Moulton Lectures

Electro-Acoustics

Lecture 1

Exploring The Fundamentals

Part 1

Presented by: Dave L Moulton

Location: Thales Acoustics

Date: 02-October-2002

Content

This first lecture explores what is probably the most fundamental relationship in acoustics.

The link between the mechanical domain and the acoustic domain.

Force = Pressure x Area

F = PA

This very innocent looking equation has a lot of hidden meaning when applied to the field of electro-acoustics, as we shall see later.

On with the Lecture

We can represent the above relationship in the form of two systems (Mechanical and Acoustic) which are tied together by the transforming factor ‘Area A’

This transformer method of linking two systems together is very common in electro-acoustics, especially when trying to model the performance of acoustic devices such as microphones, earphones and headsets etc…

We can make the following assumption about our transformer:

1. The Transformer is Ideal (No loss of Energy between systems)

Using this assumption we can state that the power transfer between systems is also ideal. This means that all the power in the mechanical system is transformed into the acoustic system, the converse also being true.

We will start by looking at the relationship used to express the power in the mechanical system:

Firstly remembering that power is defined as the rate of doing work, which can be expressed mathematically as:

In a mechanical system the work done is defined as:

Applied Force (F) multiplied by the Displacement (x) caused by the force, which can be expressed mathematically as:

Combining [1.1] and [1.2] we have:

We know that the rate of change of displacement is equal to the velocity (v) .

We can find an expression for the power in the acoustic system simply by substituting F=PA into equation 1.3

Now (Ax) represents the volume of air displaced by the pressure (usually represented by the symbol ‘U’ in acoustics) and is often called the Volume Displacement.

The rate of change of volume displacement is given the symbol U_V and is known as the Volume Velocity.

Thus expressed mathematically we have:

We can now look at expressions 1.4 and 1.7 in an analogous way by comparing them with the power relationship for the more familiar electrical system. Electrical power is related to the applied Voltage (V) and Current (I) in the following way.

Using this analogy we can modify our transformer system to show both the ‘Across Variables’ and the ‘Through Variables’.

Across variables are: F and P

Through variables are: v and U_V

Our transformer performs like any ideal electrical transformer and transforms the impedances from one domain into another by the factor A²

Thus:

and

Scalar or Vector?

Going back to the most fundamental of relationships F=PA we know that the force F must be a vector quantity since it is proportional to the acceleration which has direction. This automatically implies that the product PA must also be a vector quantity. We must now tread carefully and not naturally assume that the Pressure P is a vector and the Area A is a scalar.

In fact P is the scalar and A is the vector. This will require a bit more explanation. I have extracted an explanation that I found on the internet. It describes the situation quite clearly.

Pressure is a stress. It is a scalar given by the magnitude of the force per unit area. In a gas, it is the force per unit area exerted by the change of momentum of the molecules impinging on the surface. We know from Newton's second law that a net resultant force will cause a change of momentum in a body, and that the rate of change of momentum is equal to the applied force. It is a vector relationship, so that even if the magnitude of the momentum is unchanged, a change in the direction of motion requires a resultant force. Consider the line of action of the force due to pressure. On a molecular level, a flat solid surface is never flat. However, it is flat on average, so that, on average, for each molecule that rebounds with some component of momentum along the surface, another rebounds with the same component of the momentum in the opposite direction. The average momentum of the molecules in the direction along the surface will not change during their impact with the surface. We expect, therefore, that the average force due to pressure (on a macroscopic scale) acts in a direction which is purely normal to the surface. Furthermore, since the momentum of the molecules is randomly distributed in space, the magnitude of the force due to pressure should be independent of the direction of the surface on which it acts. For instance, a thin flat plate in air will experience a zero resultant force due to air pressure since the forces due to pressure on its two sides have the same magnitude (the pressure is independent of the orientation of the surface on which it acts) and they point in opposite directions (each force acts normal to its own surface). We say that pressure is isotropic (based on Greek words, meaning equal in all directions, or more accurately in the present case, independent of direction).

The diagram below shows a microscopic view of the impact of air particles on a surface. The air particles hit the surface with a momentum and then rebound and change direction. The change in direction represents a change in momentum, thus a force must have been exerted on an element of the surface by a particle in order for the particles momentum to change. The force exerted by the particles always acts perpendicular to the elementary surfaces.

The Ideal Flat Surface

We can also apply this principal of particle collisions to any ideal flat surface and get the same result. In the ideal case the air particles can be treated like small spherical balls, each having mass and momentum. Under the influence of a scalar pressure field these balls will strike the surface at various angles and bounce off . For a rigid surface these collisions will be perfectly elastic and there will be no loss of particle energy during the collision.

Just like in a game of snooker when the white ball hits the red ball at a certain point on its surface, the line of force exerted on the red ball will be perpendicular to the elementary area on the red ball at the point of impact. This will be the case for all possible impact directions of the white ball.

Vector Area

The general expression F=PA refers to a resultant Force caused by a Pressure acting over an Area. The Area in question is in fact a vector area which is not necessarily the total surface area.

The Vector Area is the 2 Dimensional Area that you would see if you looked directly at the 3 Dimensional object. The resultant force will be in a direction perpendicular to the vector area.

A simplistic Mathematical proof is as follows:

Consider a flat strip of material of width ‘w’ shaped so that its side view profile looks like a semicircle.

We can give this semicircle a radius ‘r’

Assume that the semicircular surface is subject to a scalar pressure field on its outer face. We will assume for this example that the inner face is not subject to the pressure field.

We now consider what is happening on a small delta area of the strip dA located at an angle q to the central axis. We can also consider that the delta area makes an arc angle of dq to the axis.

See the diagram below:

Looking at the diagram we can see that the resultant Force vector at any point on the surface can be resolved into its x and y components. The y components will all tend to cancel out across the length of the strip, leaving the x component as the direction of the acting force.

Now for the maths:

Firstly resolving the Force in the x direction