a potential other than
ground may be realized physically as an inverter, which completes all
elements of this analogy.

**Impedances:**

  Element               Impedance
  --------- ----------- --------------------------------------------------
  Spring    Inductor    $Z_c = \frac{V_m}{I_m} = \frac{j \omega}{c_m}$
  Mass      Capacitor   $Z_m = \frac{V_c}{I_c} = \frac{1}{j \omega m_m}$
  Dashpot   Resistor    $Z_d = \frac{V_m}{I_m} = r_m = \frac{1}{R_m}$

Back to Main page


# Engineering Acoustics/Solution Methods: Electro-Mechanical Analogies

After drawing the electro-mechanical analogy of a mechanical system, it
is always safe to check the circuit. There are two methods to accomplish
this:

## Review of Circuit Solving Methods

**Kirchkoff\'s Voltage law**

\"The sum of the potential drops around a loop must equal zero.\"

![](KVL.png "KVL.png")

$v_1 + v_2 + v_3 + v_4 = 0 \displaystyle$

**Kirchkoff\'s Current Law**

\"The Sum of the currents at a node (junction of more than two elements)
must be zero\"

![](KCL.png "KCL.png")

$-i_1+i_2+i_3-i_4 = 0  \displaystyle$

**Hints for solving circuits:**

Remember that certain elements can be combined to simplify the circuit
(the combination of like elements in series and parallel)

If solving a ciruit that involves steady-state sources, use impedances.
Any circuit can eventually be combined into a single impedance using the
following identities:

Impedances in series: $Z_\mathrm{eq} = Z_1 + Z_2 + \,\cdots\, + Z_n.$

Impedances in parallel:
$\frac{1}{Z_\mathrm{eq}} = \frac{1}{Z_1}  +   \frac{1}{Z_2}  + \,\cdots\, +  \frac{1}{Z_n} .$

## Dot Method: (Valid only for planar network)

This method helps obtain the dual analog (one analog is the dual of the
other). The steps for the dot product are as follows: 1) Place one dot
within each loop and one outside all the loops. 2) Connect the dots.
Make sure that only there is only one line through each element and that
no lines cross more than one element. 3) Draw in each line that crosses
an element its dual element, including the source. 4) The circuit
obtained should have an equivalent behavior as the dual analog of the
original electro-mechanical circuit.

**Example:**

![](Dotmethod.jpg "Dotmethod.jpg")

The parallel RLC Circuit above is equivalent to a series RLC driven by
an ideal current source

## Low-Frequency Limits

This method looks at the behavior of the system for very large or very
small values of the parameters and compares them with the expected
behavior of the mechanical system. For instance, you can compare the
mobility circuit behavior of a near-infinite inductance with the
mechanical system behavior of a near-infinite stiffness spring.

                  Very High Value   Very Low Value
  --------------- ----------------- ----------------
  **Capacitor**   Short Circuit     Open Circuit
  **Inductor**    Open Circuit      Closed Circuit
  **Resistor**    Open Circuit      Short Circuit

Back to Main page


# Engineering Acoustics/Primary variables of interest

## Basic Assumptions

Consider a piston moving in a tube. The piston starts moving at time t=0
with a velocity u=$u_p$. The piston fits inside the tube smoothly
without any friction or gap. The motion of the piston creates a planar
sound wave or acoustic disturbance traveling down the tube at a constant
speed c\>\>$u_p$. In a case where the tube is very small, one can
neglect the time it takes for acoustic disturbance to travel from the
piston to the end of the tube. Hence, one can assume that the acoustic
disturbance is uniform throughout the tube domain.

```{=html}
<center>
```
![](Acousticplanewave1.gif "Acousticplanewave1.gif")

```{=html}
</center>
```
### Assumptions

1\. Although sound can exist in solids or fluid, we will first consider
the medium to be a fluid at rest. The ambient, undisturbed state of the
fluid will be designated using subscript zero. Recall that a fluid is a
substance that deforms continuously under the application of any shear
(tangential) stress.

2\. Disturbance is a compressional one (as opposed to transverse).

3\. Fluid is a continuum: infinitely divisible substance. Each fluid
property assumed to have definite value at each point.

4\. The disturbance created by the motion of the piston travels at a
constant speed. It is a function of the properties of the ambient fluid.
Since the properties are assumed to be uniform (the same at every
location in the tube) then the speed of the disturbance has to be
constant. The speed of the disturbance is the speed of sound, denoted by
letter $c_0$ with subscript zero to denote ambient property.

5\. The piston is perfectly flat, and there is no leakage flow between
the piston and the tube inner wall. Both the piston and the tube walls
are perfectly rigid. Tube is infinitely long, and has a constant area of
cross section, A.

6\. The disturbance is uniform. All deviations in fluid properties are
the same across the tube for any location x. Therefore the instantaneous
fluid properties are only a function of the Cartesian coordinate x (see
sketch). Deviations from the ambient will be denoted by primed
variables.

## Variables of interest

### Pressure (force / unit area)

Pressure is defined as the normal force per unit area acting on any
control surface within the fluid.

```{=html}
<center>
```
![](Acousticcontrolsurface.gif "Acousticcontrolsurface.gif")

```{=html}
</center>
```
$p = \frac {\tilde{F}.\tilde{n}}{dS}$

For the present case,inside a tube filled with a working fluid, pressure
is the ratio of the surface force acting onto the fluid in the control
region and the tube area. The pressure is decomposed into two
components - a constant equilibrium component, $p_0$, superimposed with
a varying disturbance $p^'(x)$. The deviation $p^'$is also called the
acoustic pressure. Note that $p^'$ can be positive or negative. Unit:
$kg/ms^2$. Acoustical pressure can be measured using a microphone.

```{=html}
<center>
```
![](Acousticpressure1.gif "Acousticpressure1.gif")

```{=html}
</center>
```
### Density

Density is mass of fluid per unit volume. The density, ρ, is also
decomposed into the sum of ambient value (usually around ρ0= 1.15 kg/m3)
and a disturbance ρ'(x). The disturbance can be positive or negative, as
for the pressure. Unit: $kg/m^3$

### Acoustic volume velocity

Rate of change of fluid particles position as a function of time. Its
the well known fluid mechanics term, flow rate.

```{=html}
<center>
```
$U=\int_{s}\tilde{u}.\tilde{n}\, dS$

```{=html}
</center>
```
In most cases, the velocity is assumed constant over the entire cross
section (plug flow), which gives acoustic volume velocity as a product
of fluid velocity $\tilde{u}$ and cross section S.

```{=html}
<center>
```
$U=\tilde{u}.S$

```{=html}
</center>
```
Back to Main page


# Engineering Acoustics/Electro-acoustic analogies

## Electro-acoustical Analogies

### Acoustical Mass

Consider a rigid tube-piston system as following figure.

```{=html}
<center>
```
![](acousticalmass.gif "acousticalmass.gif")

```{=html}
</center>
```
Piston is moving back and forth sinusoidally with frequency of **f**.
Assuming $f << \frac{c}{l\ or\ \sqrt{S}}$ (where **c** is sound velocity
$c=\sqrt{\gamma R T_0}$), volume of fluid in tube is,

```{=html}
<center>
```
$\Pi_v=S\ l$

```{=html}
</center>
```
Then mass (mechanical mass) of fluid in tube is given as,

```{=html}
<center>
```
$M_M= \Pi_v \rho_0 = \rho_0 S\ l$

```{=html}
</center>
```
For sinusoidal motion of piston, fluid move as rigid body at same
velocity as piston. Namely, every point in tube moves with the same
velocity.

Applying the Newton\'s second law to the following free body diagram,

```{=html}
<center>
```
![](FBD.gif "FBD.gif")

```{=html}
</center>
```
```{=html}
<center>
```
$SP'=(\rho_0Sl)\frac{du}{dt}$

```{=html}
</center>
```
```{=html}
<center>
```
$\hat{P}=\rho_0l(j\omega)\hat{u}=j\omega(\frac{\rho_0l}{S})\hat{U}$

```{=html}
</center>
```
Where, plug flow assumption is used.

`"Plug flow" assumption:`\
`Frequently in acoustics, the velocity distribution along the normal surface of`\
`fluid flow is assumed uniform. Under this assumption, the acoustic volume velocity U is`\
`simply product of velocity and entire surface. `$U=Su$

#### Acoustical Impedance

Recalling mechanical impedance,

```{=html}
<center>
```
$\hat{Z}_M=\frac{\hat{F}}{\hat{u}}=j\omega(\rho_0Sl)$

```{=html}
</center>
```
acoustical impedance (often termed an **acoustic ohm**) is defined as,

```{=html}
<center>
```
$\hat{Z}_A=\frac{\hat{P}}{\hat{U}}=\frac{Z_M}{S^2}=j\omega(\frac{\rho_0l}{S})\quad  \left[\frac{N s}{m^5}\right]$

```{=html}
</center>
```
where, acoustical mass is defined.

```{=html}
<center>
```
$M_A=\frac{\rho_0l}{S}$

```{=html}
</center>
```
#### Acoustical Mobility

Acoustical mobility is defined as,

```{=html}
<center>
```
$\hat{\xi}_A=\frac{1}{\hat{Z}_A}=\frac{\hat{U}}{\hat{P}}$

```{=html}
</center>
```
#### Impedance Analog vs. Mobility Analog

```{=html}
<center>
```
![](Imp-mov.gif "Imp-mov.gif")

```{=html}
</center>
```
#### Acoustical Resistance

Acoustical resistance models loss due to viscous effects (friction) and
flow resistance (represented by a screen).

```{=html}
<center>
```
![](Ra_analogs.png "Ra_analogs.png")

***r~A~***is the reciprocal of***R~A~*** and is referred to as
*responsiveness*.

```{=html}
</center>
```
### Acoustical Generators

The acoustical generator components are pressure, **P** and volume
velocity, **U**, which are analogus to force, **F** and velocity, **u**
of electro-mechanical analogy respectively. Namely, for impedance
analog, pressure is analogous to voltage and volume velocity is analogus
to current, and vice versa for mobility analog. These are arranged in
the following table.

```{=html}
<center>
```
![](e-a_analogy.gif "e-a_analogy.gif")

```{=html}
</center>
```
Impedance and Mobility analogs for acoustical generators of constant
pressure and constant volume velocity are as follows:

```{=html}
<center>
```
![](acoustic_gen.png "acoustic_gen.png")

```{=html}
</center>
```
### Acoustical Compliance

Consider a piston in an enclosure.

```{=html}
<center>
```
![](Enclosed_Piston.png "Enclosed_Piston.png")

```{=html}
</center>
```
When the piston moves, it displaces the fluid inside the enclosure.
Acoustic compliance is the measurement of how \"easy\" it is to displace
the fluid.

Here the volume of the enclosure should be assumed to be small enough
that the fluid pressure remains uniform.

Assume no heat exchange 1.adiabatic 2.gas compressed uniformly , p prime
in cavity everywhere the same.

from thermo equitation ![](Equ1.jpg "Equ1.jpg") it is easy to get the
relation between disturbing pressure and displacement of the piston
![](Equ3.gif "Equ3.gif") where U is volume rate, P is pressure according
to the definition of the impendance and mobility, we can
get![](Equ4.gif "Equ4.gif")

Mobility Analog VS Impedance Analog

```{=html}
<center>
```
![](Comp.gif "Comp.gif")

```{=html}
</center>
```
### Examples of Electro-Acoustical Analogies

Example 1: Helmholtz Resonator

```{=html}
<center>
```
![](Example1holm.JPG "Example1holm.JPG")

```{=html}
</center>
```
Assumptions - (1) Completely sealed cavity with no leaks. (2) Cavity
acts like a rigid body inducing no vibrations.

Solution:

```{=html}
<center>
```
\- Impedance Analog -

```{=html}
</center>
```
```{=html}
<center>
```
![](Example2holm1sol.JPG "Example2holm1sol.JPG")

```{=html}
</center>
```
Example 2: Combination of Side-Branch Cavities

```{=html}
<center>
```
![](Exam2prob.JPG "Exam2prob.JPG")

```{=html}
</center>
```
Solution:

```{=html}
<center>
```
\- Impedance Analog -

```{=html}
</center>
```
```{=html}
<center>
```
![](Exam2sol.JPG "Exam2sol.JPG")

```{=html}
</center>
```
Back to Main page


# Engineering Acoustics/Transducers - Loudspeaker

# Acoustic transducer

The purpose of an acoustic transducer is to convert electrical energy
into acoustic energy. Many variations of acoustic transducers exist,
such as electrostatic, balanced armature and moving-coil loudspeakers.
This article focuses on moving-coil loudspeakers since they are the most
commonly used type of acoustic transducer. First, the physical
construction and principle of a typical moving coil transducer are
discussed briefly. Second, electro-mechano-acoustical modeling of each
element composing the loudspeaker is presented in a tutorial way to
reinforce and supplement the theory on electro-mechanical
analogies
and electro-acoustic
analogies
previously seen in other sections. Third, the equivalent circuit is
analyzed to introduce the theory behind Thiele-Small parameters, which
are very useful when designing loudspeaker enclosures. A method to
experimentally determine Thiele-Small parameters is also included.

## Moving-coil loudspeaker construction and principle

The classic moving-coil loudspeaker driver can be divided into three key
components:

1\) The magnet motor drive system, comprising the permanent magnet, the
center pole and the voice coil acting together to produce a mechanical
force on the diaphragm from an electrical current.

2\) The loudspeaker cone system, comprising the diaphragm and dust cap,
permitting mechanical force to be translated into acoustic pressure;

3\) The loudspeaker suspension, comprising the spider and surround,
preventing the diaphragm from breaking due to over excursion, allowing
only translational movement and tending to bring the diaphragm back to
its rest position.

The following illustration shows a cut-away view of a typical moving
coil-permanent magnet loudspeaker. A coil is mechanically coupled to a
diaphragm, also called cone, and rests in a fixed magnetic field
produced by a magnet. When an electrical current flows through the coil,
a corresponding magnetic field is emitted, interacting with the fixed
field of the magnet and thus applying a force to the coil, pushing it
away or towards the magnet. Since the cone is mechanically coupled to
the coil, it will push or pull the air it is facing, causing pressure
changes and emitting a sound wave.

```{=html}
<center>
```
![](MovingCoilLoudspeaker.png "MovingCoilLoudspeaker.png"){width="800"}

```{=html}
</center>
```
```{=html}
<center>
```
Figure 1: A cross-sectional view of a typical moving-coil loudspeaker

```{=html}
</center>
```
An equivalent circuit can be obtained to model the loudspeaker as a
lumped system. This circuit can be used to drive the design of a
complete loudspeaker system, including an enclosure and sometimes even
an amplifier that is matched to the properties of the driver. The
following section shows how such an equivalent circuit can be obtained.

## Electro-mechano-acoustical equivalent circuit

Electro-mechanico-acoustical systems such as loudspeakers can be modeled
as equivalent electrical circuits as long as each element moves as a
whole. This is usually the case at low frequencies or at frequencies
where the dimensions of the system are small compared to the wavelength
of interest. To obtain a complete model of the loudspeaker, the
interactions and properties of electrical, mechanical, and acoustical
subsystems composing the loudspeaker driver must each be modeled. The
following sections detail how the circuit may be obtained starting with
the amplifier and ending with the acoustical load presented by air. A
similar development can be found in \[1\] or \[2\].

### Electrical subsystem

The electrical part of the system is composed of a driving amplifier and
a voice coil. Most amplifiers can be approximated as a perfect voltage
source in series with the amplifier output impedance. The voice coil
exhibits an inductance and a resistance that may be directly modeled as
a circuit.

```{=html}
<center>
```
![](ElectricalSubsystemLoudspeaker.png "ElectricalSubsystemLoudspeaker.png"){width="400"}

```{=html}
</center>
```
```{=html}
<center>
```
Figure 2: The amplifier and loudspeaker electrical elements modeled as a
circuit

```{=html}
</center>
```
### Electrical to mechanical subsystem

When the loudspeaker is fed an electrical signal, the voice coil and
magnet convert current to force. Similarly, voltage is related to the
velocity. This relationship between the electrical side and the
mechanical side can be modeled by a transformer.

$\tilde{f_c} = Bl \tilde{i}$; $\tilde{u_c} = \dfrac{\tilde{e}}{Bl}$

```{=html}
<center>
```
![](_ElectricalToMechanicalLoudspeaker.png "_ElectricalToMechanicalLoudspeaker.png")

```{=html}
</center>
```
```{=html}
<center>
```
Figure 3: A transformer modeling transduction from the electrical
impedance to mechanical mobility analogy

```{=html}
</center>
```
### Mechanical subsystem

In a first approximation, a moving coil loudspeaker may be thought of as
a mass-spring system where the diaphragm and the voice coil constitute
the mass and the spider and surround constitute the spring element.
Losses in the suspension can be modeled as a resistor.

```{=html}
<center>
```
![](MechanicalSubsystemModelingLoudspeaker.png "MechanicalSubsystemModelingLoudspeaker.png"){width="1000"}

```{=html}
</center>
```
```{=html}
<center>
```
Figure 4: Mass spring system and associated circuit analogies of the
impedance and mobility type.

```{=html}
</center>
```
The equation of motion gives us :

```{=html}
<center>
```
$\tilde{f_c} = R_m \tilde{u_c} + \dfrac{\tilde{u_c}}{ j \omega C_{MS}} + j \omega M_{MD} \tilde{u_c}$

```{=html}
</center>
```
```{=html}
<center>
```
$\dfrac{\tilde{f_c} }{\tilde{u_c}}= R_m+\dfrac{1}{ j \omega C_{MS}}+ j\omega M_{MD}$

```{=html}
</center>
```
Which yields the mechanical impedance type analogy in the form of a
series RLC circuit. A parallel RLC circuit may also be obtained to get
the mobility analog following mathematical manipulation:

```{=html}
<center>
```
$\dfrac{\tilde{u_c} }{\tilde{f_c}}= \dfrac{1}{R_m+\dfrac{1}{ j \omega C_{MS}}+ j\omega M_{MD} }$

```{=html}
</center>
```
```{=html}
<center>
```
$\dfrac{\tilde{u_c} }{\tilde{f_c}}= \dfrac{1}{\dfrac{1}{G_m}+\dfrac{1}{ j \omega C_{MS}} + \dfrac{1}{\dfrac{1}{j \omega M_{MD}}}}$

```{=html}
</center>
```
Which expresses the mechanical mobility type analogy in the form of a
parallel RLC circuit where the denominator elements are respectively a
parallel conductance, inductance, and compliance.

### Mechanical to acoustical subsystem

A loudspeaker's diaphragm may be thought of as a piston that pushes and
pulls on the air facing it, converting mechanical force and velocity
into acoustic pressure and volume velocity. The equations are as follow:

$\tilde{P_d} = \dfrac{\tilde{f_c}}{\tilde{S_D}}$;
$\tilde{U_c} = \tilde{u_c}{S_D}$

These equations can be modeled by a transformer.

```{=html}
<center>
```
![](MechanicalToAcousticalLoudspeaker.png "MechanicalToAcousticalLoudspeaker.png")

```{=html}
</center>
```
```{=html}
<center>
```
Figure 5: A transformer modeling the transduction from mechanical
mobility to acoustical mobility analogy performed by a loudspeaker\'s
diaphragm

```{=html}
</center>
```
### Acoustical subsystem

The impedance presented by the air load on the loudspeaker\'s diaphragm
is both resistive due to sound radiation and reactive due to the air
mass that is being pushed radially but does not contribute to sound
radiation to the far field. The air load on the diaphragm can be modeled
as an impedance or an admittance. Specific values and approximations can
be found in \[1\], \[2\] or \[3\]. Note that the air load depends on the
mounting conditions of the loudspeaker. If the loudspeaker is mounted in
a baffle, the air load will be the same on each side of the diaphragm.
Then, if the air load on one side is $Y_{AR}$ in the admittance analogy,
then the total air load is $Y_{AR}/2$ as both loads are in parallel.

### Complete electro-mechano-acoustical equivalent circuit

Using electrical impedance, mechanical mobility and acoustical
admittance yield the following equivalent circuit, modeling the entire
loudspeaker drive unit.

```{=html}
<center>
```
![](_LoudspeakerEquivalentCircuit.png "_LoudspeakerEquivalentCircuit.png"){width="1000"}

```{=html}
</center>
```
```{=html}
<center>
```
Figure 6: A complete electro-mechano-acoustical equivalent circuit of a
loudspeaker drive unit

```{=html}
</center>
```
This circuit can be reduced by substituting the transformers and
connected loads by an equivalent loading that would present the same
impedance as the loaded transformer. An example of this is shown on
figure 7, where acoustical and electrical loads and sources have been
\"brought over\" to the mechanical side.

```{=html}
<center>
```
![](LoudspeakerEquivalentCircuitMechanical.png "LoudspeakerEquivalentCircuitMechanical.png"){width="900"}

```{=html}
</center>
```
```{=html}
<center>
```
Figure 7: Mechanical equivalent circuit modeling of a loudspeaker drive
unit

```{=html}
</center>
```
The advantage of doing such manipulations is that we can then directly
relate electrical measurements with elements in the circuit. This will
later allow us to obtain values for the different components of the
model and match this model to real loudspeaker drivers. We can further
simplify this circuit by using Norton\'s theorem and converting the
series electrical components and voltage source into an equivalent
current source and parallel electrical components. Then, using a
technique called the Dot method, presented in section Solution Methods:
Electro-Mechanical
Analogies,
we can obtain a single loop series circuit which is the dual of the
parallel circuit previously obtained with Norton\'s theorem. If we are
mainly interested in the low frequency behavior of the loudspeaker, as
should be the case when using lumped element modeling, we can neglect
the effect of the voice coil inductance, which has an effect only at
high frequencies. Furthermore, the air load impedance at low frequencies
is mass-like and can be modeled by a simple inductance $M_{M1}$. This
results in a simplified low frequency model equivalent circuit, shown of
figure 8, which is easier to manipulate than the circuit of figure 7.
Note that the analogy used for this circuit is of the impedance type.

```{=html}
<center>
```
![](LowFrequencyLoudspeakerEquivalentCircuitMechanical.png "LowFrequencyLoudspeakerEquivalentCircuitMechanical.png"){width="800"}

```{=html}
</center>
```
```{=html}
<center>
```
Figure 8: Low frequency approximation mechanical equivalent circuit of a
loudspeaker drive unit

```{=html}
</center>
```
Where $M_{M1} = 2.67 a^3 \rho$ if $a$ is the radius of the loudspeaker
and $\rho$, the density of air. Mass elements, in this case the mass of
the diaphragm and voice coil $M_{MS}$ and the air mass loading the
diaphragm $2M_{M1}$ can be regrouped in a single element:

```{=html}
<center>
```
$M_{MS} = M_{MD}+2M_{M1}$

```{=html}
</center>
```
## Thiele-Small Parameters

### Theory

The complete low frequency behavior of a loudspeaker drive unit can be
modeled with just six parameters, called Thiele-Small parameters. Most
of these parameters result from algebraic manipulation of the equations
of the circuit of figure 8. Loudspeaker driver manufacturers seldom
provide electro-mechano-acoustical parameters directly and rather
provide Thiele-Small parameters in datasheets, but conversion from one
to the other is quite simple. The Thiele-Small parameters are as follow:

1\. $R_e$, the voice coil DC resistance;

2\. $Q_{ES}$, the electrical Q factor;

3\. $Q_{MS}$, the mechanical Q factor;

4\. $f_s$, the loudspeaker resonance frequency;

5\. $S_D$, the effective surface area of the diaphragm;

6\. $V_{AS}$, the equivalent suspension volume: the volume of air that
has the same acoustic compliance as the suspension of the loudspeaker
driver.

These parameters can be related directly from the low frequency
approximation circuit of figure 8, with $R_e$ and $S_D$ being explicit.

$Q_{MS} = \dfrac{1}{R_{MS}} \sqrt{\dfrac{M_{MS}}{C_{MS}}}$;
$Q_{ES} = \dfrac{R_g + R_e }{(Bl)^2} \sqrt{\dfrac{M_{MS}}{C_{MS}}}$ ;
$f_s= \dfrac{1}{2\pi\sqrt{M_{MS}C_{MS}}}$; $V_{AS}= C_{MS}S_D^2\rho c^2$

Where $\rho c^2$ is the Bulk modulus of air. It follows that, if given
Thiele-Small parameters, one can extract the values of each component of
the circuit of figure 8 using the following equations :

$C_{MS} = \dfrac{V_{AS}}{S_D^2 \rho c^2}$;
$M_{MS}= \dfrac{1}{(2 \pi f_s)^2 C_{MS}}$;
$R_{MS} = \dfrac{1}{Q_{MS}}\sqrt{\dfrac{M_{MS}}{C_{MS}}}$;
$Bl = \sqrt{\dfrac{R_e}{2 \pi f_s Q_{ES} C_{MS} } }$;
$M_{MD}= M_{MS}-2M_{M1}$;

### Measurement

Many methods can be used to measure Thiele-Small parameters of drivers.
Measurement of Thiele-Small parameters is sometimes necessary if a
manufacturer does not provide them. Also, the actual Thiele-Small
parameters of a given loudspeaker can differ from nominal values
significantly. The method described in this section comes from \[2\].
Note that for this method, the loudspeaker is considered to be mounted
in an infinite baffle. In practice, a baffle with a diameter of four
times that of the loudspeaker is sufficient. Measurements without a
baffle are also possible: the air mass loading will simply be halved and
can be easily accounted for. The setup for this method includes an FFT
analyzer or a mean to obtain an impedance curve. A signal generator of
variable frequency and an AC meter can also be used.

```{=html}
<center>
```
![](Setup_to_measure_impedance_of_loudspeakers.png "Setup_to_measure_impedance_of_loudspeakers.png"){width="300"}

```{=html}
</center>
```
```{=html}
<center>
```
Figure 9: Simple experimental setup to measure the impedance of a
loudspeaker drive unit

```{=html}
</center>
```
```{=html}
<center>
```
$Z_{spk}= R\dfrac{V_{spk}}{V_s \left( 1-\dfrac{V_{spk}}{V_s} \right)}$

```{=html}
</center>
```
```{=html}
<center>
```
![](ImpedanceCurveLoudspeaker.png "ImpedanceCurveLoudspeaker.png"){width="600"}

```{=html}
</center>
```
```{=html}
<center>
```
Figure 10: A typical loudspeaker drive unit impedance curve

```{=html}
</center>
```
Once the impedance curve of the loudspeaker is measured, $R_e$ and $f_s$
can be directly identified by looking at the low frequency asymptote of
the impedance value and the center frequency of the resonance peak. If
the frequencies where $Z_{spk}=\sqrt{R_e R_c}$ are identified as $f_l$
and $f_h$, Q factors can be calculated.

```{=html}
<center>
```
$Q_{MS}= \dfrac{f_s}{f_h-f_l} \sqrt{\dfrac{R_c}{R_e}}$

```{=html}
</center>
```
```{=html}
<center>
```
$Q_{ES} = \dfrac{Q_{MS}}{\dfrac{R_c}{R_e}-1}$

```{=html}
</center>
```
$S_D$ can simply be approximated by $\pi a^2$, where $a$ is the radius
of the loudspeaker driver. The last remaining Thiele-Small parameter,
$V_{AS}$ is slightly trickier to measure. The idea is to either increase
mass or reduce compliance of the loudspeaker drive unit and note the
shift in resonance frequency. If a known mass $M_x$ is added to the
loudspeaker diaphragm, the new resonance frequency will be:

```{=html}
<center>
```
$f^'_s= \dfrac{1}{2 \pi \sqrt{(M_{MS} + M_x) C_{MS} } }$

```{=html}
</center>
```
And the equivalent suspension volume may be obtained with:

```{=html}
<center>
```
$V_{AS} =  \left( 1- \dfrac{f^{'2}_s}{f_s^2} \right) \dfrac{S_D^2 \rho c^2}{(2 \pi f^'_s)^2 M_x}$

```{=html}
</center>
```
Hence, all Thiele-Small parameters modeling the low frequency behavior
of the loudspeaker drive unit can be obtained from a fairly simple
setup. These parameters are of tremendous help in loudspeaker enclosure
design.

### Numerical example

This section presents a numerical example of obtaining Thiele-Small
parameters from impedance curves. The impedance curves presented in this
section have been obtained from simulations using nominal Thiele-Small
parameters of a real woofer loudspeaker. Firsy, these Thiele-Small
parameters have been transformed into an electro-mechano-acoustical
circuit using the equation presented before. Second, the circuit was
treated as a black box and the method to extract Thiele-Small parameters
was used. The purpose of this simulation is to present the method, step
by step, using realistic values so that the reader can get more familiar
with the process, the magnitude of the values and with what to expect
when performing such measurements.

For this simulation, a loudspeaker of radius $a=6.55cm$ is mounted on a
baffle sufficiently large to act as an infinite baffle. Its impedance is
obtained and plotted in figure 11, where important cursors have already
been placed.

```{=html}
<center>
```
![](SimulatedImpedanceWoofer.png "SimulatedImpedanceWoofer.png"){width="1000"}

```{=html}
</center>
```
```{=html}
<center>
```
Figure 11: Simulated measurement of an impedance curve for a woofer
loudspeaker

```{=html}
</center>
```
The low frequency asymptote is immediately identified as $Re=6.6\Omega$.
The resonance is clear and centered at $fs=33Hz$. The value of the
impedance at this frequency is about $66\Omega$. This yields
$\sqrt{R_eR_c}=20.8\Omega$, which occurs at $f_l=19.5 Hz$ and
$f_h= 52.5 Hz$. With this information, we can compute some of the
Thiele-Small parameters.

```{=html}
<center>
```
$Q_{MS}= \dfrac{f_s}{f_h-f_l}\sqrt{\dfrac{R_c}{R_e}}=\dfrac{33}{52.5-19.5}*\sqrt{\dfrac{66}{6.6}}= 3.1$

```{=html}
</center>
```
```{=html}
<center>
```
$Q_{ES} = \dfrac{Q_{MS}}{\dfrac{R_c}{R_e}-1} = \dfrac{3.1}{\dfrac{66}{6.6}-1} = 0.35$

```{=html}
</center>
```
As a next step, a mass of $M_x=10g$ is fixed to the loudspeaker
diaphragm. This shifts the resonance frequency and yields a new
impedance curve, as shown on figure 12.

```{=html}
<center>
```
![](SimulatedImpedanceWooferAddedMass.png "SimulatedImpedanceWooferAddedMass.png"){width="1000"}

```{=html}
</center>
```
```{=html}
<center>
```
Figure 12: Simulated measurement of an impedance curve for a woofer
loudspeaker

```{=html}
</center>
```
```{=html}
<center>
```
$S_{D} =  \pi a^2 = 0.0135 m^2$

```{=html}
</center>
```
```{=html}
<center>
```
$V_{AS} =  \left( 1- \dfrac{27.5^2}{33^2} \right) \dfrac{0.0135^2*1.18*344^2}{(2 \pi 27.5)^2*0.01}= 0.0272 m^3$

```{=html}
</center>
```
Once all six Thiele-Small parameters have been obtained, it is possible
to calculate values for the electro-mechano-acoustical circuit modeling
elements of figure 6 or 7. From then, the design of an enclosure can
start. This is discussed in application sections Sealed box subwoofer
design
and Bass reflex enclosure
design.

# References

\[1\] Kleiner, Mendel. Electroacoustics. CRC Press, 2013.

\[2\] Beranek, Leo L., and Tim Mellow. Acoustics: sound fields and
transducers. Academic Press, 2012.

\[3\] Kinsler, Lawrence E., et al. Fundamentals of Acoustics, 4th
Edition. Wiley-VCH, 1999.

\[4\] Small, Richard H. \"Direct radiator loudspeaker system analysis.\"
Journal of the Audio Engineering Society 20.5 (1972): 383-395.


# Engineering Acoustics/Moving Resonators

## Moving Resonators

Consider the situation shown in the figure below. We have a typical
Helmholtz resonator driven by a massless piston which generates a
sinusoidal pressure $P_G$, however the cavity is not fixed in this case.
Rather, it is supported above the ground by a spring with compliance
$C_M$. Assume the cavity has a mass $M_M$.

Recall the Helmholtz resonator (see Module
#9).
The difference in this case is that the pressure in the cavity exerts a
force on the bottom of the cavity, which is now not fixed as in the
original Helmholtz resonator. This pressure causes a force that acts
upon the cavity bottom. If the surface area of the cavity bottom is
$S_C$, then Newton\'s Laws applied to the cavity bottom give

```{=html}
<center>
```
$\sum{F} = p_CS_C - \frac{x}{C_M} = M_M\ddot{x} \Rightarrow p_CS_C = \left[\frac{1}{j\omega C_M} + j\omega M_M\right]u$

```{=html}
</center>
```
In order to develop the equivalent circuit, we observe that we simply
need to use the pressure (potential across $C_A$) in the cavity to
generate a force in the mechanical circuit. The above equation shows
that the mass of the cavity and the spring compliance should be placed
in series in the mechanical circuit. In order to convert the pressure to
a force, the transformer is used with a ratio of $1:S_C$.

## Example

A practical example of a moving resonator is a marimba. A marimba is a
similar to a xylophone but has larger resonators that produce deeper and
richer tones. The resonators (seen in the picture as long, hollow pipes)
are mounted under an array of wooden bars which are struck to create
tones. Since these resonators are not fixed, but are connected to the
ground through a stiffness (the stand), it can be modeled as a moving
resonator. Marimbas are not tunable instruments like flutes or even
pianos. It would be interesting to see how the tone of the marimba
changes as a result of changing the stiffness of the mount.

For more information about the acoustics of marimbas see
<http://www.mostlymarimba.com/techno1.html>

Back to main page


# Engineering Acoustics/Speed of sound

When sound waves propagate in a medium, they cause fluctuations in the
pressure, density, temperature and particle velocity in the medium. The
total pressure, $P$, in the medium can be expressed as:

$P=P_o+p'$

where $P_o$ is the hydrostatic or ambient pressure and $p'$ is the
acoustic pressure or pressure fluctuation.

The hydrostatic pressure can be thought as the mean pressure while the
acoustic pressure represents fluctuations around the mean pressure.
Similarly, the density, temperature, and particle velocity are separated
into mean and fluctuating components.

$\rho=\rho_o+\varrho '$

$T=T_o + T'$

$\vec{U} =\vec{u}_o+\vec{u}'$

where $\rho_o$ is the ambient density; $\varrho '$, the density
fluctuation; $T_o$, the ambient temperature; $T'$, the temperature
fluctuation; $\vec{u}_o$, the mean velocity; and $\vec{u}'$, the
particle velocity. Notice that pressure, density, and temperature are
scalar quantities and the particle velocity is a vectorial quantity.
!Planar wave traveling inside a
tube\|440x440px
Let\'s consider a planar wave traveling in the x-direction inside a tube
filled with a fluid at rest ($\vec{u}_o=0$) with constant pressure,
density and temperature ($P_o$, $\rho_o$, and $T_o$, respectively). As
the wave moves through the fluid at a speed, $c_o$,  it creates
infinitesimally small fluctuations in the initially stagnant fluid in
front. All four quantities describing the fluid vary around their mean
value, increasing or decreasing depending on whether the fluid is being
compressed or expanded. To obtain a relation for the propagating speed,
$c_o$, an inertial frame of reference is used and a control volume is
drawn around the wave. !Control volume around a traveling planar
wave\|400x400px
Applying continuity,

$\rho_o c_o A = (\rho_o+\varrho')(c_o-u') A$

and neglecting higher order terms,

$\rho_o u'= c_o \varrho'$

Applying conservation of momentum,

$P_o+ \rho_o c_o^2 = P_o + p' + (\rho_o+\varrho')(c_o-u')^2$

neglecting higher order terms,

$\rho_o c^2_o = p' + (\rho_o+\varrho')(c_o^2-2 c_o u')$

$0 = p' + -2c_o \rho_o u' +\varrho' c_o^2$,

and using the continuity equation,

$c_o^2=\frac{p'}{\varrho'}$

The speed of sound is related to the ratio of pressure fluctuation
(acoustic pressure) to density fluctuation. Given that the speed of
sound is always a positive quantity, an increase in the fluid pressure
implies an increase in the fluid density and vice versa. The total
pressure is expressed as a Taylor series expansion about the ambient
density to relate its infinitesimally small fluctuations to the total
pressure and density.

$P(\rho_o+\varrho')= P_o + p' = P(\rho_o)+ \frac{\partial P(\rho_o)}{\partial \rho} \varrho' + \frac{1}{2} \frac{\partial^2 P(\rho_o)}{\partial \rho^2} \varrho'^2 + ...$

Neglecting second order terms and higher, the speed of sound can be
related to the total pressure and density.

$c_o^2=\frac{p'}{\varrho'}= \frac{\partial P(\rho_o)}{\partial \rho}$

As a sound wave moves through a fluid, the fluid is usually assumed to
follow an adiabatic and reversible thermodynamic path. So, the heat
transfer between fluid particles is negligible and the changes caused by
the sound wave onto the fluid can be reverse to their original state
without changing the entropy of the system. For an isentropic process,
the total pressure and density are related by the thermodynamic
relation,

$P=C \rho^\gamma$.

Taking the partial derivative and using the ideal gas law,
$P=\rho R_g T$,

$\frac{\partial P}{\partial \rho} = C \gamma \rho ^{\gamma -1 } = \frac{\gamma P}{\rho} = \gamma R_g T$.

The speed of sound can be expressed in terms of the ambient pressure,
density and temperature using,

$c_o^2= \frac{\partial P(\rho_o)}{\partial \rho} =  \frac{\gamma P_o}{\rho_o} = \gamma R_g T_o$.

Using the definition of the Bulk modulus,

$B_o=\rho_o  \frac{\partial P(\rho_o)}{\partial \rho}= \gamma P_o$

the speed of sound can written as,

$c_o = \sqrt{\frac{\gamma P_o}{\rho_o}} = \sqrt{\frac{B_o}{\rho_o}} = \sqrt{ \gamma R_g T_o}$.

## See also

Original Document [^1]

Acoustics Wikibooks: Sound Speed

Speed of sound in
water

```{=html}
<references />
```

[^1]: 1


# Engineering Acoustics/Acoustic wave equation

To derive the wave equation, three relations are combined: the equation
of state, ideal gas law; continuity equation, conservation of mass; and
Newton\'s law, conservation of momentum. From the speed of sound, a
relation between the acoustic pressure and the bulk modulus can be
derived,

$c_o^2=\frac{\partial P}{\partial \rho} = \frac{p'}{\varrho'}= \frac{p'}{\rho-\rho_o}.$

Using the definition of the condensation, relative change of density in
a fluid, denoted by $s$,

$\frac{p'}{\rho_o}= c_o^2 \Big( \frac{\rho-\rho_o}{\rho_o} \Big) = c_o^2s,$

and introducing the bulk modulus which makes implicit use of the ideal
gas law,

$p'= \rho_o c_o^2 s = \gamma P_o s = B_o s.$

The general form of the continuity equation for a control volume from
fluid dynamics is simplified to its one-dimensional form in Cartesian
coordinates.

$\frac{\partial \rho}{\partial t} + \vec{\nabla} (\rho \vec{U}) =0$

$\frac{\partial \rho}{\partial t} + \rho_o \frac{\partial U}{\partial x} =0$

$\frac{\partial U}{\partial x} = \frac{-1}{\rho_o} \frac{\partial \rho}{\partial t} = \frac{-1}{\rho_o c_o^2} \frac{\partial P}{\partial t}$
!Pressure on fluid element inside a
tube\|left\|440x440px

Using Newton\'s law on the fluid element, the net force acting on the
fluid boundaries causes an acceleration on the fluid proportional to its
mass,

$A (P(x)-P(x+dx))= (A dx \rho_o) \frac{dU}{dt}.$

Noticing that

$\frac{P(x)-P(x+dx)}{dx} = \frac{\partial P}{\partial x}$

as $dx$ approaches zero, evaluating the derivative, and neglecting small
terms,

$\frac{\partial P}{\partial x} dx= (\rho_o dx) \frac{dU}{dt} = (\rho_o dx) \Big( \frac{\partial U}{\partial t} + \frac{\partial U}{\partial x} \frac{dx}{dt}\Big)$

$\frac{\partial U}{\partial t}= \frac{-1}{\rho_o}\frac{\partial P}{\partial x}.$

To obtain the wave equation, the partial derivative with respect to time
is taken for the continuity equation and  the partial derivative with
respect to space for the conservation of momentum equation. By equating
the two results and eliminating the density term on both sides,

$\frac{\partial^2 U}{\partial t \partial x}= \frac{-1}{\rho_o c_o^2} \frac{\partial^2 P}{\partial t^2}= \frac{-1}{\rho_o}\frac{\partial^2 P}{\partial x^2}$

the acoustic wave equation is recovered

$\frac{\partial^2 P}{\partial x^2}= \frac{1}{c_o^2} \frac{\partial^2 P}{\partial t^2}$

$\frac{\partial^2 P}{\partial x^2}- \frac{1}{c_o^2} \frac{\partial^2 P}{\partial t^2}=0.$

The equation above is the acoustic wave equation in its one-dimensional
form. It can be generalized to 3-D Cartesian
coordinates$\frac{\partial^2 P}{\partial x^2}+\frac{\partial^2 P}{\partial y^2}+\frac{\partial^2 P}{\partial z^2}- \frac{1}{c_o^2} \frac{\partial^2 P}{\partial t^2}=0.$

Using the Laplace operator, it can be generalized to other coordinate
systems

$\nabla^2 P- \frac{1}{c_o^2} \frac{\partial^2 P}{\partial t^2}=0$

### One-dimensional wave solution in Cartesian coordinates

The one-dimensional acoustic wave equation is described by the second
order partial differential equation,

$\frac{\partial^2 P}{\partial x^2}= \frac{1}{c_o^2} \frac{\partial^2 P}{\partial t^2}.$

It can be solved using separation of variables. Suppose the pressure is
the product of one function only dependent on space and another function
only dependent on time,

$P(x,t)=X(x)T(t).$

Substituting back into the wave equation,

$X''T=\frac{1}{c_o^2}XT''$

$\frac{X''}{X}=\frac{1}{c_o^2}\frac{T''}{T}=-\lambda^2.$

This substitution leads to two homogeneous second order ordinary
differential equations, one in time and one in space.

$T'' + c_o^2 \lambda^2 T=0$

$X'' + \lambda^2 X=0$

The time function is expected to be dependent on the angular frequency
of the wave

$T=Ce^{j \omega t}.$

Substituting and solving for the constant which is define as the wave
number, $K$,

$-\omega^2 + c_o^2 \lambda^2=0$

$K^2=\lambda^2=\frac{\omega^2}{c_o^2}.$

The wave number relates the angular velocity of the wave to its
propagation speed in the medium. It can be expressed in different forms,

$K=\frac{\omega}{c_o}=\frac{2 \pi f}{c_o}=\frac{2 \pi}{\lambda},$

where $f$ is the frequency in hertz and $\lambda$ is the wavelength.The
second differential equation can be solved using the wave number. The
spacial function is given a general form,

$X=Ce^{jrx}.$

Substituting and solving for $r$,

$-r^2+K^2=0$

$r=\pm K.$

The solution of the 1-D acoustic wave equation is obtained,

$P(x,t)=(C_1e^{j K x} + C_2e^{-j K x})e^{j \omega t}.$

The real and imaginary parts of the solution are also solutions to the
1-D wave equation.

$P(x,t)=C_1cos(\omega t + Kx) + C_2cos(\omega t - Kx)$

$P(x,t)=C_1sin(\omega t + Kx) + C_2sin(\omega t - Kx)$

Using phasor notation, the solution is written in more compact form,

$\mathbf{P(x,t)}=\mathbf{P}e^{j(\omega t \pm Kx)}.$

The actual solution is recovered by taking the real part of the above
complex form. The value of the constants above is determined by applying
initial and boundary conditions. In general, any function of the
following form is a solution for periodic waves,

$P(x,t)=f_1(\omega t + Kx)+f_2(\omega t - Kx),$

and similarly, for progressive waves,

$P(x,t)=f(ct + x)+g(ct - x)$

$P(x,t)=f(\xi)+g(\eta),$

where $f$ and $g$ are arbitrary functions, that represent two waves
traveling in opposing directions. These are known as the d\'Alembert
solutions. The form of these two functions can be found by applying
initial and boundary conditions.

Original document[^1]

[^1]: 1


# Engineering Acoustics/Reflection and transmission of planar waves

## Specific acoustic impedance

Before discussing the reflection and transmission of planar waves, the
relation between particle velocity and acoustic pressure is
investigated.

$\frac{\partial u}{\partial t}= \frac{-1}{\rho_o}\frac{\partial p}{\partial x}$

The acoustic pressure and particle velocity can be described in complex
form.

$\mathbf{p}=\mathbf{P}e^{j(\omega t -Kx)}$

$\mathbf{u}=\mathbf{u_o}e^{j(\omega t -Kx)}$

Differentiating and substituting,

$-j \omega \mathbf{u}=\frac{-j K \mathbf{p}}{\rho_o}=\frac{-j \omega \mathbf{p}}{\rho_o c_o}$

$\mathbf{u}=\frac{\mathbf{p}}{\rho_o c_o}$

The specific acoustic impedance for planar waves is defined.

$\mathbf{z}=\frac{\mathbf{p}}{\mathbf{u}}=\rho_o c_o=r_o$

## Planar wave: Normal incidence

Consider an incident planar wave traveling in an infinite medium with
specific impedance $r_1=\rho_1 c_1$ which encounters the boundary
between medium 1 and medium 2. Part of the wave is reflected back into
medium 1 and the remaining part is transmitted to medium 2 with specific
impedance $r_2=\rho_2 c_2$. The pressure field in medium 1 is described
by the sum of the incident and reflected components of the wave.

$\mathbf{p_1}=\mathbf{p_i}+\mathbf{p_r}=\mathbf{P_i}e^{j(\omega t -K_1x)}+\mathbf{P_r}e^{j(\omega t +K_1x)}$

The pressure field in medium 2 is composed only of the transmitted
component of the wave.

$\mathbf{p_2}=\mathbf{p_t}=\mathbf{P_t}e^{j(\omega t -K_2x)}$
!Reflection and transmission of normally incident planar
wave.\|440x440px
Notice that the frequency of the wave remains constant across the
boundary, however the specific acoustic impedance changes across the
boundary. The propagation speed in each medium is different, so the wave
number of each medium is also different. There are two boundary
conditions to be satisfied:

1.  The acoustic pressure must be continuous at the boundary
2.  The particle velocity must be continuous at the boundary

Imposition of the first boundary condition yields

$\mathbf{p_1}(x=0)=\mathbf{p_2}(x=0),$

$\mathbf{P_i} + \mathbf{P_r} = \mathbf{P_t}.$

Imposition of second boundary condition yields

$\mathbf{u_1}(x=0)=\mathbf{u_2}(x=0),$

$\mathbf{u_i}(x=0)+ \mathbf{u_r}(x=0)= \mathbf{u_t}(x=0),$

and using the definition of specific impedance, the equations are
expressed in terms of pressure

$\frac{\mathbf{P_i}}{r_1} - \frac{\mathbf{P_r}}{r_1} = \frac{\mathbf{P_t}}{r_2}.$

The pressure reflection coefficient is the ratio of the reflected
acoustic pressure over the incident acoustic pressure,
$\mathbf{R}=\frac{\mathbf{P_r}}{\mathbf{P_i}}$. The pressure
transmission coefficient is the ratio of the transmitted acoustic
pressure over the incident acoustic pressure,
$\mathbf{T}=\frac{\mathbf{P_t}}{\mathbf{P_i}}$. The specific acoustic
impedance ratio is also defined as: $\zeta=\frac{r_2}{r_1}$. Applying
the above definitions, the boundary conditions can be rewritten as:

$1+\mathbf{R}=\mathbf{T}$

$1-\mathbf{R}=\frac{\mathbf{T}}{\zeta}.$

Solving for the pressure reflection coefficient yields:

$\mathbf{R}=\mathbf{T}-1=\frac{\zeta-1}{\zeta+1}=\frac{r_2-r_1}{r_2+r_1}.$

Solving for the pressure transmission coefficient yields:

$\mathbf{T}=\mathbf{R}+1=\frac{2 \zeta}{\zeta +1}=\frac{2r_2}{r_2+r_1}.$

Solving for the specific acoustic impedance ratio yields:

$\zeta = \frac{1+\mathbf{R}}{1-\mathbf{R}} = \frac{\mathbf{T}}{2-\mathbf{T}}.$

## Case 1: Rigid boundary

Consider an incident planar wave that encounters a rigid boundary. This
is the case if the specific impedance of medium 2 is significantly
larger than the specific impedance of medium 1. Thus, the specific
acoustic impedance ratio is very large, the reflection coefficient
approaches 1 and the transmission coefficient approaches 2.

$\mathbf{R}=1=\frac{\mathbf{P_r}}{\mathbf{P_i}} \Rightarrow \mathbf{P_r} = \mathbf{P_i} \Rightarrow \mathbf{u}(x=0)=0$

$\mathbf{T}=2=\frac{\mathbf{P_t}}{\mathbf{P_i}} \Rightarrow \mathbf{P_t} = 2 \mathbf{P_i} \Rightarrow \mathbf{p}(x=0)= 2 \mathbf{P_i}$

The amplitudes of the incident and reflected waves are equal. The
reflected wave is in phase with the incident wave. The particle velocity
at the boundary is zero. The acoustic pressure amplitude at the boundary
is equal to twice the pressure amplitude of the incident wave and it is
maximum.

## Case 2: Resilient boundary

Consider an incident planar wave that encounters a resilient boundary.
This is the case if the specific impedance of medium 2 is significantly
smaller than the specific impedance of medium 1. Thus, the specific
acoustic impedance ratio approaches zero, the reflection coefficient
approaches -1 and the transmission coefficient approaches zero.

$\mathbf{R}=-1=\frac{\mathbf{P_r}}{\mathbf{P_i}} \Rightarrow \mathbf{P_r} = - \mathbf{P_i} \Rightarrow \mathbf{u}(x=0)= \frac{\mathbf{2P_i}}{r_1}$

$\mathbf{T}=0=\frac{\mathbf{P_t}}{\mathbf{P_i}} \Rightarrow \mathbf{P_t} = 0 \Rightarrow \mathbf{p}(x=0)= 0$

The amplitudes of the incident and reflected waves are equal. The
reflected wave is 180\\degree out of phase with the incident wave. The
particle velocity at the boundary is a maximum. The acoustic pressure at
the boundary is zero.

## Case 3: Equal impedance in both media

Consider two media with the same specific acoustic impedance so that the
specific acoustic impedance ratio is unity, the reflection coefficient
is zero and the transmission coefficient is unity. Therefore, the wave
is not reflected, only transmitted. It behaves as if there was no
boundary.


# Engineering Acoustics/Transverse vibrations of strings

## Introduction

This section deals with the wave nature of vibrations constrained to one
dimension. Examples of this type of wave motion are found in objects
such a pipes and tubes with a small diameter (no transverse motion of
fluid) or in a string stretched on a musical instrument.

Stretched strings can be used to produce sound (e.g. music instruments
like guitars). The stretched string constitutes a mechanical system that
will be studied in this chapter. Later, the characteristics of this
system will be used to help to understand by analogies acoustical
systems.

## What is a wave equation?

There are various types of waves (i.e. electromagnetic, mechanical,
etc.) that act all around us. It is important to use wave equations to
describe the time-space behavior of the variables of interest in such
waves. Wave equations solve the fundamental equations of motion in a way
that eliminates all variables but one. Waves can propagate longitudinal
or parallel to the propagation direction or perpendicular (transverse)
to the direction of propagation. To visualize the motion of such waves
click
here
(Acoustics animations provided by Dr. Dan Russell,Kettering University)

## One dimensional Case

Assumptions :

\- the string is uniform in size and density

\- stiffness of string is negligible for small deformations

\- effects of gravity neglected

\- no dissipative forces like frictions

\- string deforms in a plane

\- motion of the string can be described by using one single spatial
coordinate

Spatial representation of the string in vibration:

![](1Dwave_graph1.png "1Dwave_graph1.png"){width="300"}

The following is the free-body diagram of a string in motion in a
spatial coordinate system:

![](string_dwg.jpg "string_dwg.jpg")

From the diagram above, it can be observed that the tensions in each
side of the string will be the same as follows:

![](equations1.jpg "equations1.jpg")

Using Taylor series to expand we obtain:

![](equations2.jpg "equations2.jpg")

## Characterization of the mechanical system

A one dimensional wave can be described by the following equation
(called the wave equation):

$\left( \frac{\partial^2 y}{\partial x^2} \right)=\left( \frac{1}{c^2} \right)\left( \frac{\partial^2 y}{\partial t^2} \right)$

where,

$y(x,t)= f(\xi)+g(\eta)\,$ is a solution,

With $\xi=ct-x\,$ and $\eta=ct+x\,$

This is the D\'Alambert solution, for more information see:
1

Another way to solve this equation is the Method of separation of
variables. This is useful for modal analysis. This assumes the solution
is of the form:

$y(x,t)= f(x)g(t)\$

The result is the same as above, but in a form that is more convenient
for modal analysis.

For more information on this approach see: Eric W. Weisstein et al.
\"Separation of Variables.\" From MathWorld---A Wolfram Web Resource.
2

Please see Wave
Properties
for information on variable c, along with other important properties.

For more information on wave equations see: Eric W. Weisstein. \"Wave
Equation.\" From MathWorld---A Wolfram Web Resource.
3

Example with the function $f(\xi)$ :

Example: Java String
simulation

This show a simple simulation of a plucked string with fixed ends.

Back to Main page


# Engineering Acoustics/Time-Domain Solutions

## d\'Alembert Solutions

In 1747, Jean Le Rond
d\'Alembert
published a solution to the one-dimensional wave equation.

The general solution, now known as the d\'Alembert method, can be found
by introducing two new variables:

$\xi=ct-x\,$ and $\eta=ct+x\,$

and then applying the chain rule to the general form of the wave
equation.

From this, the solution can be written in the form:

$y(\xi,\eta)= f(\xi)+g(\eta)\,=f(x+ct)+g(x-ct)$

where f and g are arbitrary functions, that represent two waves
traveling in opposing directions.

A more detailed look into the proof of the d\'Alembert solution can be
found here.

## Example of Time Domain Solution

If f(ct-x) is plotted vs. x for two instants in time, the two waves are
the same shape but the second displaced by a distance of c(t2-t1) to the
right.

The two arbitrary functions could be determined from initial conditions
or boundary values.

Back to Main page


# Engineering Acoustics/Boundary Conditions and Forced Vibrations

## Boundary Conditions

The functions representing the solutions to the wave equation previously
discussed,

i.e. $y(x,t)= f(\xi)+g(\eta)\,$ with $\xi=ct-x\,$ and $\eta=ct+x\,$

are dependent upon the boundary and initial conditions. If it is assumed
that the wave is propagating through a string, the initial conditions
are related to the specific disturbance in the string at t=0. These
specific disturbances are determined by location and type of contact and
can be anything from simple oscillations to violent impulses. The
effects of boundary conditions are less subtle.

The most simple boundary conditions are the Fixed Support and Free End.
In practice, the Free End boundary condition is rarely encountered since
it is assumed there are no transverse forces holding the string (e.g.
the string is simply floating).

### For a Fixed Support

The overall displacement of the waves travelling in the string, at the
support, must be zero. Denoting x=0 at the support, This requires:

$y(0,t)= f(ct-0)+g(ct+0) = 0\,$

Therefore, the total transverse displacement at x=0 is zero.

The sequence of wave reflection for incident, reflected and combined
waves are illustrated below. Please note that the wave is traveling to
the left (negative **x** direction) at the beginning. The reflected wave
is ,of course, traveling to the right (positive **x** direction).

**t=0** ![](Fixed_t0.gif "Fixed_t0.gif")

**t=t1** ![](Fixed_t1.gif "Fixed_t1.gif")

**t=t2** ![](Fixed_t2.gif "Fixed_t2.gif")

**t=t3** ![](Fixed_t3.gif "Fixed_t3.gif")

### For a Free Support

Unlike the Fixed Support boundary condition, the transverse displacement
at the support does not need to be zero, but must require the sum of
transverse forces to cancel. If it is assumed that the angle of
displacement is small,

$\sin(\theta)\approx\theta=\left( \frac{\partial y}{\partial x} \right)\,$

and so,

$\sum F_{y} = T\sin(\theta)\approx T\left( \frac{\partial y}{\partial x} \right)=0\,$

But of course, the tension in the string, or T, will not be zero and
this requires the slope at x=0 to be zero:

i.e. $\left( \frac{\partial y}{\partial x} \right)=0\,$

Again for free boundary, the sequence of wave reflection for incident,
reflected and combined waves are illustrated below:

**t=0** ![](Fixed_t0.gif "Fixed_t0.gif")

**t=t1** ![](Free_t1.gif "Free_t1.gif")

**t=t2** ![](Free_t2.gif "Free_t2.gif")

**t=t3** ![](Free_t3.gif "Free_t3.gif")

### Other Boundary Conditions

There are many other types of boundary conditions that do not fall into
our simplified categories. As one would expect though, it isn\'t
difficult to relate the characteristics of numerous \"complex\" systems
to the basic boundary conditions. Typical or realistic boundary
conditions include mass-loaded, resistance-loaded, damping-loaded, and
impedance-loaded strings. For further information, see Kinsler,
Fundamentals of Acoustics, pp 54--58.

Here is a website with nice movies of wave reflection at different
BC\'s: Wave
Reflection

## Wave Properties

To begin with, a few definitions of useful variables will be discussed.
These include; the wave number, phase speed, and wavelength
characteristics of wave travelling through a string.

The speed that a wave propagates through a string is given in terms of
the phase speed, typically in m/s, given by:

$c = \sqrt{T/\rho_{L}}\,$ where $\rho_{L}\,$ is the density per unit
length of the string.

The wave number is used to reduce the transverse displacement equation
to a simpler form and for simple harmonic motion, is multiplied by the
lateral position. It is given by:

$k=\left( \frac{\omega}{c} \right)\,$ where $\omega=2\pi f\,$

Lastly, the wavelength is defined as:

$\lambda=\left( \frac{2\pi}{k} \right)=\left( \frac{c}{f} \right)\,$

and is defined as the distance between two points, usually peaks, of a
periodic waveform.

These \"wave properties\" are of practical importance when calculating
the solution of the wave equation for a number of different cases. As
will be seen later, the wave number is used extensively to describe wave
phenomenon graphically and quantitatively.

For further information: Wave
Properties

## Forced Vibrations

1.forced vibrations of infinite string suppose there is a string very
long , at x=0 there is a force exerted on it.

F(t)=Fcos(wt)=Real{Fexp(jwt)}

use the boundary condition at x=0,

neglect the reflected wave

it is easy to get the wave form

where w is the angular velocity, k is the wave number.

according to the impedance definition

it represents the characteristic impedance of the string. obviously, it
is purely resistive, which is like the resistance in the mechanical
system.

------------------------------------------------------------------------

The dissipated power

Note: along the string, all the variables propagate at same speed.

link title a useful link to show
the time-space property of the wave.

------------------------------------------------------------------------

Some interesting animations of the wave at different boundary
conditions.

1.hard boundary (which is like a fixed end)

2.soft boundary (which is like a free end)

3.from low density to high density string

4.from high density to low density string

Back to Main page


# Engineering Acoustics/Boundary Conditions and Wave Properties

## Boundary Conditions

The functions representing the solutions to the wave equation previously
discussed,

i.e. $y(x,t)= f(\xi)+g(\eta)\,$ with $\xi=ct-x\,$ and $\eta=ct+x\,$

are dependent upon the boundary and initial conditions. If it is assumed
that the wave is propagating through a string, the initial conditions
are related to the specific disturbance in the string at t=0. These
specific disturbances are determined by location and type of contact and
can be anything from simple oscillations to violent impulses. The
effects of boundary conditions are less subtle.

The most simple boundary conditions are the Fixed Support and Free End.
In practice, the Free End boundary condition is rarely encountered since
it is assumed there are no transverse forces holding the string (e.g.
the string is simply floating).

`- For a Fixed Support:`

The overall displacement of the waves travelling in the string, at the
support, must be zero. Denoting x=0 at the support, This requires:

$y(0,t)= f(ct-0)+g(ct+0) = 0\,$

Therefore, the total transverse displacement at x=0 is zero.

`- For a Free Support:`

Unlike the Fixed Support boundary condition, the transverse displacement
at the support does not need to be zero, but must require the sum of
transverse forces to cancel. If it is assumed that the angle of
displacement is small,

$\sin(\theta)\approx\theta=\left( \frac{\partial y}{\partial x} \right)\,$

and so,

$\sum F_{y} = T\sin(\theta)\approx T\left( \frac{\partial y}{\partial x} \right)=0\,$

But of course, the tension in the string, or T, will not be zero and
this requires the slope at x=0 to be zero:

i.e. $\left( \frac{\partial y}{\partial x} \right)=0\,$

`- Other Boundary Conditions:`

There are many other types of boundary conditions that do not fall into
our simplified categories. As one would expect though, it isn\'t
difficult to relate the characteristics of numerous \"complex\" systems
to the basic boundary conditions. Typical or realistic boundary
conditions include mass-loaded, resistance-loaded, damping-loaded, and
impedance-loaded strings. For further information, see Kinsler,
Fundamentals of Acoustics, pp 54--58.

## Wave Properties

To begin with, a few definitions of useful variables will be discussed.
These include; the wave number, phase speed, and wavelength
characteristics of wave travelling through a string.

The speed that a wave propagates through a string is given in terms of
the phase speed, typically in m/s, given by:

$c = \sqrt{T/\rho_{L}}\,$ where $\rho_{L}\,$ is the density per unit
length of the string.

The wave number is used to reduce the transverse displacement equation
to a simpler form and for simple harmonic motion, is multiplied by the
lateral position. It is given by:

$k=\left( \frac{\omega}{c} \right)\,$ where $\omega=2\pi f\,$

Lastly, the wavelength is defined as:

$\lambda=\left( \frac{2\pi}{k} \right)=\left( \frac{c}{f} \right)\,$

and is defined as the distance between two points, usually peaks, of a
periodic waveform.

These \"wave properties\" are of practical importance when calculating
the solution of the wave equation for a number of different cases. As
will be seen later, the wave number is used extensively to describe wave
phenomenon graphically and quantitatively.

For further information: Wave
Properties

Back to Main page

------------------------------------------------------------------------

Edited by: Mychal Spencer


# Engineering Acoustics/Attenuation of Sound Waves

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## Introduction

When sound travels through a medium, its intensity diminishes with
distance. This weakening in the energy of the wave results from two
basic causes, scattering and absorption. The combined effect of
scattering and absorption is called attenuation. For small distances or
short times the effects of attenuation in sound waves can usually be
ignored. Yet, for practical reasons it should be considered. So far in
our discussions, sound has only been dissipated by the spreading of the
wave, such as when we consider spherical and cylindrical waves. However
this dissipation of sound in these cases is due to geometric effects
associated with energy being spread over an increasing area and not
actually to any loss of total energy.

## Types of Attenuation

As mentioned above, attenuation is caused by both absorption and
scattering. Absorption is generally caused by the media. This can be due
to energy loss by both viscosity and heat conduction. Attenuation due to
absorption is important when the volume of the material is large.
Scattering, the second cause of attenuation, is important when the
volume is small or in cases of thin ducts and porous materials.

#### Viscosity and Heat conduction

Whenever there is a relative motion between particles in a media, such
as in wave propagation, energy conversion occurs. This is due to stress
from viscous forces between particles of the medium. The energy lost is
converted to heat. Because of this, the intensity of a sound wave
decreases more rapidly than the inverse square of distance. Viscosity in
gases is dependent upon temperature for the most part. Thus as you
increase the temperature you increase the viscous forces.

#### Boundary Layer Losses

A special type of absorption occurs when a sound wave travels over a
boundary, such as a fluid flowing over a solid surface. In such a
situation, the fluid in immediate contact with the surface must be at
rest. Subsequent layers of fluid will have a velocity that increases as
the distance from the solid surface increases such as in the figure
below.

The velocity gradient causes an internal stress associated with
viscosity, that leads to a loss of momentum. This loss of momentum leads
to a decrease in the amplitude of a wave close to the surface. The
region over which the velocity of the fluid decreases from its nominal
velocity to that of zero is called the acoustic boundary layer. The
thickness of the acoustic boundary layer due to viscosity can be
expressed as

$\delta_{visc}=\sqrt{\left(\frac{ 2*\mu}{\omega*\rho_o}\right)}$

Where $\mu \,$ is the shear viscosity number. Ideal fluids would not
have a boundary layer thickness since $\mu=0 \,$ .

### Relaxation

Attenuation can also occur by a process called relaxation. One of the
basic assumptions prior to this discussion on attenuation was that when
a pressure or density of a fluid or media depended only on the
instantaneous values of density and temperature and not on the rate of
change in these variables. However, whenever a change occurs,
equilibrium is upset and the media adjusts until a new local equilibrium
is achieved. This does not occur instantaneously, and pressure and
density will vary in the media. The time it takes to achieve this new
equilibrium is called the relaxation time, $\theta \,$ . As a
consequence the speed of sound will increase from an initial value to
that of a maximum as frequency increases. Again the losses associated
with relaxation are due to mechanical energy being transformed into
heat.

## Modeling of losses

The following is done for a plane wave. Losses can be introduced by the
addition of a complex expression for the wave number

$k=\ \beta-j\alpha$

which when substituted into the time-solution yields

$\ p = A e^{-\alpha x}e^{jwt-j\beta x}$

with a new term of $\ e^{-\alpha x}$ which resulted from the use of a
complex wave number. Note the negative sign preceding $\alpha$ to denote
an exponential decay in amplitude with increase values of $x$.

$\ \alpha$ is known as the absorption coefficient with units of nepers
per unit distance and $\ \beta$ is related to the phase speed. The
absorption coefficient is frequency dependent and is generally
proportional to the square of sound frequency. However, its relationship
does vary when considering the different absorption mechanisms as shown
below.

The velocity of the particles can be expressed as

$\ u=\frac{k}{w*\rho_o}p= \frac{1}{\rho_o c}\left(1-j\frac{\alpha}{k}\right)p$

The impedance for this traveling wave would be given by

$\ z= \rho_o c\frac{1}{1-j\frac{\alpha}{k}}$

From this we can see that the rate of decrease in intensity of an
attenuated wave is $\ a=8.7\alpha$

## References

Wood, A. A Textbook of Sound. London: Bell, 1957.

Blackstock, David. Fundamentals of Physical Acoustics. New York: Wiley,
2000.

Attenuation considerations in
Ultrasound

Back to main page


# Engineering Acoustics/Sound Propagation in a Cylindrical Duct With Compliant Wall

There have been many applications that deals with sound propagation in a
compliant wall duct. These range from hydraulic applications(i.e. water
hammer) to biomechanical
applications (i.e. pressure pulse in an
artery). In working with sound
propagation in a circular duct, the duct wall is often assumed to be
rigid so that any pressure disturbance in the fluid has no effect on the
wall. However, if the wall is assumed to be compliant, i.e. wall
deformation is possible when a pressure disturbance is encountered,
then, this will change the speed of the sound propagation. In reality,
the rigid wall assumption will be valid if the pressure disturbance in
the fluid, which is a function of the fluid density, is very small so
that the deformation of the wall is insignificant. However, if the duct
wall is assumed to be thin, i.e. \~ 1/20 of the radius or smaller, or if
the wall is made of plastic type of material with low Young\'s modulus
and density, or if the fluid contained is "heavy", the rigid wall
approximation is no longer true. In this case, the wall is assumed to be
compliant.

In the book by Morse & Ingard \[1\], the wall stiffness is defined as
K~w~, and this is the ratio between the pressure disturbances, p, to the
fractional change in cross-sectional area of the duct, produced by p. Of
course this pressure disturbance p is not static and the inertial of the
wall has to be considered. Because the deformation of the wall is due to
the pressure disturbances in the fluid, this is a typical
fluid-structure interaction problem, where the pressure disturbances in
the fluid cause the structural deformation, which in turn, modifies the
pressure disturbances.

Unlike sound propagate in a duct with a rigid wall where sound pressure
travels down the tube axially; part of the pressure is used to stretch
the tube radially. Clearly, because of the inclusion of tube wall
displacement, this becomes a fluid-structure interaction problem.

# Analysis

In this analysis, it is expected that the speed of the propagation will
depend on the material properties of the tube wall, i.e. Young\'s
modulus and the density. Also, as the analysis unfolds, it will become
apparently clear that the speed of propagation will vary with excitation
frequency, unlike wave propagation in a rigid-wall tube. Keep in mind
that the analysis presented here is a very simple one. Of course
depending on the fluid & structure model, one can reach a more accurate
results. In the analysis presented here, the goal is to provide a basic
physical understanding of sound propagation in a cylindrical duct with
compliant wall.

Assumptions:

1.  One-dimensional analysis
2.  All viscous & thermal dissipation are neglected
3.  Analysis are per unit length
4.  No mean flow velocity is considered

## Fluid

Two simplified fluid equations will be considered here:

$-\frac{\partial u}{\partial x}=\kappa \frac{\partial p}{\partial t}\dots \dots .Eq(1)$

and

$-\frac{\partial p}{\partial x}={\rho }_f\frac{\partial u}{\partial t}\dots \dots .Eq(2)$

where $\kappa =\frac{1}{{\rho }_fc^2}$, $u$ is the fluid particle
velocity and $p$ is the fluid pressure.

The first equation is the continuity equation, where the density term is
replaced by the pressure term by applying the ideal gas law $p=\rho RT$
and the isentropic gas law $c^2=\gamma RT$.

Because of the compliant wall, the fluid experiences an additional
compressibility effects, and according to Morse & Ingard \[1\] , this
additional compressibility is derived based on the wall stiffness (K).
The wall stiffness is defined as the ratio between the pressure and the
fractional change in cross-sectional area. By introducing K into Eq. 1,
it yields

$-\frac{\partial u}{\partial x}=\left(\kappa +\frac{1}{K}\right)\frac{\partial p}{\partial t}\dots \dots Eq\left(3\right)$

If the mass of the tube is considered, then additional mass parameter,
M~w~, must be included.

The total stiffness impedance of the wall is then:

$Z_w=j\omega M_w+\frac{K_w}{j\omega }=j\frac{M_w}{\omega }\left({\omega }^2-{\omega }^2_n\right)$

Treating this wall impedance as a compliance term (i.e.
$K=j\omega Z_w$), and substitute back into Eq 3 yields

$-\frac{\partial u}{\partial x}=\left(\kappa +\frac{1}{M_w({{\omega }^2_n-\omega }^2)}\right)\frac{\partial p}{\partial t}$

Here, ${\omega }_n=\sqrt{\frac{K_w}{M_w}}$.

Furthermore, by taking out ${\kappa }$, the expression becomes

$-\frac{\partial u}{\partial x}=\kappa \left(\frac{M_w\left({{\omega }^2_n-\omega }^2\right)+{\rho }_fc^2}{M_w({{\omega }^2_n-\omega }^2)}\right)\frac{\partial p}{\partial t}$

After some manipulation, it yields

$-\frac{\partial u}{\partial x}=\kappa \left(\frac{{{\omega }^2_1-\omega }^2}{{{\omega }^2_n-\omega }^2}\right)\frac{\partial p}{\partial t}$

where ${\omega }^2_1={\omega }^2_n+\frac{{\rho }_fc^2}{M_w}$

For rigid wall, as $K_w\to \infty$, ${\omega }^2_n\to \infty$,
${\omega }^2_1\to \infty$, hence,
$-\frac{\partial u}{\partial x}=\kappa \frac{\partial p}{\partial t}$,
which leads back to Equation 1 .

If the impedance analogy is used, i.e. pressure is the voltage and the
velocity the current, then

$C=\ \kappa \left(\frac{{{\omega }^2_1-\omega }^2}{{{\omega }^2_n-\omega }^2}\right)$
,where C is the compliance of the wall per unit length.

The speed of the sound is then determined by $\sqrt{\frac{1}{LC}}$,
hence

$c_p=c\sqrt{\left(\frac{{{\omega }^2_n-\omega }^2}{{{\omega }^2_1-\omega }^2}\right)}$

Here, $c_p$ is the phase velocity and it depends on the excitation
frequency,${\omega }$, the acoustic wave is dispersive. When the
excitation frequency is below the natural frequency, ${\omega }_n$, the
phase speed is lower than that of the free wave speed in fluid.

The next step is to identify K~w~ and M~w~, which will be determined
through structural response.

## Structure

Assume the undeformed tube has a diameter of D and after deformation, it
becomes $D+\triangle D$. The area change in this case is
$\frac{\pi }{4}\left(2D\triangle D+{\left(\triangle D\right)}^2\right)$.
The ratio between the two is
$\frac{\triangle A}{A}=\frac{2\triangle D}{D}+{\left(\frac{\triangle D}{D}\right)}^2$.
Hence, the wall stiffness is then

$K_w=\frac{\triangle pD}{2\triangle D}$.

where the inverse of this is known as the compliance or
distensibility(a
bio-medical term).

To determine the term $\frac{\triangle D}{D}$, it is necessary to look
at the structural response by using Newtown\'s Law and Hook\'s Law.

Consider a cross section of a half tube with diameter D, thickness h and
tension T,

!A cut away cylindrical
tube{width="300"}

The hoop stress in a cylindrical tube is given by,

$\sigma =p\frac{D}{2h}$ Applying Hook\'s Law, it is possible to
determine the strain, $\varepsilon$, by

$\varepsilon =\frac{\sigma}{E}$.

With some substitutions,

$\varepsilon =\frac{pD}{2hE}$

For small strain,

$\Delta \varepsilon =\frac{\Delta pD}{2hE}=\frac{\Delta D}{D}$ Hence,

$K_w=E\frac{h}{D}$

This is the wall stiffness, a function of only the tube elastic
properties.

Mass of the tube per unit length is considered, then

$M_w={\rho}_s\pi h(D+h)$.

Finally, it is possible to plot the phase speed and the wall impedance
verse the excitation frequency.

# Discussions

In the simulation the thickness to diameter ratio,$\frac{h}{D}$ is 0.1,
the material is steel with $\rho = 7800 \frac{kg}{m^3}$ and
$E = 2x10^{11} Pa$. The fluid contained inside is assumed to be air with
$\rho = 1.21 \frac{kg}{m^3}$ and the free wave speed of
$c = 343 \frac{m}{s}$.

![](Tws1.jpg "Tws1.jpg"){width="600"}

In this diagram, the \'o\' denotes the real part of the phase speed and
the \'+\' denotes the imaginary part of the phase speed. The straight
line shows the sound speed in air with a numerical value of
$c = 343 \frac{m}{s}$. In this plot, propagation of wave is possible
only if the phase speed is real. There are two important frequencies
that deserves a close attention. The first is the natural frequency of
the empty structure, i.e. ${\omega }_n$ and the natural frequency of the
fluid loaded structure, ${\omega }_1$. In this plot,
${\omega }_n = 450 Hz$ while ${\omega }_1 = 550 Hz$.

Unlike 1-D wave propagation in a rigid duct where the propagation speed
is a constant, the phase speed depends on the excitation frequency. It
shows that the propagation speed decreases as the excitation frequency
approaches to ${\omega }_n$. Between ${\omega }_n$ and ${\omega }_1$,
the phase speed is imaginary, which means no wave can propagate in
between these two frequencies. As soon as the frequency increases pass
${\omega }_1$, the phase speed is greater than the free wave speed of
343 m/s. As the excitation frequency increases, the phase speed
approaches to the free wave speed.

When the excitation frequency is increased, the parts of the fluid
energy is used to excite the tube, until the excitation frequency
matches ${\omega }_n$. Beyond ${\omega }_n$, no true wave propagation is
possible because in between these two frequencies.

For a very rigid tube, i.e. $E = 2x10^{20} Pa$ , the phase speed is
exactly the free wave speed in air, which is a constant. This agrees
with what have been discussed before for a 1-D wave propagation in a
rigid duct.

![](Tws2.jpg "Tws2.jpg"){width="600"}

When the stiffness is reduced to 1/100 of the steel, there are numerous
differences than the steel tube. First, at the low frequency, the phase
speed is slower. This is because the lower wall stiffness, the more the
wall can be stretched, hence can absorb more energy. Also, the system
has a much lower ${\omega }_n$ and ${\omega }_1$.

![](Tws3.jpg "Tws3.jpg"){width="600"}

From the above analysis, it is possible to conclude the following:

1\. The stiffness, the wall thickness and the density of the tube
affects the phase speed dramatically.

2\. A reduction in stiffness, reduces the propagation speed at low
frequencies. The wave also becomes evanescent at much lower frequency.
This is because the natural frequency is reduced. As the stiffness
increases, the propagation speed approaches to that of the free wave
speed regardless of the frequency region. This is the case of rigid
wall.

3\. The propagation speed of a wave in a duct with compliant wall is
dispersive as it depends greatly on the frequency. The phase speed
differs significantly than that of in a rigid wall.

4\. The non-propagating zone is known as the stop band. Here, the wall
properties can be modified to create a larger stop band. Hence, a duct
with compliant wall can be considered as a type of band filter.

# References

\[1\]. Morse & Ingard (1968), \"Theoretical Acoustics\", Princeton
University Press, Princeton, New Jersey