# Sound Waves

## Sound Waves

Sound waves in air involve longitudinal oscillations of the molecules. These waves are characterised as pressure or density fluctuations. The pressure fluctuations are of the order of 1 Pa, whereas atmospheric pressure is about 105 Pa. Figure (14.16) shows a loudspeaker which produces compressions and rarefactions in the air in a tube. As the diaphragm of the loudspeaker moves forward, it produces a compression, that is, an increase in pressure Δp above the equilibrium value po. When the diaphragm moves backward, it produces a rarefaction, a decrease in pressure, +Δp, below po. The collisions between the molecules cause these compressions and rarefactions to propagate as a sound wave down the pipe.

 At a given instant there are points such as b and d toward which molecules converge and therefore increase the local pressure to a maximum of po + Δpo. On either side of points a and c, the molecules are moving away and so local pressure at these points drop to a minimum of po - Δpo. Note that the points (a, b, c and d) of maximum pressure deviation (±Δpo) have zero displacement. Moreover, pressure fluctuations are 90o out of phase with the displacements. The converse is also true: zeroes in pressure fluctuation correspond to maxima or minima in the displacement. The speed of longitudinal waves in a fluid is given by v = √B/ρ (14.31)

where B is the bulk modulus defined as

B = -Δp/ΔV/ V

where ΔV / V  is the fractional change in volume produced by the change in pressure Δp. The SI unit of B is N/m2. The negative sign is needed to make B positive since a positive change in pressure results in a negative change in volume V. One may visualize bulk modulus as spring constant of air.

The propagation of sound in air is an adiabatic process. For air the bulk modulus in adiabatic condition is given by

B = γp(14.32)

where γ is an adiabatic exponent whose numerical value for air is 1.4.

Thus, velocity of sound in air is given by

v = √γp/ρ     (14.33)

Using gas equation p/ρ = RT/M

∴v = √γRT/M   (14.34)

Substituting the values of γ, R and M we obtain the approximate value of speed of sound in terms of absolute temperature T, as

v ≈ 20√T (14.35)

Example 14.12

Determine the speed of sound waves in water, and find the wavelength of a wave having a frequency of 242 Hz. Take Bwater  = 2 × 109 Pa.

Solution

Speed of sound wave, v = √B/ρ = √(2 x 109)/103  =  1414 m/s

Wavelength λ = v/ f = 5.84 m

Example 14.13

If the average molar mass of air is 28.8 × 10-3 kg/mol, then find the speed of sound in air at 300K.

Solution

Hereγ = 1.4 (diatomic), M = 28.8 × 10-3 kg/mol

T = 300 K, R = 8.31 J mol1 K−1.

v = √γRT/M = √(1.4)(8.31)(300)/28.8 x 10 -3 = 348 m/s

Relationship between Pressure Waves and Displacement Waves

For a harmonic wave, the longitudinal displacement (y) is given by

y = A sin(kx - wt)(14.36)

We know p = −B ⧊v / v

Since change in volume is produced by the displacement of the particles, therefore,

ΔV/ V = δy/δx

Thus,p = −Bδy/δx  = −BAk cos(kx - wt)

orp = −po cos(kx - wt)(14.37)

where po = BAk is the pressure amplitude

Note that displacement and pressure amplitudes are π/2 out of phase.

### Standing Sound Waves

Standing waves can be produced in air columns, for example, in organ pipes, flutes, and other wind instruments because sound waves are reflected both at a closed end and at an open end of a pipe. In an open pipe both the ends are open, whereas in a closed pipe one end is closed.

### Closed pipes

In fig. (14.17) the motion of the molecules is represented by solid arrows at t = 0 and by dashed arrows at t = T/2, where T is the period. At the closed end the displacement is permanently zero; that is, it is a displacement node. From the previous section we know that this corresponds to a pressure antinode. (At the instant depicted the pressure is a maximum at the closed end. Half a period later, the density and pressure will be a minimum.)

The open end has to be at atmospheric pressure, so it is a pressure node and a displacement antinode. The fundamental mode occurs when L = λ/4, which means
f1 = v/(4L). The higher harmonics are easily constructed if we keep in mind that the closed end is a displacement node, whereas the open end is a displacement antinode. (Traditionally the displacement is depicted rather than the pressure.) Figure (14.18) shows that for a closed pipe only the odd harmonics are possible:

(Closed pipe)fn = (nv)/(4l) (n = 1, 3, 5….)(14.38)

### Open Pipes

The knowledge that each open end is a displacement antinode allows us to draw the various modes right away (see fig.(14.19)). Note that the fundamental frequency of an open pipe is double that of a closed pipe of the same length. In an open pipe, all the harmonics are possible:

fn =(nv)/(2l)(14.39)

Standing waves in pipes decay very quickly after the source of excitation is removed. The interaction between the air in the pipe and the surrounding air also means that The above equations are not quite correct. It is found that pipes of the same length but of different diameters have slightly different resonant frequencies. The effective length is approximately the real length plus 0.6 of the radius (R) of each open end.

leff = l + 0.6 R

Example 14.14

An air column at 51°C and a tuning fork produce 4 beats per second when sounded together. As the temperature of the air column is gradually decreased to 16°C, they again produce 1 beat per second. Find the frequency of the tuning fork. It is observed they again produce one beat per second If the temperature is decreased further. It implies that the frequency of air column is more than that of the tuning fork.

Solution

Let f be the frequency of the tuning fork.

The number of beats per second decreases as the frequency of the air column decreases. Hence the frequency of air column is initially (f + 4) at 51°C and (f + 1) at 16oC

Since the velocity of sound is proportional to the square root of temperature i.e.,

v ∝ √T

∴ f + 4/f + 1 = √273 + 51/273+16

Thus f = 50 Hz

 Example 14.15 AB is cylinder of length 1.0 m, fitted with a thin flexible diaphragm C at the middle and two other thin flexible diaphragms A and B at the ends. The chamber 1 contains hydrogen and Chamber 2 oxygen. The diaphragms A and B are set into vibrations of same frequency and behave as antinodes. What is the minimum frequency of these vibrations for which the diaphragm C is a node? Given, the velocity of sound in hydrogen is 1100 m/s and in oxygen is 300 m/s. Determine the fundamental frequency of the system of two air columns.

Solution

Both the chamber act like closed organ pipe.

The tubes AC and BC will be two closed tubes with antinodes at A and B and nodes at C.

 Chamber 1 v1 = CH = 1100 m/s Fundamental frequency, f1 = v1/4l = 1100/4(0.5)  = 550Hz ⇒v1/v2 = 550/150 ⇒3v1 = 11 v2 Chamber 2 v2 = 300 m/s f2 = v1/4l = 300/4(0.5)  = 150  Hz vmin = 3v1 = 3(550) = 1650 Hz orvmin = 11v2 = 11(150) = 1650 Hz Let n1th harmonic of chamber 1 coincides with the n2th harmonic of chamber 2. Thus n1f1 = n2f2 or n1/n2 = f2/f1 = 150/550 = 3/11

The third harmonic of chamber 1 coincides with eleventh harmonic of chamber 2.

Hence, the fundamental frequency of the system is f = 3f1 = 3(550) = 1650 Hz