Elementary Physics, Mechanics, and Acoustics
This appendix derives some basic results from the field of physics, particularly mechanics and acoustics, which are referenced elsewhere in this book, or which are known to be needed in practical synthesis models.Newton's Laws of Motion
Perhaps the most heavily used equation in physics is Newton's second law of motion:In this formulation, the applied force is considered positive in the direction of positive massposition . The force and acceleration are, in general, vectors in threedimensional space . In other words, force and acceleration are generally vectorvalued functions of time . The mass is a scalar quantity, and can be considered a measure of the inertia of the physical system (see §B.1.1 below).
Newton's Three Laws of Motion
Newton's three laws of motion may be stated as follows: Every object in a state of uniform motion will remain in that state of motion unless an external force acts on it.
 Force equals mass times acceleration [ ].
 For every action there is an equal and opposite reaction.
Mass
Mass is an intrinsic property of matter. From Newton's second law, , we have that the amount of force required to accelerate an object, by a given amount, is proportional to its mass. Thus, the mass of an object quantifies its inertiaits resistance to a change in velocity. We can measure the mass of an object by measuring the gravitational force between it and another known mass, as described in the next section. This is a special case of measuring its acceleration in response to a known force. Whatever the force , the mass is given by divided by the resulting acceleration , again by Newton's second law . The usual mathematical model for an ideal mass is a dimensionless point at some location in space. While no real objects are dimensionless, they can often be treated mathematically as dimensionless points located at their center of mass, or centroid (§B.4.1). The physical state of a mass at time consists of its position and velocity in 3D space. The amount of mass itself, , is regarded as a fixed parameter that does not change. In other words, the state of a physical system typically changes over time, while any parameters of the system, such as mass , remain fixed over time (unless otherwise specified).Gravitational Force
We are all familiar with the force of gravity. It is a fundamental observed property of our universe that any two masses and experience an attracting force given by the formulawhere is the distance between the centroids of the masses and at time , and is the gravitation constant.^{B.2} The law of gravitation Eq.(B.2) can be accepted as an experimental fact which defines the concept of a force.^{B.3} The giant conceptual leap taken by Newton was that the law of gravitation is universalapplying to celestial bodies as well as objects on earth. When a mass is ``dropped'' and allowed to ``fall'' in a gravitational field, it is observed to experience a uniform acceleration proportional to its mass. Newton's second law of motion (§B.1) quantifies this result.
Hooke's Law
Consider an ideal spring suspending a mass from a rigid ceiling, as depicted in Fig.B.1. Assume the mass is at rest, and that its distance from the ceiling is fixed. If denotes the mass of the earth, and is the distance of mass 's center from the earth's center of mass, then the downward force on the mass is given by Eq.(B.2) aswhere is the displacement of the spring from its natural length. We call the spring constant, or stiffness of the spring. In terms of our previous notation, we have
Applying Newton's Laws of Motion
As a simple example, consider a mass driven along a frictionless surface by an ideal spring , as shown in Fig.B.2. Assume that the mass position corresponds to the spring at rest, i.e., not stretched or compressed. The force necessary to compress the spring by a distance is given by Hooke's law (§B.1.3):where we have defined as the initial displacement of the mass along . This is a differential equation whose solution gives the equation of motion of the massspring junction for all time:^{B.7}
where denotes the frequency of oscillation in radians per second. More generally, the complete space of solutions to Eq.(B.4), corresponding to all possible initial displacements and initial velocities , is the set of all sinusoidal oscillations at frequency :
Work = Force times Distance = Energy
Work is defined as force times distance. Work is a measure of the energy expended in applying a force to move an object.^{B.8} The work required to compress a spring through a displacement of meters, starting from rest, is thenWork can also be negative. For example, when uncompressing an ideal spring, the (positive) work done by the spring on its moving end support can be interpreted also as saying that the end support performs negative work on the spring as it allows the spring to uncompress. When negative work is performed, the driving system is always accepting energy from the driven system. This is all simply accounting. Physically, one normally considers the driver as the agent performing the positive work, i.e., the one expending energy to move the driven object. Thus, when allowing a spring to uncompress, we consider the spring as performing (positive) work on whatever is attached to its moving end. During a sinusoidal massspring oscillation, as derived in §B.1.4, each period of the oscillation can be divided into equal sections during which either the mass performs work on the spring, or vice versa. Gravity, spring forces, and electrostatic forces are examples of conservative forces. Conservative forces have the property that the work required to move an object from point to point , either with or against the force, depends only on the locations of points and in space, not on the path taken from to .
Potential Energy in a Spring
When compressing an ideal spring, work is performed, and this work is stored in the spring in the form of what we call potential energy. Equation (B.6) above gives the quantitative formula for the potential energy stored in an ideal spring after it has been compressed meters from rest.Kinetic Energy of a Mass
Kinetic energy is energy associated with motion. For example, when a spring uncompresses and accelerates a mass, as in the configuration of Fig.B.2, work is performed on the mass by the spring, and we say that the potential energy of the spring is converted to kinetic energy of the mass. Suppose in Fig.B.2 we have an initial spring compression by meters at time , and the mass velocity is zero at . Then from the equation of motion Eq.(B.5), we can calculate when the spring returns to rest (). This first happens at the first zero of , which is time . At this time, the velocity, given by the timederivative of Eq.(B.5),Mass Kinetic Energy from Virtual Work
From Newton's second law, (introduced in Eq.(B.1)), we can use d'Alembert's idea of virtual work to derive the formula for the kinetic energy of a mass given its speed . Let denote a small (infinitesimal) displacement of the mass in the direction. Then we have, using the calculus of differentials,Energy in the MassSpring Oscillator
Summarizing the previous sections, we say that a compressed spring holds a potential energy equal to the work required to compress the spring from rest to its current displacement. If a compressed spring is allowed to expand by pushing a mass, as in the system of Fig.B.2, the potential energy in the spring is converted to kinetic energy in the moving mass. We can draw some inferences from the oscillatory motion of the massspring system written in Eq.(B.5): From a global point of view, we see that energy is conserved, since the oscillation never decays.
 At the peaks of the displacement (when is either or ), all energy is in the form of potential energy, i.e., the spring is either maximally compressed or stretched, and the mass is momentarily stopped as it is changing direction.
 At the zerocrossings of , the spring is momentarily relaxed, thereby holding no potential energy; at these instants, all energy is in the form of kinetic energy, stored in the motion of the mass.
 Since total energy is conserved (§B.2.5), the kinetic
energy of the mass at the displacement zerocrossings is exactly the
amount needed to stretch the spring to displacement (or compress
it to ) before the mass stops and changes direction. At all
times, the total energy is equal to the sum of the potential
energy stored in the spring, and the kinetic energy
stored in the mass:
Energy Conservation
It is a remarkable property of our universe that energy is conserved under all circumstances. There are no known exceptions to the conservation of energy, even when relativistic and quantum effects are considered.^{B.9} Energy may be defined as the ability to do work, where work may be defined as force times distance (§B.2).Energy Conservation in the MassSpring System
Recall that Newton's second law applied to a massspring system, as in §B.1.4, yieldsMomentum
The momentum of a mass is usually defined byConservation of Momentum
Like energy, momentum is conserved in physical systems. For example, when two masses collide and recoil from each other, the total momentum after the collision equals that before the collision. Since the momentum is a threedimensional vector in Euclidean space, momentum conservation provides three simultaneous equations, in general.^{B.10}RigidBody Dynamics
Below are selected topics from rigidbody dynamics, a subtopic of classical mechanics involving the use of Newton's laws of motion to solve for the motion of rigid bodies moving in 1D, 2D, or 3D space.^{B.11} We may think of a rigid body as a distributed mass, that is, a mass that has length, area, and/or volume rather than occupying only a single point in space. Rigid body models have application in stiff strings (modeling them as disks of mass interconnect by ideal springs), rigid bridges, resonator braces, and so on. We have already used Newton's to formulate mathematical dynamic models for the ideal pointmass (§B.1.1), spring (§B.1.3), and a simple massspring system (§B.1.4). Since many physical systems can be modeled as assemblies of masses and (normally damped) springs, we are pretty far along already. However, when the springs interconnecting our pointmasses are very stiff, we may approximate them as rigid to simplify our simulations. Thus, rigid bodies can be considered massspring systems in which the springs are so stiff that they can be treated as rigid massless rods (infinite springconstants , in the notation of §B.1.3). So, what is new about distributed masses, as opposed to the pointmasses considered previously? As we will see, the main new ingredient is rotational dynamics. The total momentum of a rigid body (distributed mass) moving through space will be described as the sum of the linear momentum of its center of mass (§B.4.1 below) plus the angular momentum about its center of mass (§B.4.13 below).Center of Mass
The center of mass (or centroid) of a rigid body is found by averaging the spatial points of the body weighted by the mass of those points:^{B.12}Linear Momentum of the Center of Mass
Consider a system of pointmasses , each traveling with vector velocity , and not necessarily rigidly attached to each other. Then the total momentum of the system is
Whoops, No Angular Momentum!
The previous result might be surprising since we said at the outset
that we were going to decompose the total momentum into a sum
of linear plus angular momentum. Instead, we found that the total
momentum is simply that of the center of mass, which means any angular
momentum that might have been present just went away. (The center of
mass is just a point that cannot rotate in a measurable way.) Angular
momentum does not contribute to linear momentum, so it provides three
new ``degrees of freedom'' (three new energy storage dimensions, in 3D
space) that are ``missed'' when considering only linear momentum.
To obtain the desired decomposition of momentum into linear plus
angular momentum, we will choose a fixed reference point in space
(usually the center of mass) and then, with respect to that reference
point, decompose an arbitrary massparticle travel direction into the
sum of two mutually orthogonal vector components: one will be the
vector component pointing radially with respect to the fixed
point (for the ``linear momentum'' component), and the other will be
the vector component pointing tangentially with respect to the
fixed point (for the ``angular momentum''), as shown in
Fig.B.3. When the reference point is the center of mass, the
resultant radial force component gives us the force on the center of
mass, which creates linear momentum, while the net tangential
component (times distance from the centerofmass) give us a resultant
torque about the reference point, which creates angular
momentum. As we saw above, because the tangential force component
does not contribute to linear momentum, we can simply sum the external
force vectors and get the same result as summing their radial
components. These topics will be discussed further below, after some
elementary preliminaries.
Translational Kinetic Energy
The translational kinetic energy of a collection of masses is given byRotational Kinetic Energy
where
is called the mass moment of inertia.
Mass Moment of Inertia
The mass moment of inertia (or simply moment of inertia), plays the role of mass in rotational dynamics, as we saw in Eq.(B.7) above. The mass moment of inertia of a rigid body, relative to a given axis of rotation, is given by a weighted sum over its mass, with each masspoint weighted by the square of its distance from the rotation axis. Compare this with the center of mass (§B.4.1) in which each masspoint is weighted by its vector location in space (and divided by the total mass). Equation (B.8) above gives the moment of inertia for a single pointmass rotating a distance from the axis to be . Therefore, for a rigid collection of pointmasses , ,^{B.14} the moment of inertia about a given axis of rotation is obtained by adding the component moments of inertia:where is the distance from the axis of rotation to the th mass. For a continuous mass distribution, the moment of inertia is given by integrating the contribution of each differential mass element:
(B.10) 
where is the distance from the axis of rotation to the mass element . In terms of the density of a continuous mass distribution, we can write
Circular Disk Rotating in Its Own Plane
For example, the moment of inertia for a uniform circular disk of total mass and radius , rotating in its own plane about a rotation axis piercing its center, is given byCircular Disk Rotating About Its Diameter
The moment of inertia for the same circular disk rotating about an axis in the plane of the disk, passing through its center, is given byPerpendicular Axis Theorem
In general, for any 2D distribution of mass, the moment of inertia about an axis orthogonal to the plane of the mass equals the sum of the moments of inertia about any two mutually orthogonal axes in the plane of the mass intersecting the first axis. To see this, consider an arbitrary mass element having rectilinear coordinates in the plane of the mass. (All three coordinate axes intersect at a point in the massdistribution plane.) Then its moment of inertia about the axis orthogonal to the mass plane is while its moment of inertia about coordinate axes within the massplane are respectively and . This, the perpendicular axis theorem is an immediate consequence of the Pythagorean theorem for right triangles.Parallel Axis Theorem
Let denote the moment of inertia for a rotation axis passing through the center of mass, and let denote the moment of inertia for a rotation axis parallel to the first but a distance away from it. Then the parallel axis theorem says thatStretch Rule
Note that the moment of inertia does not change when masses are moved along a vector parallel to the axis of rotation (see, e.g., Eq.(B.9)). Thus, any rigid body may be ``stretched'' or ``squeezed'' parallel to the rotation axis without changing its moment of inertia. This is known as the stretch rule, and it can be used to simplify geometry when finding the moment of inertia. For example, we saw in §B.4.4 that the moment of inertia of a pointmass a distance from the axis of rotation is given by . By the stretch rule, the same applies to an ideal rod of mass parallel to and distance from the axis of rotation. Note that mass can be also be ``stretched'' along the circle of rotation without changing the moment of inertia for the mass about that axis. Thus, the point mass can be stretched out to form a mass ring at radius about the axis of rotation without changing its moment of inertia about that axis. Similarly, the ideal rod of the previous paragraph can be stretched tangentially to form a cylinder of radius and mass , with its axis of symmetry coincident with the axis of rotation. In all of these examples, the moment of inertia is about the axis of rotation.Area Moment of Inertia
The area moment of inertia is the second moment of an area around a given axis:Radius of Gyration
For a planar distribution of mass rotating about some axis in the plane of the mass, the radius of gyration is the distance from the axis that all mass can be concentrated to obtain the same mass moment of inertia. Thus, the radius of gyration is the ``equivalent distance'' of the mass from the axis of rotation. In this context, gyration can be defined as rotation of a planar region about some axis lying in the plane. For a bar crosssection with area , the radius of gyration is given bywhere is the area moment of inertia (§B.4.8) of the crosssection about a given axis of rotation lying in the plane of the crosssection (usually passing through its centroid):
Rectangular CrossSection
For a rectangular crosssection of height and width , area , the area moment of inertia about the horizontal midline is given byCircular CrossSection
For a circular crosssection of radius , Eq.(B.11) tells us that the squared radius of gyration about any line passing through the center of the crosssection is given byTwo Masses Connected by a Rod
As an introduction to the decomposition of rigidbody motion into translational and rotational components, consider the simple system shown in Fig.B.5. The excitation force density^{B.15} can be applied anywhere between and along the connecting rod. We will deliver a vertical impulse of momentum to the mass on the right, and show, among other observations, that the total kinetic energy is split equally into (1) the rotational kinetic energy about the center of mass, and (2) the translational kinetic energy of the total mass, treated as being located at the center of mass. This is accomplished by defining a new frame of reference (i.e., a moving coordinate system) that has its origin at the center of mass. First, note that the drivingpoint impedance (§7.1) ``seen'' by the driving force varies as a function of . At , The excitation sees a ``point mass'' , and no rotation is excited by the force (by symmetry). At , on the other hand, the excitation only sees mass at time 0, because the vertical motion of either pointmass initially only rotates the other pointmass via the massless connecting rod. Thus, an observation we can make right away is that the driving point impedance seen by depends on the striking point and, away from , it depends on time as well. To avoid dealing with a timevarying drivingpoint impedance, we will use an impulsive force input at time . Since momentum is the timeintegral of force ( ), our excitation will be a unit momentum transferred to the twomass system at time 0.Striking the Rod in the Middle
First, consider . That is, we apply an upward unitforce impulse at time 0 in the middle of the rod. The total momentum delivered in the neighborhood of and is obtained by integrating the applied force density with respect to time and position:Striking One of the Masses
Now let . That is, we apply an impulse of vertical momentum to the mass on the right at time 0. In this case, the unit of vertical momentum is transferred entirely to the mass on the right, so thatHowever, after time zero, things get more complicated, because the mass on the left gets dragged into a rotation about the center of mass. To simplify ongoing analysis, we can define a bodyfixed frame of reference^{B.16} having its origin at the center of mass. Let denote a velocity in this frame. Since the velocity of the center of mass is , we can convert any velocity in the bodyfixed frame to a velocity in the original frame by adding to it, viz.,
Angular Velocity Vector
When working with rotations, it is convenient to define the angularvelocity vector as a vector pointing along the axis of rotation. There are two directions we could choose from, so we pick the one corresponding to the righthand rule, i.e., when the fingers of the right hand curl in the direction of the rotation, the thumb points in the direction of the angular velocity vector.^{B.18} The length should obviously equal the angular velocity . It is convenient also to work with a unitlength variant . As introduced in Eq.(B.8) above, the mass moment of inertia is given by where is the distance from the (instantaneous) axis of rotation to the mass located at . In terms of the angularvelocity vector , we can write this as (see Fig.B.6)where
Vector Cross Product
The vector cross product (or simply vector product, as opposed to the scalar product (which is also called the dot product, or inner product)) is commonly used in vector calculusa basic mathematical toolset used in mechanics [270,258], acoustics [349], electromagnetism [356], quantum mechanics, and more. It can be defined symbolically in the form of a matrix determinant:^{B.19}where denote the unit vectors in . The crossproduct is a vector in 3D that is orthogonal to the plane spanned by and , and is oriented positively according to the righthand rule.^{B.20} The second and third lines of Eq.(B.15) make it clear that . This is one example of a host of identities that one learns in vector calculus and its applications.
CrossProduct Magnitude
It is a straightforward exercise to show that the crossproduct magnitude is equal to the product of the vector lengths times the sine of the angle between them:^{B.21}where
Mass Moment of Inertia as a Cross Product
In Eq.(B.14) above, the mass moment of inertia was expressed in terms of orthogonal projection as , where . In terms of the vector cross product, we can now express it asTangential Velocity as a Cross Product
Referring again to Fig.B.4, we can write the tangential velocity vector as a vector cross product of the angularvelocity vector (§B.4.11) and the position vector :To see this, let's first check its direction and then its magnitude. By the righthand rule, points up out of the page in Fig.B.4. Crossing that with , again by the righthand rule, produces a tangential velocity vector pointing as shown in the figure. So, the direction is correct. Now, the magnitude: Since and are mutually orthogonal, the angle between them is , so that, by Eq.(B.16),
Angular Momentum
The angular momentum of a mass rotating in a circle of radius with angular velocity (rad/s), is defined byRelation of Angular to Linear Momentum
Recall (§B.3) that the momentum of a mass traveling with velocity in a straight line is given byThus, the angular momentum is times the linear momentum . Linear momentum can be viewed as a renormalized special case of angular momentum in which the radius of rotation goes to infinity.
Angular Momentum Vector
Like linear momentum, angular momentum is fundamentally a vector in . The definition of the previous section suffices when the direction does not change, in which case we can focus only on its magnitude . More generally, let denote the 3space coordinates of a pointmass , and let denote its velocity in . Then the instantaneous angular momentum vector of the mass relative to the origin (not necessarily rotating about a fixed axis) is given bywhere denotes the vector cross product, discussed in §B.4.12 above. The identity was discussed at Eq.(B.17). For the special case in which is orthogonal to , as in Fig.B.4, we have that points, by the righthand rule, in the direction of the angular velocity vector (up out of the page), which is satisfying. Furthermore, its magnitude is given by
Angular Momentum Vector in Matrix Form
The two crossproducts in Eq.(B.19) can be written out with the help of the vector analysis identity^{B.23}where
The matrix is the Cartesian representation of the mass moment of inertia tensor, which will be explored further in §B.4.15 below. The vector angular momentum of a rigid body is obtained by summing the angular momentum of its constituent mass particles. Thus,
Mass Moment of Inertia Tensor
As derived in the previous section, the moment of inertia tensor, in 3D Cartesian coordinates, is a threebythree matrix that can be multiplied by any angularvelocity vector to produce the corresponding angular momentum vector for either a point mass or a rigid mass distribution. Note that the origin of the angularvelocity vector is always fixed at in the space (typically located at the center of mass). Therefore, the moment of inertia tensor is defined relative to that origin. The moment of inertia tensor can similarly be used to compute the mass moment of inertia for any normalized angular velocity vector asSince rotational energy is defined as (see Eq.(B.7)), multiplying Eq.(B.22) by gives the following expression for the rotational kinetic energy in terms of the moment of inertia tensor:
We can show Eq.(B.22) starting from Eq.(B.14). For a pointmass located at , we have
which agrees with Eq.(B.20). Thus we have derived the moment of inertia in terms of the moment of inertia tensor and the normalized angular velocity for a pointmass at . For a collection of masses located at , we simply sum over their masses to add up the moments of inertia:
Simple Example
Consider a mass at . Then the mass moment of inertia tensor isExample with Coupled Rotations
Now let the mass be located at so thatOffDiagonal Terms in Moment of Inertia Tensor
This all makes sense, but what about those offdiagonal terms in ? Consider the vector angular momentum (§B.4.14):Principal Axes of Rotation
A principal axis of rotation (or principal direction) is an eigenvector of the mass moment of inertia tensor (introduced in the previous section) defined relative to some point (typically the center of mass). The corresponding eigenvalues are called the principal moments of inertia. Because the moment of inertia tensor is defined relative to the point in the space, the principal axes all pass through that point (usually the center of mass). As derived above (§B.4.14), the angular momentum vector is given by the moment of inertia tensor times the angularvelocity vector:Positive Definiteness of the Moment of Inertia Tensor
From the form of the moment of inertia tensor introduced in Eq.(B.24)
Rotational Kinetic Energy Revisited
If a pointmass is located at
and is rotating about an
axisofrotation
with angular velocity , then the
distance from the rotation axis to the mass is
,
or, in terms of the vector cross product,
. The tangential velocity of the mass is
then , so that the kinetic energy can be expressed as
(cf. Eq.(B.23))
where
Torque
as depicted in Fig.B.7. The moment arm is the distance from the applied force to the point being twisted. For example, in the case of a wrench turning a bolt, is the force applied at the end of the wrench by one's hand, orthogonal to the wrench, while the moment arm is the length of the wrench. Doubling the length of the wrench doubles the torque. This is an example of leverage. When is increased, a given twisting angle is spread out over a larger arc length , thereby reducing the tangential force required to assert a given torque . For more general applied forces , we may compute the tangential component by projecting onto the tangent direction. More precisely, the torque about the origin applied at a point may be defined by
where is the applied force (at ) and denotes the cross product, introduced above in §B.4.12. Note that the torque vector is orthogonal to both the lever arm and the tangentialforce direction. It thus points in the direction of the angular velocity vector (along the axis of rotation). The torque magnitude is
Newton's Second Law for Rotations
The rotational version of Newton's law iswhere denotes the angular acceleration. As in the previous section, is torque (tangential force times a moment arm ), and is the mass moment of inertia. Thus, the net applied torque equals the time derivative of angular momentum , just as force equals the timederivative of linear momentum :
Equations of Motion for Rigid Bodies
We are now ready to write down the general equations of motion for rigid bodies in terms of for the center of mass and for the rotation of the body about its center of mass. As discussed above, it is useful to decompose the motion of a rigid body into (1)
 the linear velocity of its center of mass, and
 (2)
 its angular velocity about its center of mass.
where is the vector torque defined in Eq.(B.27), is the angular momentum, is the mass moment of inertia tensor, and is the angular velocity of the rigid body about its center of mass. Note that if the center of mass is moving, we are in a moving coordinate system moving with the center of mass (see next section). We may call the intrinsic momentum of the rigid body, i.e., that in a coordinate system moving with the center of the mass. We will translate this to the nonmoving coordinate system in §B.4.20 below. The driving torque is given by the resultant moment of the external forces, using Eq.(B.27) for each external force to obtain its contribution to the total moment. In other words, the external moments (tangential forces times moment arms) sum up for the net torque just like the radial force components summed to produce the net driving force on the center of mass.
BodyFixed and SpaceFixed Frames of Reference
Rotation is always about some (instantaneous) axis of rotation that is free to change over time. It is convenient to express rotations in a coordinate system having its origin ( ) located at the centerofmass of the rigid body (§B.4.1), and its coordinate axes aligned along the principal directions for the body (§B.4.16). This bodyfixed frame then moves within a stationary spacefixed frame (or ``star frame''). In Eq.(B.29) above, we wrote down Newton's second law for angular motion in the bodyfixed frame, i.e., the coordinate system having its origin at the center of mass. Furthermore, it is simplest ( is diagonal) when its axes lie along principal directions (§B.4.16). As an example of a local bodyfixed coordinate system, consider a spinning top. In the bodyfixed frame, the ``vertical'' axis coincides with the top's axis of rotation (spin). As the top loses rotational kinetic energy due to friction, the top's rotationaxis precesses around a circle, as observed in the spacefixed frame. The other two bodyfixed axes can be chosen as any two mutually orthogonal axes intersecting each other (and the spin axis) at the center of mass, and lying in the plane orthogonal to the spin axis. The spacefixed frame is of course that of the outside observer's inertial frame^{B.28}in which the top is spinning.Angular Motion in the SpaceFixed Frame
Let's now consider angular motion in the presence of linear motion of the center of mass. In general, we have [270]Euler's Equations for Rotations in the BodyFixed Frame
Suppose now that the bodyfixed frame is rotating in the spacefixed frame with angular velocity . Then the total torque on the rigid body becomes [270]Similarly, the total external forces on the center of mass become
Examples
For a uniform sphere, the crossterms disappear and the moments of inertia are all the same, leaving , for . Since any three orthogonal vectors can serve as eigenvectors of the moment of inertia tensor, we have that, for a uniform sphere, any three orthogonal axes can be chosen as principal axes. For a cylinder that is not spinning about its axis, we similarly obtain two uncoupled equations , for , given (no spin). Note, however, that if we replace the circular crosssection of the cylinder by an ellipse, then and there is a coupling term that drives (unless happens to cancel it).Properties of Elastic Solids
Young's Modulus
Young's modulus can be thought of as the spring constant for solids. Consider an ideal rod (or bar) of length and crosssectional area . Suppose we apply a force to the face of area , causing a displacement along the axis of the rod. Then Young's modulus is given byYoung's Modulus as a Spring Constant
Recall (§B.1.3) that Hooke's Law defines a spring constant as the applied force divided by the spring displacement , or . An elastic solid can be viewed as a bundle of ideal springs. Consider, for example, an ideal bar (a rectangular solid in which one dimension, usually its longest, is designated its length ), and consider compression by along the length dimension. The length of each spring in the bundle is the length of the bar, so that each spring constant must be inversely proportional to ; in particular, each doubling of length doubles the length of each ``spring'' in the bundle, and therefore halves its stiffness. As a result, it is useful to normalize displacement by length and use relative displacement . We need displacement per unit length because we have a constant spring compliance per unit length. The number of springs in parallel is proportional to the crosssectional area of the bar. Therefore, the force applied to each spring is proportional to the total applied force divided by the crosssectional area . Thus, Hooke's law for each spring in the bundle can be writtenString Tension
The tension of a vibrating string is the force used to stretch it. It is therefore directed along the axis of the string. A force must be applied at the endpoint on the right, and a force is applied at the endpoint on the left. Each point interior to the string is pulled equally to the left and right, i.e., the net force on an interior point is . (A nonzero force on a massless point would produce an infinite acceleration.) If the crosssectional area of the string is , then the tension is given by the stress on the string times .Wave Equation for the Vibrating String
Consider an elastic string under tension which is at rest along the dimension. Let , , and denote the unit vectors in the , , and directions, respectively. When a wave is present, a point originally at along the string is displaced to some point specified by the displacement vectorwhere has been canceled on both sides of the equation. Note that no approximations have been made so far. The next step is to express the force in terms of the tension of the string at rest, the elastic constant of the string, and geometrical factors. The displaced string element is the vector
having magnitude
NonStiff String
Let's now assume the string is perfectly flexible (zero stiffness) so that the direction of the force vector is given by the unit vector tangent to the string. (To accommodate stiffness, it would be necessary to include a force component at right angles to the string which depends on the curvature and stiffness of the string.) The magnitude of at any position is the rest tension plus the incremental tension needed to stretch it the fractional amountwhere no geometrical limitations have yet been placed on the magnitude of and , other than to prevent the string from being stretched beyond its elastic limit. The four equations (B.31) through (B.35) can be combined into a single vector wave equation that expresses the propagation of waves on the string having three displacement components. This differential equation is nonlinear, so that superposition no longer holds. Furthermore, the three displacement components of the wave are coupled together at all points along the string, so that the wave equation is no longer separable into three independent 1D wave equations. To obtain a linear, separable wave equation, it is necessary to assume that the strains , , and be small compared with unity. This is the same assumption ( ) necessary to derive the usual wave equation for transverse vibrations only in the  plane. When (B.35) is expanded into a Taylor series in the strains, and when only the firstorder terms are retained, we obtain
(B.36) 
This is the linearized wave equation for the string, based only on the assumptions of elasticity of the string, and strain magnitudes much less than unity. Using this linearized equation for the force , it is found that (B.31) separates into the three wave equations
(B.37)  
(B.38)  
(B.39) 
where is the longitudinal wave velocity, and is the transverse wave velocity. In summary, the two transverse wave components and the longitudinal component may be considered independent (i.e., ``superposition'' holds with respect to vibrations in these three dimensions of vibration) provided powers higher than 1 of the strains (relative displacement) can be neglected, i.e.,
and
Wave Momentum
The physical forward momentum carried by a transverse wave along a string is conveyed by a secondary longitudinal wave [391]. A less simplified wave equation which supports longitudinal wave momentum is given by [391, Eqns. 38ab](B.40)  
(B.41)  
(B.42) 
where and denote longitudinal and transverse displacement, respectively, and the commonly used ``dot'' and ``prime'' notation for partial derivatives has been introduced, e.g.,
(B.43)  
(B.44) 
(See also Eq.(C.1).) We see that the term in the first equation above provides a mechanism for transverse waves to ``drive'' the generation of longitudinal waves. This coupling cannot be neglected if momentum effects are desired. Physically, the rising edge of a transverse wave generates a longitudinal displacement in the direction of wave travel that propagates ahead at a much higher speed (typically an order of magnitude faster). The falling edge of the transverse wave then cancels this forward displacement as it passes by. See [391] for further details (including computer simulations).
Properties of Gases
Particle Velocity of a Gas
The particle velocity of a gas flow at any point can be defined as the average velocity (in meters per second, m/s) of the air molecules passing through a plane cutting orthogonal to the flow. The term ``velocity'' in this book, when referring to air, means ``particle velocity.'' It is common in acoustics to denote particle velocity by lowercase .Volume Velocity of a Gas
The volume velocity of a gas flow is defined as particle velocity times the crosssectional area of the flow, orPressure is Confined Kinetic Energy
According the kinetic theory of ideal gases [180], air pressure can be defined as the average momentum transfer per unit area per unit time due to molecular collisions between a confined gas and its boundary. Using Newton's second law, this pressure can be shown to be given by one third of the average kinetic energy of molecules in the gas.Proof: This is a classical result from the kinetic theory of gases [180]. Let be the total mass of a gas confined to a rectangular volume , where is the area of one side and the distance to the opposite side. Let denote the average molecule velocity in the direction. Then the total net molecular momentum in the direction is given by . Suppose the momentum is directed against a face of area . A rigidwall elastic collision by a mass traveling into the wall at velocity imparts a momentum of magnitude to the wall (because the momentum of the mass is changed from to , and momentum is conserved). The average momentumtransfer per unit area is therefore at any instant in time. To obtain the definition of pressure, we need only multiply by the average collision rate, which is given by . That is, the average velocity divided by the roundtrip distance along the dimension gives the collision rate at either wall bounding the dimension. Thus, we obtain
Bernoulli Equation
In an ideal inviscid, incompressible flow, we have, by conservation of energy,
constant
where
Bernoulli Effect
The Bernoulli effect provides that, when a gas such as air flows, its pressure drops. This is the basis for how aircraft wings work: The crosssectional shape of the wing, called an aerofoil (or airfoil), forces air to follow a longer path over the top of the wing, thereby speeding it up and creating a net upward force called lift. Figure B.8 illustrates the Bernoulli effect for the case of a reservoir at constant pressure (``mouth pressure'') driving an acoustic tube. Any flow inside the ``mouth'' is neglected. Within the acoustic channel, there is a flow with constant particle velocity . To conserve energy, the pressure within the acoustic channel must drop down to . That is, the flow kinetic energy subtracts from the pressure kinetic energy within the channel. For a more detailed derivation of the Bernoulli effect, see, e.g., [179]. Further discussion of its relevance in musical acoustics is given in [144,197].Air Jets
Referring again to Fig.B.8, the gas flow exiting the acoustic tube is shown as forming a jet. The jet ``carries its own pressure'' until it dissipates in some form, such as any combination of the following: heat (now allowing for ``friction'' in the flow),
 vortices (angular momentum),
 radiation (sound waves), or
 pressure recovery.
Acoustic Intensity
Acoustic intensity may be defined byAcoustic Energy Density
The two forms of energy in a wave are kinetic and potential. Denoting them at a particular time and position by and , respectively, we can write them in terms of velocity and wave impedance as follows:Energy Decay through Lossy Boundaries
Since the acoustic energy density is the energy per unit volume in a 3D sound field, it follows that the total energy of the field is given by integrating over the volume:Ideal Gas Law
The ideal gas law can be written aswhere
Isothermal versus Isentropic
If air compression/expansion were isothermal (constant temperature ), then, according to the ideal gas law , the pressure would simply be proportional to density . It turns out, however, that heat diffusion is much slower than audio acoustic vibrations. As a result, air compression/expansion is much closer to isentropic (constant entropy ) in normal acoustic situations. (An isentropic process is also called a reversible adiabatic process.) This means that when air is compressed by shrinking its volume , for example, not only does the pressure increase (§B.7.3), but the temperature increases as well (as quantified in the next section). In a constantentropy compression/expansion, temperature changes are not given time to diffuse away to thermal equilibrium. Instead, they remain largely frozen in place. Compressing air heats it up, and relaxing the compression cools it back down.Adiabatic Gas Constant
The relative amount of compression/expansion energy that goes into temperature versus pressure can be characterized by the heat capacity ratiowhere for dry air at normal temperatures. Thus, if a volume of ideal gas is changed from to , the pressure change is given by
Heat Capacity of Ideal Gases
In statistical thermodynamics [175,138], it is derived that each molecular degree of freedom contributes to the molar heat capacity of an ideal gas, where again is the ideal gas constant. An ideal monatomic gas molecule (negligible spin) has only three degrees of freedom: its kinetic energy in the three spatial dimensions. Therefore, . This means we expectSpeed of Sound in Air
The speed of sound in a gas depends primarily on the temperature, and can be estimated using the following formula from the kinetic theory of gases:^{B.33}Air Absorption
This section provides some further details regarding acoustic air absorption [318]. For a plane wave, the decline of acoustic intensity as a function of propagation distance is given by


There is also a (weaker) dependence of air absorption on temperature [183]. Theoretical models of energy loss in a gas are developed in Morse and Ingard [318, pp. 270285]. Energy loss is caused by viscosity, thermal diffusion, rotational relaxation, vibration relaxation, and boundary losses (losses due to heat conduction and viscosity at a wall or other acoustic boundary). Boundary losses normally dominate by several orders of magnitude, but in resonant modes, which have nodes along the boundaries, interior losses dominate, especially for polyatomic gases such as air.^{B.34} For air having moderate amounts of water vapor () and/or carbon dioxide (), the loss and dispersion due to and vibration relaxation hysteresis becomes the largest factor [318, p. 300]. The vibration here is that of the molecule itself, accumulated over the course of many collisions with other molecules. In this context, a diatomic molecule may be modeled as two masses connected by an ideal spring. Energy stored in molecular vibration typically dominates over that stored in molecular rotation, for polyatomic gas molecules [318, p. 300]. Thus, vibration relaxation hysteresis is a loss mechanism that converts wave energy into heat. In a resonant mode, the attenuation per wavelength due to vibration relaxation is greatest when the sinusoidal period (of the resonance) is equal to times the timeconstant for vibrationrelaxation. The relaxation timeconstant for oxygen is on the order of one millisecond. The presence of water vapor (or other impurities) decreases the vibration relaxation time, yielding loss maxima at frequencies above 1000 rad/sec. The energy loss approaches zero as the frequency goes to infinity (wavelength to zero). Under these conditions, the speed of sound is approximately that of dry air below the maximumloss frequency, and somewhat higher above. Thus, the humidity level changes the dispersion crossover frequency of the air in a resonant mode.
Wave Equation in Higher Dimensions
The wave equation in 1D, 2D, or 3D may be written aswhere, in 3D, denotes the amplitude of the wave at time and position , and
Plane Waves in Air
Figure B.9 shows a 2D crosssection of a snapshot (in time) of the sinusoidal plane waveVector Wavenumber
Mathematically, a sinusoidal plane wave, as in Fig.B.9 or Fig.B.10, can be written aswhere p(t,x) is the pressure at time (seconds) and position (3D Euclidean space). The amplitude , phase , and radian frequency are ordinary sinusoid parameters [451], and is the vector wavenumber:
 (unit) vector of direction cosines
 (scalar) wavenumber along travel direction
 wavenumber along the travel direction in its magnitude
 travel direction in its orientation
constant
where , , and are real constants, and ,
, and are 3D spatial coordinates. Thus, the set of all points
yielding the same value
define a
plane of constant phase in .
As we know from elementary vector calculus, the direction of maximum
phase advance is given by the gradient of the phase
:
Solving the 2D Wave Equation
Since solving the wave equation in 2D has all the essential features of the 3D case, we will look at the 2D case in this section. Specializing Eq.(B.47) to 2D, the 2D wave equation may be written aswhich is a standing wave along .
2D Boundary Conditions
We often wish to find solutions of the 2D wave equation that obey certain known boundary conditions. An example is transverse waves on an ideal elastic membrane, rigidly clamped on its boundary to form a rectangle with dimensions meters. Similar to the derivation of Eq.(B.49), we can subtract the second sinusoidal traveling wave from the first to yieldwhere
3D Sound
The mathematics of 3D sound is quite elementary, as we will see below. The hard part of the theory of practical systems typically lies in the mathematical approximation to the ideal case. Examples include Ambisonics [158] and wave field synthesis [49]. Consider a point source at position . Then the acoustic complex amplitude at position is given byNext Section:
Digital Waveguide Theory
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