Chapter

Waves and Sound

Blinn College - Physics 1401 - Terry Honan

. - Mechanical Waves

A mechanical wave is a disturbance in a medium that propagates through the medium. A wave carries energy as it propagates. We will generically label this disturbance by u. Consider a stretched string. Take x to be the position along the string and t is time. The equilibrium position of the string is its relaxed position. The disturbance y (u=y for a string) is the perpendicular distance of a point on the string from equilibrium. When a pulse is put in the string it maintains its shape and travels the length of the string at a fixed speed of v=(Sqrt[(Subscript[F, T])/μ]), where (Subscript[F, T]) is the tension in the string and μ is the linear density of the string, its mass per length.

A wave is said to be transverse when the direction of the disturbance is perpendicular to the direction of propagation. Waves on a string are examples of transverse waves. Electromagnetic waves are also transverse; in the electromagnetic case the disturbance is the electric field, which is perpendicular to the direction of propagation.  Note that electromagnetic waves are not mechanical waves; there is no medium and it can propagate in a vacuum. There is a plane of possible directions perpendicular to a direction of propagation. Choosing such a direction is choosing a polarization.  Transverse waves can be polarized.

Now consider a stretched spring. As with the string, perpendicular pulse can also be put into this stretched spring and that will propagate as a transverse wave. If a compression pulse is put into the spring then that pulse will also propagate as a wave. Here a point on the spring moves back and forth a distance u from equilibrium but parallel to the direction of propagation x. When the disturbance is parallel to the direction of propagation we call the wave longitudinal.  Longitudinal waves cannot be polarized.

Another example of a longitudinal wave is sound in a fluid.  Here the molecules move back and forth parallel to the direction of wave propagation.  We can view the disturbance of sound waves either in terms of displacement or in terms of pressure.  Sound waves travel at the speed of sound.  This varies with temperature; at 20 °C it is 343 m/s

({{Wave Type, Disturbance - u, ({{Transverse or}, {Longitudinal}}), Wave Speed}, {Waves on a string, u=y(x,t), Transverse, v=(Sqrt[((Subscript[F, T])/μ)])}, {({{Electromagnetic}, {Waves}}), ({{u=E                  }, {       =Electric field}}), Transverse, ({{v=c                         }, {   =3.00×(10^8)(m/s)}})}, {({{Compression}, {Waves on a Spring}}), u=Parallel disp., Longitudinal, no formula given}, {({{Sound Waves}, {in a Fluid}}), ({{u=Pressure  or    }, {u=Displacement}}), Longitudinal, ({{v=(Subscript[v, sound])                 }, {  =343(m/s) (in air)}})}})

. - Waves in One Dimension

Left- and Right-moving Pulses

Consider one-dimensional wave with a disturbance u(x,t) of the following form

u(x,t)=f(x-v t)+g(x+v t),

where f and g are arbitrary functions. To understand this, consider the function  u(x)=f(x). If we shift this by a in the positive direction we get  u(x)=f(x-a).

We can now see that  u(x,t)=f(x-v t) describes a pulse of arbitrary shape  u(x)=f(x) moving in the positive direction with speed v.  u(x,t)=g(x+v t) corresponds to a pulse of a different arbitrary shape moving in the opposite direction at the same speed.

Sinusoidal Waves

We often consider waves where the shape of the pulse f (or g) are sinusoidal.

f(x)=A cos(k x)

A is called the amplitude. k is called the wave number; this is related to the wavelength λ, which is the spatial period of the function. Since the period of sine is 2 π and the period of the function f is λ we get  k λ=2 π or

k=((2 π)/λ).

If we take this function f and move it in the positive or negative direction we get f(x∓v t)=A cos[k (x∓v t)] or

u(x,t)=A cos(k x∓ω t),

where the angular frequency ω and wave number are related to the wave speed by k v=ω.

If we choose some point on the string (Subscript[x, 0]) then at that position moves as

A cos(ω t+ϕ).

This is our expression for simple harmonic motion.  Since the angular frequency is related to the frequency by ω=2 π f the wave speed can also be written in terms of the frequency and wavelength.

v=(ω/k)=f λ

Dispersion and Generalized Waves

With the wave examples we are discussing in this chapter, the wave speed is independent of frequency. More generally, we can have different physical situations which are described by similar mathematics where we still get sinusoidal waves. These generalized waves are dispersive, meaning that different frequency waves will have different speeds. Although we have sinusoidal waves, we no longer have simple pulses that maintain their shape as solutions. If we begin with a pulse then the pulse will spread out with time.

Light in a vacuum is a three-dimensional wave that is not dispersive. Light passing through a medium does have dispersion. A consequence of different frequencies, colors, having different speeds is the splitting of light into its spectrum by a prism.

Another example of dispersive waves is surface water waves; these, it turns out, are a mixture of longitudinal and transverse displacements. In quantum mechanics matter waves are dispersive.

. - Waves on a String and Power

A rope or a string can be described as having a linear density μ. This is the mass per length of the rope or string.

μ=(m/L)=(mass/length)   (linear density)

A heavy rope has a large linear density and a light string has a small one. Along a rope of string the tension will stay constant. We will use (Subscript[F, T]) to denote tension, instead of just T to avoid confusion with the period of harmonic motion, which recall is related to the frequency T=1/f. The speed of waves on a string can be shown to be

v=(Sqrt[((Subscript[F, T])/μ)]).

Waves carry energy. When a sinusoidal wave travels down a string the flow of energy is constant.  If some quantity of energy flows in a time then the energy per time or power is constant.  Now consider some point on the string.  If the wave is moving in some direction at the speed v, then the energy moves past the point at the same rate.  In a small time Δ t all the energy in an small segment of width Δ x=v Δ t will pass the point.

The energy of a particle in simple harmonic motion satisfies

E=(1/2)m (Subscript[v, max]^2)

Since a point on the string moves in simple harmonic motion we can apply this formula to Δ x.  The small mass of the segment is

Δ m=μ Δ x=μ v Δ t

and the small energy in the segment is

Δ E=(1/2)Δ m (Subscript[v, max]^2)=(1/2)μ v Δ t  (Subscript[v, max]^2).

The power is given by 𝒫=Δ E/Δ t.  The maximum speed of a point in simple harmonic motion is (Subscript[v, max])=ω A.  It follow that the power travelling down a string due to a sinusoidal wave is

𝒫=(1/2)μ (A^2) (ω^2) v.

Example . - Waves on a Steel Wire

y(x,t)=(0.020`(QuantityUnit)) sin[(105 (((QuantityUnit))^(-1))) t+(3.0 (((QuantityUnit))^(-1))) x]

Solution

A=0.020`(QuantityUnit) ,  ω=105 (((QuantityUnit))^(-1)),  k=3.0 (((QuantityUnit))^(-1))  and   μ=0.014226(QuantityUnit)/(QuantityUnit)

f=(ω/(2π))=16.7(QuantityUnit),  λ=((2π)/k)=2.09(QuantityUnit)   and  v=(ω/k)=35(((QuantityUnit))/((QuantityUnit)))=f λ

Solution

(Subscript[v, max])=ω A=2.1(QuantityUnit)/(QuantityUnit) .

Solution

v=(Sqrt[((Subscript[F, T])/μ)])   ⟹   (Subscript[F, T])=μ (v^2)=17.4(QuantityUnit)

Solution

𝒫=(1/2)μ (A^2) (ω^2) v=1.10(QuantityUnit)

. - Waves in Three Dimensions

Plane Waves and Spherical Waves

In three dimensions we define a surface of constant phase, typically the crest of the wave, to be a wave front.  An important notion is that of a plane wave; take the wave to move in the positive x-direction with the disturbance u being uniform along the yz-plane.  This turns our disturbance from a genuinely three dimensional function u(x,y,z,t) into being a function of only x, spatially.

u(x,t)=A cos(k x-ω t)     (Plane Wave)

Wave fronts correspond to planes parallel to the yz-plane, separated by one wavelength and moving in the positive x-direction at speed v.

A point source produces a spherical wave.  We can write the disturbance u as a function of r, the distance form the source and time t.

u(r,t)=A(r) cos(k r-ω t)     (Spherical Wave)

The wave fronts are concentric spheres separated by λ.  The waves fronts move away from the source at the speed v.

Intensity

We saw in the discussion of waves on a string that the average power transmitted down the string was proportional to the wave amplitude squared, (Subscript[𝒫, ave])∝(Amplitude^2).  For a wave that moves though three dimensions, what is analogous to power in the string case is the intensity, which we define as power per area.

I=((Subscript[𝒫, ave])/A)=((Subscript[E, ave])/(A Δt))    (Intensity=(Power/Area)=((Average Energy)/(Area×time)))

The intensity will always be proportional to the amplitude squared.

With a plane wave the area stays uniform so the intensity is uniform.  For a spherical wave the relevant area is the surface area of a sphere, where A=4 π (r^2).  It follows that the intensity varies with r, the distance form the source, by an inverse square law.

I=((Subscript[𝒫, ave])/(4 π (r^2)))

(Subscript[𝒫, ave]) is the total power output of the source, the rate at which it emits energy.

Sound Waves in a Fluid

Displacement and Pressure

We now consider sound waves in a fluid (a liquid or a gas.)  Our one-dimensional model consists of a plane wave, which is three dimensional but where nothing varies along the yz-plane, so it is a function of the one spatial variable x.  Alternatively, we can describe it as a fluid in a pipe with frictionless walls, making the cross-sectional area A unimportant.

We may view sound waves in a fluid as pressure waves or as displacement waves.  Consider the motion of particles (molecules) relative to their equilibrium positions in a fluid.  The displacement s is the position of particles relative to their equilibrium position.  In our one-dimensional model s(x,t) is the longitudinal displacement of particles as a function of position x and time t.  Longitudinal means that the displacement is along the direction of propagation, the ±x direction.  Similarly, we can write the pressure as a function of x and t, P(x,t).  By pressure P we mean the gauge pressure, which is the difference between (Subscript[P, absolute]), the absolute pressure, and (Subscript[P, 0]), the atmospheric pressure or the ambient pressure in the fluid.  Although the absolute pressure cannot be negative, the gauge pressure P can.

P=(Subscript[P, gauge])=(Subscript[P, absolute])-(Subscript[P, 0])

Intensity and Sound Level

We can perceive sound over a wide range of intensities; this makes a logarithmic scale convenient for measuring loudness.  We will call our measure of loudness β, the sound level, and measure it in decibels dB.  If we define

(Subscript[I, 0])=(10^(-12)) (W/(m^2))

as the threshold of hearing, then we can define the sound level (or decibel level) β, in dB, by:

β=(10 dB) log((I/(Subscript[I, 0]))) .

Example . - A Very Loud Loudspeaker

Solution

110 dB=(10 dB) log((I/(Subscript[I, 0])))    ⟹  11=log((I/(Subscript[I, 0])))    ⟹ (10^11)=(I/(Subscript[I, 0]))   ⟹ I=(10^11)(Subscript[I, 0])=(10^11)×(10^(-12)) (W/(m^2))=0.1 (W/(m^2))

Solution

r=6(QuantityUnit)  and  I=0.1 (W/(m^2))=((Subscript[𝒫, ave])/(4 π (r^2)))  ⟹  (Subscript[𝒫, ave])=4 π (r^2) I==(452.4(((QuantityUnit))^2)) I=45.2(QuantityUnit)

The Doppler Effect and Shock Waves

When a train is moving toward you with its whistle blowing the observed pitch of the whistle, its frequency, increases.  When moving away its pitch decreases.  This is known as the Doppler effect; it is a general property of waves but we will discuss it in the context of sound waves. Take the speed of sound to be v; this is the speed of sound with respect to the stationary atmosphere.  Consider (Subscript[v, s]) and (Subscript[v, o]) to be the approaching velocities of the source, producing the sound, and the observer, listening to it.  By “approaching velocities” we mean we will take their signs to be positive when moving toward the other. If the source or observer is moving away from the other, we will take the sign of its velocity to be negative.

λ' is the wavelength as heard by the observer. The wavelength at the observer λ' is the distance between wave fronts; it is smaller than λ=v/f by (Subscript[v, s]) T, λ'=λ-(Subscript[v, s]) T, where T=1/f is the period.  λ' is smaller since the second wavefront left the source after it moved (Subscript[v, s]) T after the first wavefront left the source. The wave moves at speed v relative to the stationary atmosphere, so relative to the moving observer the wave speed is v+(Subscript[v, o]).

λ'=λ-(Subscript[v, s])T=(v/f)-((Subscript[v, s])/f)=((v-(Subscript[v, s]))/f)  and  v'=v+(Subscript[v, o])

Combining these two expressions we can find the wave frequency as heard by the observer.

f'=((v')/(λ'))=((v+(Subscript[v, o]))/((v-(Subscript[v, s]))/f))=((v+(Subscript[v, o]))/(v-(Subscript[v, s])))f

When the source travels faster than the speed of sound, the wave fronts meet to form a conical shock wave trailing the source.  This shock wave is the sonic boom from a supersonic jet. Another example is the wedge-shaped wake behind a boat; the boat is moving faster than the surface water waves to create a shock wave.  The angle of the shock wave behind the source is called the Mach angle, (Subscript[θ, Mach]).  In a time t the spherical wave front moves by v t  and the source moves by (Subscript[v, s]) t ; the Mach angle is found by simple trigonometry.

sin (Subscript[θ, Mach])=(v/(Subscript[v, s]))

Example . - A Train Whistle

Solution

f=720(QuantityUnit)  and  v=343(QuantityUnit)/(QuantityUnit)

(Subscript[v, s])=+42(((QuantityUnit))/((QuantityUnit)))  and  (Subscript[v, 0])=+29(((QuantityUnit))/((QuantityUnit)))  ⟹  f'=((v+(Subscript[v, o]))/(v-(Subscript[v, s])))f=890(QuantityUnit)

Solution

(Subscript[v, s])=-42(((QuantityUnit))/((QuantityUnit)))  and  (Subscript[v, 0])=-29(((QuantityUnit))/((QuantityUnit)))  ⟹  f'=((v+(Subscript[v, o]))/(v-(Subscript[v, s])))f=587(QuantityUnit)

Solution

(Subscript[v, s])=+42(((QuantityUnit))/((QuantityUnit)))  and  (Subscript[v, 0])=-29(((QuantityUnit))/((QuantityUnit)))  ⟹  f'=((v+(Subscript[v, o]))/(v-(Subscript[v, s])))f=751(QuantityUnit)

Solution

(Subscript[v, s])=-42(((QuantityUnit))/((QuantityUnit)))  and  (Subscript[v, 0])=+29(((QuantityUnit))/((QuantityUnit)))  ⟹  f'=((v+(Subscript[v, o]))/(v-(Subscript[v, s])))f=695(QuantityUnit)

Example . - A Hypersonic Sonic Boom

Solution

sin (Subscript[θ, Mach])=(v/(Subscript[v, s]))  ⟹  (Subscript[θ, Mach])=(sin^(-1))((v/(Subscript[v, s])))=(sin^(-1))((v/(5v)))=(sin^(-1))((1/5))=11.5°

. - Superposition

Superposition in One Dimension

For any waves, including generalized dispersive waves, we have the principle of superposition. This means that the sum of two wave solutions is a solution. For strings in one dimension the only way that we can combine different waves is if they are traveling in opposite directions since the general expression for waves is

u(x,t)=f(x-v t)+g(x+v t)

and two pulses moving in the same direction move at the same speed.  Suppose the f and g solutions are pulses moving toward each other.  Our general solution shows that when a pulse moving in the positive x-direction, f(x-v t), meets a pulse moving in the opposite direction,  g(x+v t), they add where they overlap and move off unchanged by the interaction.

Reflection of Waves

Reflection from a Fixed Point

If one end of a string is held fixed then a pulse will flip while being reflected.

Reflection from a Free End with Tension

Suppose one end of a string is attached to a ring that slides without friction on a post. In this case, the pulse will reflect without being flipped.

Standing Waves

Standing Waves

When a mechanical system, say a mass-spring system, is driven by a periodic force it will oscillate at the driving frequency.  An extended system, like a building, a bridge or a musical instrument will have many natural frequencies of vibration.  We will now study these natural frequencies for several simple cases that involve standing waves.

To describe a standing wave consider a one-dimensional case with two sinusoidal waves with the same amplitude and frequency but moving in opposite directions.

u(x,t)=A cos(k x-ω t)+A cos(k x+ω t)

To add these two waves we will use trig. identities.  Start with the formulas for cosine of the sum and difference of two angles and add the results.

cos(α±β)=cos α cos β∓sin α sin β    ⟹   cos(α+β)+cos(α-β)=2cos α cos β.

Using  α=k x  and  β=ω t  we get our result.

u(x,t)=2A cos(k x) cos(ω t)

The positions where the disturbance u is zero at all times are called nodes.  We refer to the positions where the disturbance has its largest amplitude oscillation as antinodes.

cos(k x)=0      ⟺   nodes

cos(k x)=±1  ⟺   antinodes

Note that the distance between nodes is half a wavelength.  The wavelength and frequency are the same as for the two left and right moving sinusoidal waves, so it follows the we still have f λ=v.

Standing Waves on a String

Suppose a string under tension is fixed at either end.  If you pluck this string it will vibrate in a fairly complex pattern.  This complex pattern can be understood in terms of linear superpositions of some simple vibrational modes, called harmonics.  We will now describe these harmonics.

The string has tension (Subscript[F, T]) and linear density μ so the wave speed is v=(Sqrt[(Subscript[F, T])/μ]).  It is fixed at x=0 and x=L.  We insist on a standing wave solution that is zero at x=0; to do this shift the spatial part of the trig. function to sin(k x), since sin 0=0.  Recall that u=y for a string. A general expression for the standing wave becomes:

y(x,t)=(Subscript[y, max]) sin(k x) cos(ω t+ϕ)

where (Subscript[y, max]) is the amplitude of the resulting standing waves and not the amplitudes of the left-right moving waves in the previous subsection.  For the solution to be zero at L we have sin(k L)=0.  This implies k L=m π.  Since k=2 π/λ we can find the wavelengths of our harmonic modes.

(Subscript[λ, m])=((2 L)/m)  where m=1,2,3,...

Let us get at this expression above in a less mathematical way.  The distance between nodes of the resulting standing wave pattern is half a wavelength.  m of these half-wavelengths fit into L,  so m λ/2=L.

The frequency follows from f λ=v.

(Subscript[f, m])=m (Subscript[f, 1])  where  (Subscript[f, 1])=(v/(2 L)) and  m=1,2,3,…

(Subscript[f, 1]) is the fundamental frequency or the first harmonic.  The higher m values are the higher harmonics.  An instrument will always produce higher harmonics, integer multiples of the fundamentals, but the frequencies we associate with an instrument are the fundamental frequencies.  We tune to correct the fundamental frequencies.  When we tune a stringed instrument we vary the tension in the string, thus varying the wave speed; given the string’s length, we have a unique fundamental frequency for a string.  A piano has a separate string for each note, thus tuning a piano is a very tedious procedure. A standard guitar has just six strings but the player’s fingers moving on the frets creates many effective lengths for each string, producing many different notes.

Example . - Guitars Strings

Solution

(Subscript[f, 1])=329.63(QuantityUnit), L=66(QuantityUnit)=0.66(QuantityUnit)  and  (Subscript[F, T])=43.6(QuantityUnit)

(Subscript[f, 1])=(v/(2 L))  ⟹  v=(Subscript[f, 1])2L=435.11(((QuantityUnit))/((QuantityUnit)))

v=(Sqrt[((Subscript[F, T])/μ)])   ⟹   μ =((Subscript[F, T])/(v^2))=2.30×(10^(-4))(((QuantityUnit))/((QuantityUnit)))

Solution

(Subscript[f, 2])=2(Subscript[f, 1])=659.26(QuantityUnit),  (Subscript[f, 3])=3(Subscript[f, 1])=988.89(QuantityUnit)  and  (Subscript[f, 4])=4(Subscript[f, 1])=1318.5(QuantityUnit)

Solution

(Subscript[f, 1])=82.41(QuantityUnit)  ⟹  (Subscript[f, 2])=2(Subscript[f, 1])=164.82(QuantityUnit),  (Subscript[f, 3])=3(Subscript[f, 1])=247.22(QuantityUnit)  and  (Subscript[f, 4])=4(Subscript[f, 1])=329.63(QuantityUnit)

Solution

(Subscript[f, 1])=(v/(2 L))  and  (Subscript[f, 1]^′)=((v')/(2 L))  ⟹  ((v')/v)= ((Subscript[f, 1]^′)/(Subscript[f, 1]))=(1/4)

v=(Sqrt[((Subscript[F, T])/μ)])   ⟹   (Subscript[F, T])=μ (v^2)  ⟹  ((Subscript[F, T]^′)/(Subscript[F, T]))=(((((v')/v)))^2)=(1/16)

Standing Sound Waves in a Pipe

We will now consider standing waves formed by sound in a pipe.  The boundary conditions at the ends of the pipe depend on whether the end is open or closed.  At an open end the pressure P is fixed at zero at the end, since the pressure we use in our wave discussion is the gauge pressure which is the pressure difference from that outside the pipe.  The displacement s at the end of the pipe is unconstrained.  The result is at an open end we have a pressure node and a displacement antinode.

Open end at (Subscript[x, 0]):  s((Subscript[x, 0]),t)=±(Subscript[s, max]) (antinode),  P((Subscript[x, 0]),t)=0  (node)

At a closed end of a pipe the displacement is forced to be zero, a node and the pressure is unconstrained, an antinode.

Closed end at (Subscript[x, 0]):  s((Subscript[x, 0]),t)=0 (node),  P((Subscript[x, 0]),t)=±(Subscript[P, max])  (antinode)

When sound waves are viewed as pressure wave, the sine and cosines are swapped and that them swaps nodes and antinodes.

Now consider a pipe of length L with both ends open. From the pressure perspective there is a node at either end, just as the case of a string with both ends fixed.  The displacement perspective looks different but the counting is still the same; we need an integer number of half wavelengths over the length L.

(Subscript[λ, m])=((2 L)/m)  and  (Subscript[f, m])=m (Subscript[f, 1])  with  (Subscript[f, 1])=(v/(2 L))    (m=1,2,3,...)

Now we turn to a pipe of length L with one end open and the other closed.  From the pressure perspective there is a node at the open end and an antinode at the closed end.  The displacement point of view gives the reverse, an antinode at the open end and a node at the closed end.  To get the standing waves to fit these boundary conditions we must have an odd number of quarter wavelengths spanning the length L, so L=m λ/4 where m is odd.

(Subscript[λ, m])=((4 L)/m)  and  (Subscript[f, m])=m (Subscript[f, 1])  with  (Subscript[f, 1])=(v/(4 L))    (m=1,3,5,...)

Example . - Organ Pipes

Solution

(Subscript[f, 1])=329.63(QuantityUnit) , v=(Subscript[v, sound])=343(((QuantityUnit))/((QuantityUnit)))  and   (Subscript[f, 1])=(v/(2 L))  ⟹  L=(v/(2 (Subscript[f, 1])))=0.520(QuantityUnit)

Solution

(Subscript[f, 2])=2(Subscript[f, 1])=659.26(QuantityUnit),  (Subscript[f, 3])=3(Subscript[f, 1])=988.89(QuantityUnit)  and  (Subscript[f, 4])=4(Subscript[f, 1])=1318.5(QuantityUnit)

Solution

(Subscript[f, 1])=329.63(QuantityUnit) , v=(Subscript[v, sound])=343(((QuantityUnit))/((QuantityUnit)))  and   (Subscript[f, 1])=(v/(4 L))  ⟹  L=(v/(4 (Subscript[f, 1])))=0.260(QuantityUnit)

Solution

(Subscript[f, 3])=3(Subscript[f, 1])=988.26(QuantityUnit),  (Subscript[f, 5])=5(Subscript[f, 1])=1648.2(QuantityUnit)  and  (Subscript[f, 7])=7(Subscript[f, 1])=2307.4(QuantityUnit)

Beats

Consider interference between two sound waves that differ slightly in frequency.  The resulting sound is heard to pulse at a frequency much smaller than that of the component waves.  At some position from some source a wave will have the form:  u(t)=A cos(ω t+ϕ).  To make the result its most dramatic we will assume that at the position where the combined waves are heard, the amplitudes of the two waves are the same.  Also, for simplicity, we will choose t=0 to be when both waves are at a phase ϕ=0.  The two waves become:

(Subscript[u, 1])(t)=A cos((Subscript[ω, 1]) t)   and   (Subscript[u, 2])(t)=A cos((Subscript[ω, 2]) t).

To add these two waves we will use the same trig. identity we used with standing waves.

cos(α+β)+cos(α-β)=2cos α cos β.

Using α=(1/2)((Subscript[ω, 1])+(Subscript[ω, 2])) t  and β=(1/2)((Subscript[ω, 1])-(Subscript[ω, 2])) t we get:

u(t)=(Subscript[u, 1])(t)+(Subscript[u, 2])(t)=A cos((Subscript[ω, 1]) t)+A cos((Subscript[ω, 2]) t)
        =2 A cos( (((Subscript[ω, 1])+(Subscript[ω, 2]))/2)t )  cos( (((Subscript[ω, 1])-(Subscript[ω, 2]))/2)t ) .

This can be written in terms of the average (angular) frequency (Subscript[ω, ave]) and the beat (angular) frequency (Subscript[ω, beat]).

u(t)=2 A cos( (Subscript[ω, ave]) t ) cos( (1/2)(Subscript[ω, beat]) t )    where  (Subscript[ω, ave])=(((Subscript[ω, 1])+(Subscript[ω, 2]))/2)  and  (Subscript[ω, beat])=| (Subscript[ω, 1])-(Subscript[ω, 2])|

Using  ω=2 π f we get the average frequency and beat frequency:

u(t)=2 A cos(2 π (Subscript[f, ave]) t ) cos( π (Subscript[f, beat]) t )    where  (Subscript[f, ave])=(((Subscript[f, 1])+(Subscript[f, 2]))/2)  and  (Subscript[f, beat])=| (Subscript[f, 1])-(Subscript[f, 2])|

The frequency (Subscript[f, ave]) oscillates rapidly.  Since the function cos(2 π (Subscript[f, ave]) t ) varies between ±1, the function u(t) stays between ±2 A cos( π (Subscript[f, beat]) t ), this is called the envelope function,