 # Longitudinal sound/pressure waves in a tube

1.8.1 General theory

Consider a tube of cylindrical cross section and length L inside which there is only air. Assume that the length of the tube is along the x axis. We consider pressure waves within this tube in the form of longitudinal oscillations of the air column. At time 0, the density of the air column within the length [a;, x + dx] is p(Q,x). The mass of the air within this column at time 0 is therefore p(0, x)dx. At time t, this infinitesimal air column gets displaced to the interval [a? + u(t, x),x + dx + u(t, x + da?)] and hence by mass conservation, the density of air within this displaced column is

p(0. 3’)

pit, x + u) = p(0, x)dx/(dx + u(t, x + dx)) = -----?

1 + u,x ,x)

Assuming an equation of state p = F(p), the pressure within this displaced column is

p(t,x + u) = F(p(0,a?)(l + u x(t, a;))-1)

The difference between the pressures at x + u and at x + dx + u + u Xdx = x + dx + u + du is

p(t,x--u)— p(t,x+dx+u+du) = p(t,x+u)— p(t, x--dx--u+u ,xdx)

= p(t, x--u(t, x))—p(t, x+dx+u(t, x+dx)) =

F(p(0,a?)(l + «.a:)-1) - F(p(0,a? + da?)(l + u,x + u^xxdx)~r)

= -dx-^-F(p(0,x)(l + u.k)-1) ax

This equation leads us immediately to the Newton’s second law of motion:

p(Q,x)dx.uM(t,x) = -da?—F(p(0,a?)(l + ax

or equivalently,

+ ¿r( ) = 0

This is the exact equation of motion for the longitudinal sound/pressure wave u(t, x).

Simplifying approximations: Assume that p(0,a?) = po, ie, the initial density is a constant. Further, assume that |uja;| << 1, ie, the oscillations of the air column are small. Then, we have approximately,

F( p(M.) = F(po(1 _ )} = F(po) _ F'Mpou

1 + u x

and we get with such an approximation, the one dimensional wave equation for u:

uitt(t,x) - Fpa)u,xx(t,x) = 0

The velocity of the sound wave is read off immediately from this equation as c= / F'(po) = x/dp(po)/dp

1.8.2 Exercises

 Solve approximately the exact nonlinear sound propagation wave equation using perturbation theory by attaching a small perturbation parameter to the nonlinearity.

 If the gas within a pipe is charged with charge density proportional to the mass density p(t, x) and if an external electric field E(t, x) is applied across the x axis, then what will be the modification to the wave equation for the gas ?

 If the cross section of the pipe varies with time in accord with the radius of the pipe at (i, x) being R(t,x), then what will the equations of motion of the gas through the pipe look like assuming that the length of the pipe is parallel to the z direction and that the non-vanishing components of the gas velocity are vz(t,z,r), vr(t,z,r) where r = y/a;2 + y'2? Assume that the density of the gas within the pipe also has the same form p(t,z,r).

1.8.3 Points to remember

 By applying Newton’s second law of motion and the equation of mass conservation along with the equation of state of a gas inside a tube, we can derive after approximations, the one dimensional wave equation for the pressure or equivalently the gas displacement inside the pipe. The force on an infinitesimal element of the gas is obtained as a pressure gradient .

 If the gas is charged and an external electric field is present, then we can modify the wave equation by taking into account apart from the forces induced by pressure gradients the force of the electric field on the charges.

# The difference between transverse and longitudinal waves in terms of wave polarization

When a wave p(t, r) € R3 is vector valued, and it propagates along the direction n, we can express it as

ip(t, r) = •¡/’(i — n.r/c, 0)

It is readily verified that this satisfies the three dimensional wave equation. We say that this wave field is transversely polarized ifand that it is longitudinally polarized if

û x V>(t, r) = 0

If both of these are not satisfied, then it is partially longitudinal and partially transverse polarized. For example, the TEM waves in empty space are transversely polarized as follows from the Maxwell equations V.E’(t, r) = 0, V.H(t, r) = 0. Soundwaves in a tube are longitudinally polarized while the electromagnetic field within a waveguide is partially longitudinal and partially transverse polarized. For example, the TE modes have the electric field transversely polarized while the magnetic field partially longitudinal and partially transverse while the TM modes have the magnetic field transversely polarized while the electric field partially longitudinal and partially transversely polarized.

1.9.1 Exercises

 Give examples of electromagnetic waves that are partially transverse and partially longitudinally polarized.

 If <¡>(t,r) = .exp(-ik.r) in phasor form is the electric scalar potential and if the magnetic vector potential can be neglected, then show that the electric field is longitudinally polarized.

 If a wave has wave vector k and can be expressed as a sum of a longitudinally polarized component and a parallely polarized component as

ip(t, r) = F(cjt — k.r)k + G(wt — k.r)(fi — (k, n)k/k2}

where ñ is an arbitrary constant unit vector, then calculate curhp(t, r) and divù>(t,r). Specifically, show that the former is expressible completely in terms of G only while the latter is expressible in terms of F only.

1.9.2 Points to remember

 Transverse waves as in electromagnetics usually follow from the wave equation by a condition like the divergence of the wave vector field vanishes.

 Sound waves in a pipe are examples of longitudinal waves, as also are waves in spring.

 There may exist waves which have both longitudinal and transverse components as for example, the electromagnetic waves in a waveguide or a cavity resonator.