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Part IV: Solutions of the wave equation in one dimension

2.3. General solutions of the one-dimensional wave equation

The one-dimensional wave equation was derived with either sound pressure or particle velocity as the dependent variable. Particle displacement, or the variational density, may also be used as the dependent variable. This can be seen from Eqs. (2.4a) and (2.13a) and the conservation of mass, which requires that the product of the density and the volume of a small box of gas remain constant. That is,

ρ^{'} V = ρ_{0} V_{0} = constant

$ρ^{'} V = ρ_{0} V_{0} = constant$

(2.28)

and so

ρ^{'} d V = - V d ρ^{'} .

$ρ^{'} d V = - V d ρ^{'} .$

(2.29)

Let

ρ^{'} = ρ_{0} + ρ,

$ρ^{'} = ρ_{0} + ρ,$

(2.30)

where ρ is the incremental change in density. Then, approximately, from Eqs. (2.8) and (2.29),

ρ_{0} τ = - V_{0} ρ .

$ρ_{0} τ = - V_{0} ρ .$

(2.31)

Differentiating,

\frac{\partial τ}{\partial t} = - \frac{V_{0}}{ρ_{0}} \frac{\partial ρ}{\partial t}

$\frac{\partial τ}{\partial t} = - \frac{V_{0}}{ρ_{0}} \frac{\partial ρ}{\partial t}$

so that, from Eq. (2.13a),

\frac{\partial ρ}{\partial t} = - ρ_{0} \frac{\partial u}{\partial x} .

$\frac{\partial ρ}{\partial t} = - ρ_{0} \frac{\partial u}{\partial x} .$

(2.32)

Also, we know that the particle velocity is the time rate of change of the particle displacement:

u = \frac{\partial ξ_{x}}{\partial t} .

$u = \frac{\partial ξ_{x}}{\partial t} .$

(2.33)

Inspection of Eqs. (2.4a),((2.13a)), (2.32), and (2.33) shows that the pressure, particle velocity, particle displacement, and variational density are related to each other by derivatives and integrals in space and time. These operations performed on the wave equation do not change the form of the solution, as we shall see shortly. Because the form of the solution is not changed, the same wave equation may be used for determining density, displacement, or particle velocity as well as sound pressure by substituting p, or ξ _x, or u for p in Eq. (2.20a) or ρ, ξ, or q for p in Eq. (2.20b), assuming, of course, that there is no rotation in the medium.

2.3.1. General solution

With pressure as the dependent variable, the wave equation is

\frac{\partial^{2} p}{\partial x^{2}} = \frac{1}{c^{2}} \frac{\partial^{2} p}{\partial t^{2}} .

$\frac{\partial^{2} p}{\partial x^{2}} = \frac{1}{c^{2}} \frac{\partial^{2} p}{\partial t^{2}} .$

(2.34)

The general solution to this equation is a sum of two terms,

p = f_{1} (t - \frac{x}{c}) + f_{2} (t + \frac{x}{c}),

$p = f_{1} (t - \frac{x}{c}) + f_{2} (t + \frac{x}{c}),$

(2.35)

where f ₁ and f ₂ are arbitrary functions. We assume only that they have continuous derivatives of the first and second order. Note that because t and x occur together, the first derivatives with respect to x and t are exactly the same except for a factor of ± c.

The ratio x/c must have the dimensions of time, so that c is a speed. From

c^{2} = γ P_{0} / ρ_{0} [Eq. (2.19)]

$c^{2} = γ P_{0} / ρ_{0} [Eq. (2.19)]$

we find that

c = {(1.4 \times \frac{10^{5}}{1.18})}^{1 / 2} = 344.4 m / s

$c = {(1.4 \times \frac{10^{5}}{1.18})}^{1 / 2} = 344.4 m / s$

in air at an ambient pressure of 10⁵ Pa and at 22°C. This quantity is nearly the same as the experimentally determined value of the speed of sound, 344.8 see Eq. (1.8), so that we recognize c as the speed at which a sound wave is propagated through the air.

From the general solution to the wave equation given in Eq. (2.35) we observe two very important facts:

1. The sound pressure at any point x in space can be separated into two components: an outgoing wave, f ₁(t − x/c), and a backward-traveling wave, f ₂(t + x/c).
2. Regardless of the shape of the outward-going wave (or of the backward-traveling wave), it is propagated without change of shape. To show this, let us assume that, at t = t ₁, the sound pressure at x = 0 is f ₁(t ₁). At a time t + t ₁ + t ₂ the sound wave will have traveled a distance x equal to t ₂ c m. At this new time the sound pressure is equal to
$p = f_{1} (t_{1} + t_{2} - t_{2} c) = f_{1} (t_{1}) .$ $p = f_{1} (t_{1} + t_{2} - t_{2} c) = f_{1} (t_{1}) .$
In other words the sound pressure has propagated without change. The same argument can be made for the backward-traveling wave that goes in the − x direction.

It must be understood that inherent in Eqs. (2.34) and (2.35) are two assumptions. First, the wave is a plane wave, i.e., it does not expand laterally. Thus the sound pressure is not a function of the y and z ordinates but is a function of distance only along the x ordinate. Second, it is assumed that there are no losses or dispersion (scattering of the wave by turbulence or temperature gradients, etc.) in the air, so that the wave does not lose energy as it is propagated.

2.3.2. Steady-state solution

In nearly all the studies that we make in this text we are concerned with the steady state. Let us first consider the time-dependent part of the solution at a fixed point in space so that the pressure is only dependent on time. As is well known from the theory of Fourier series, a steady-state periodic wave of arbitrary shape can be represented by a linear summation of sine-wave functions, each of which is of the form

p (t) = \sum_{n = - \infty}^{\infty} p_{n} (t),

$p (t) = \sum_{n = - \infty}^{\infty} p_{n} (t),$

(2.36)

where

p_{n} (t) = c_{n} e^{j ω_{n} t} = c_{n} (\cos ω_{n} t + j \sin ω_{n} t),

$p_{n} (t) = c_{n} e^{j ω_{n} t} = c_{n} (\cos ω_{n} t + j \sin ω_{n} t),$

(2.37)

where ω _n = nω = 2π nf is the angular frequency and c _n is the peak amplitude of the nth component of the wave given by

c_{n} = \frac{1}{T} \int_{0}^{T} p (t) e^{- j ω_{n} t} d t,

$c_{n} = \frac{1}{T} \int_{0}^{T} p (t) e^{- j ω_{n} t} d t,$

(2.38)

where T = 1/f is the period of the wave. Taking the second time derivative of p _n yields

\frac{\partial^{2}}{\partial t^{2}} p_{n} (t) = \frac{\partial^{2}}{\partial t^{2}} c_{n} e^{j ω_{n} t} = - ω_{n}^{2} c_{n} e^{j ω_{n} t} = - ω_{n}^{2} p_{n} (t),

$\frac{\partial^{2}}{\partial t^{2}} p_{n} (t) = \frac{\partial^{2}}{\partial t^{2}} c_{n} e^{j ω_{n} t} = - ω_{n}^{2} c_{n} e^{j ω_{n} t} = - ω_{n}^{2} p_{n} (t),$

(2.39)

which gives the identities

\frac{\partial}{\partial t} = j ω_{n},

$\frac{\partial}{\partial t} = j ω_{n},$

(2.40)

\frac{\partial^{2}}{\partial t^{2}} = - ω_{n}^{2} .

$\frac{\partial^{2}}{\partial t^{2}} = - ω_{n}^{2} .$

(2.41)

Hence the steady-state plane-wave equation for any point in space can be written in the form

(\frac{\partial^{2}}{\partial x^{2}} + \frac{ω_{n}^{2}}{c^{2}}) p_{n} (x, t) = 0,

$(\frac{\partial^{2}}{\partial x^{2}} + \frac{ω_{n}^{2}}{c^{2}}) p_{n} (x, t) = 0,$

(2.42)

which is generally known as the Helmholtz wave equation. Because the wave is propagated without change of shape, we need to consider, in the steady state, only those solutions to the wave equation for which the time dependence at each point in space is sinusoidal and which have the same angular frequencies nω as the source. A general solution that satisfies this equation is given by

p_{n} (x, t) = (p_{n +} e^{- j ω_{n} x / c} + p_{n -} e^{j ω_{n} x / c}) e^{j ω_{n} t},

$p_{n} (x, t) = (p_{n +} e^{- j ω_{n} x / c} + p_{n -} e^{j ω_{n} x / c}) e^{j ω_{n} t},$

(2.43)

where the + and − subscripts indicate the forward and backward traveling waves respectively. In the steady state, therefore, we may replace f ₁ and f ₂ of Eq. (2.35) by a sum of functions each having a particular angular driving frequency ω _n so that

p (x, t) = \sum_{n = - \infty}^{\infty} p_{n} (x, t) = \sum_{n = - \infty}^{\infty} ℜ ((p_{n +} e^{- j ω_{n} x / c} + p_{n -} e^{j ω_{n} x / c}) e^{j ω_{n} t}) .

$p (x, t) = \sum_{n = - \infty}^{\infty} p_{n} (x, t) = \sum_{n = - \infty}^{\infty} ℜ ((p_{n +} e^{- j ω_{n} x / c} + p_{n -} e^{j ω_{n} x / c}) e^{j ω_{n} t}) .$

(2.44)

Generally we omit writing ℜ although it always must be remembered that the real part must be taken when using the final expression for the sound pressure that would actually be observed, for example, when making an animated plot of a sound field.

It is customary in texts on acoustics to define a wave-number k where

k = \frac{ω}{c} = \frac{2 π f}{c} = \frac{2 π}{λ},

$k = \frac{ω}{c} = \frac{2 π f}{c} = \frac{2 π}{λ},$

(2.45)

which can be considered as the spatial angular frequency in rad/m. When k is multiplied by a characteristic dimension such as the length of a tube or the radius of a circular radiator, it forms a useful dimensionless parameter that is proportional to the frequency. Let us now drop ℜ and the subscript n for convenience. Also, we will replace the factor e ^{j
ω
^t} with a tilde. Any one term of Eq. (2.44), with these changes, becomes

\tilde{p} (x) = {\tilde{p}}_{+} e^{- j k x} + {\tilde{p}}_{-} e^{j k x} .

$\tilde{p} (x) = {\tilde{p}}_{+} e^{- j k x} + {\tilde{p}}_{-} e^{j k x} .$

(2.46)

Eq. (2.46) represents two traveling waves: one with amplitude

{\tilde{p}}_{+}

${\tilde{p}}_{+}$

traveling in the positive x direction and the other with amplitude

{\tilde{p}}_{-}

${\tilde{p}}_{-}$

traveling in the negative x direction, where the amplitudes are independent of position x. The appearance of these two solutions occurs because in solving the wave equation we have not specified the direction of travel or any boundary conditions and so the result simply tells us that these solutions can occur. The complex values of

{\tilde{p}}_{+}

${\tilde{p}}_{+}$

and

{\tilde{p}}_{-}

${\tilde{p}}_{-}$

are determined from the boundary conditions. The real parts of the forward and reverse traveling solutions are represented in Fig. 2.3 (a) and (b) respectively, which shows the waveforms in space at a snapshot in time, whereas if the plots were animated, they would be moving in the directions of the arrows. At any fixed point, the pressure or velocity would oscillate as the wave passed through it, with the oscillations having the same shape versus time as versus distance. This is a property of plane waves where the waves propagate without changing shape. Similarly, the solution to Eq. (2.22a) for velocity, assuming steady-state conditions is

\tilde{u} (x) = {\tilde{u}}_{+} e^{- j k x} + {\tilde{u}}_{-} e^{j k x} .

$\tilde{u} (x) = {\tilde{u}}_{+} e^{- j k x} + {\tilde{u}}_{-} e^{j k x} .$

(2.47)

Figure 2.3 Solutions to the steady-state one-dimensional wave equation. (a) Forward and (b) reverse traveling waves.

A similar expression for the velocity can also be obtained from the expression for the pressure by applying Eq. (2.4a) to Eq. (2.46):

\begin{matrix} \tilde{u} (x) = \frac{1}{- j ω ρ_{0}} \frac{\partial}{\partial x} \tilde{p} (x) \\ = \frac{1}{ρ_{0} c} ({\tilde{p}}_{+} e^{- j k x} - {\tilde{p}}_{-} e^{j k x}), \end{matrix}

$\begin{matrix} \tilde{u} (x) = \frac{1}{- j ω ρ_{0}} \frac{\partial}{\partial x} \tilde{p} (x) \\ = \frac{1}{ρ_{0} c} ({\tilde{p}}_{+} e^{- j k x} - {\tilde{p}}_{-} e^{j k x}), \end{matrix}$

(2.48)

the real part of which is also shown in Fig. 2.3. The wave Eq. (2.48) for velocity is similar to Eq. (2.46) for pressure except for one important difference, which is the minus sign preceding

{\tilde{p}}_{-}

${\tilde{p}}_{-}$

. The reason for this is fairly simple. During a positive pressure half-cycle, the resulting velocity is always in the direction of travel. Therefore, in the case of the wave with amplitude

{\tilde{p}}_{+}

${\tilde{p}}_{+}$

traveling in the positive x direction, positive pressure produces positive velocity because it is in the positive x direction, as shown by the dashed arrows in Fig. 2.3a. However, in the case of the wave with amplitude

{\tilde{p}}_{-}

${\tilde{p}}_{-}$

traveling in the negative x direction, positive pressure produces negative velocity because it is in the negative x direction, as shown by the dashed arrows in Fig. 2.3b. Of course, the converse applies during a negative pressure half cycle. The ratio of pressure to particle velocity is the specific acoustic impedance Z _s of the medium, which is obtained by dividing the pressure from Eq. (2.46) by the velocity from Eq. (2.48) to give

Z_{s} = \frac{\tilde{p} (x)}{\tilde{u} (x)} = ρ_{0} c .

$Z_{s} = \frac{\tilde{p} (x)}{\tilde{u} (x)} = ρ_{0} c .$

(2.49)

It is worth noting that in the case of freely traveling waves, which are also known as progressive waves, the pressure and particle velocity are in phase and hence the impedance has a real value. This is very much a characteristic of traveling longitudinal waves, a class that includes sound pressure waves because the particles oscillate in the direction of propagation as opposed to transverse waves whereby the medium oscillates in a direction at right angles to the direction of propagation. An example of the latter is the wave motion of a plucked string.

Example 2.2. Determine the power flow in a freely traveling wave at a fixed point as a function of time.

Answer:

\begin{matrix} p (t) = K \cos ω t \\ u (t) = p (t) / ρ c \\ P o w e r f l o w = p^{*} u = (K^{2} / ρ c) \cos^{2} ω t = (K^{2} / ρ c) (1 - \sin^{2} ω t) \end{matrix}

$\begin{matrix} p (t) = K \cos ω t \\ u (t) = p (t) / ρ c \\ P o w e r f l o w = p^{*} u = (K^{2} / ρ c) \cos^{2} ω t = (K^{2} / ρ c) (1 - \sin^{2} ω t) \end{matrix}$

Thus the power flows by a point in a freely traveling wave like a series of “sausages.” This is explained by referring back to Fig. 1.1. The vibrating surface sends power into the wave when it is moving either to the right or the left. At the instant whenever the surface changes direction, the power drops to zero.

Example 2.3. Assume that for the steady state, at a point x = 0, the sound pressure in a one-dimensional outward-traveling wave has the recurrent form shown by the dotted curve in Fig. Ex. 2.3a. This wave form is given by the real part of the equation

p (0, t) = 4 e^{j 628 t} + 2 e^{j 1884 t} .

$p (0, t) = 4 e^{j 628 t} + 2 e^{j 1884 t} .$

(a) What are the particle velocity and the particle displacement as a function of time at x = 5 m? (b) What are the rms values of these two quantities? (c) Are the rms values dependent upon x?

Solution

a. We have for the solution of the wave equation giving both x and t [see Eq. 2.46]
$p (x, t) = 4 e^{j 628 (t - x / c)} + 2 e^{j 1884 (t - x / c)} .$ $p (x, t) = 4 e^{j 628 (t - x / c)} + 2 e^{j 1884 (t - x / c)} .$
From Eq. (2.4a) we see that
$u (x, t) = - \frac{1}{j ω ρ_{0}} \frac{\partial p (x, t)}{\partial x}$ $u (x, t) = - \frac{1}{j ω ρ_{0}} \frac{\partial p (x, t)}{\partial x}$
or
$u (x, t) = \frac{1}{ρ_{0} c} p (x, t) .$ $u (x, t) = \frac{1}{ρ_{0} c} p (x, t) .$
And from Eq. (2.33) we have
$ξ (x, t) = \frac{1}{j ρ_{0} c} (\frac{4}{628} e^{j 628 (t - x / c)} + \frac{2}{1884} e^{j 1884 (t - x / c)}) .$ $ξ (x, t) = \frac{1}{j ρ_{0} c} (\frac{4}{628} e^{j 628 (t - x / c)} + \frac{2}{1884} e^{j 1884 (t - x / c)}) .$
At x = 5 m, x/c = 5/344.8 = 0.0145 s,
$u (5, t) = \frac{1}{407} (4 e^{j 628 (t - 0.0145)} + 2 e^{j 1884 (t - 0.0145)})$ $u (5, t) = \frac{1}{407} (4 e^{j 628 (t - 0.0145)} + 2 e^{j 1884 (t - 0.0145)})$
and
$ξ (5, t) = \frac{1}{407} (\frac{4}{628} e^{j [628 (t - 0.0145) - (π / 2)]} + \frac{2}{1884} e^{j [1884 (t - 0.0145) - (π / 2)]}) .$ $ξ (5, t) = \frac{1}{407} (\frac{4}{628} e^{j [628 (t - 0.0145) - (π / 2)]} + \frac{2}{1884} e^{j [1884 (t - 0.0145) - (π / 2)]}) .$
Taking the real parts of the two preceding equations,
$u (5, t) = \frac{1}{407} (4 \cos (628 t - 9.1) + 2 \cos (1884 t - 27.3))$ $u (5, t) = \frac{1}{407} (4 \cos (628 t - 9.1) + 2 \cos (1884 t - 27.3))$
$ξ (5, t) = \frac{1}{407} (\frac{4}{628} \sin (628 t - 9.1) + \frac{2}{1884} \sin (1884 t - 27.3)) .$ $ξ (5, t) = \frac{1}{407} (\frac{4}{628} \sin (628 t - 9.1) + \frac{2}{1884} \sin (1884 t - 27.3)) .$
Note that each term in the particle displacement is 90 degrees out of time phase with the velocity and that the wave shape is different. As might be expected, integration diminishes the higher frequencies. These equations are plotted in Fig. Ex. 2.3b.
b. The rms magnitude of a sine wave is equal to its peak amplitude divided by $\sqrt{2}$ $\sqrt{2}$ . This may be verified by squaring the sine wave and finding the average value over one cycle and then taking the square root of the result. If two sine waves of different frequencies are present at one time, the rms value of the combination is equal to the square root of the sums of the squares of the individual peak amplitudes divided by $\sqrt{2}$ $\sqrt{2}$ , so that
$p = \frac{1}{\sqrt{2}} \sqrt{4^{2} + 2^{2}} = 3.16 Pa,$ $p = \frac{1}{\sqrt{2}} \sqrt{4^{2} + 2^{2}} = 3.16 Pa,$
$u = \frac{1}{407 \sqrt{2}} \sqrt{4^{2} + 2^{2}} = 7.77 \times 10^{- 3} m / s,$ $u = \frac{1}{407 \sqrt{2}} \sqrt{4^{2} + 2^{2}} = 7.77 \times 10^{- 3} m / s,$
$ξ_{x} = \frac{1}{407 \sqrt{2}} \sqrt{{(\frac{4}{628})}^{2} + {(\frac{2}{1884})}^{2}} = 1.12 \times 10^{- 5} m .$ $ξ_{x} = \frac{1}{407 \sqrt{2}} \sqrt{{(\frac{4}{628})}^{2} + {(\frac{2}{1884})}^{2}} = 1.12 \times 10^{- 5} m .$
c. The rms values of u and ξ _x are independent of x for a plane progressive sound wave.