1   Sound waves

Part 2

The velocity of sound waves

This varies slightly with the temperature of the air but not, as is often supposed, with the air pressure, at least not over all normally encountered ranges of temperature and pressure.

At 0°C, the velocity of sound waves in air can be taken as 331 m/s, and this increases by 0.6 m/s for every 1°C rise in temperature.

Thus:

at –10°C the velocity is 325 m/s

at +10°C it is 337 m/s

at +20°C it is 343 m/s.

340 m/s is a reasonable figure to use for normal room temperatures.

In substances other than air, sound waves travel at quite different velocities (Table 1.1).

The high velocity of sound in helium, about three times that in ordinary air, accounts for the high-pitched voices of divers and others whose lungs and vocal tracts contain the gas. (Not an experiment to be tried for more than a very few seconds!)

Table 1.1

Units

1.   The bar. This is another unit of pressure but is most commonly used in meteorology as it is approximately equal to normal atmospheric pressure. It is related to the pascal by:

1 bar = 100 000 Pa

or 10 μbar = 1 Pa

(Television weather charts show air pressures in millibars.)

2.   When dealing with sound wave pressures we must of course remember that pressures are alternating. Use then has to be made of the concept of a steady pressure which is equivalent to the alternating one. A similar problem is met in the a.c. (alternating current) mains, which in the UK is quoted as having a voltage of 230 V. In fact, the mains voltage goes through a cycle of a maximum of about 340 V, falling to zero and then reaching a maximum of 340 V in the opposite direction. It then goes back to zero and repeats the operation, taking one fiftieth of a second for each cycle (hence the frequency of 50 Hz). This fluctuation can be regarded as equivalent in power to a d.c. (direct current) of 230 V. Engineers use the term root mean square (r.m.s.) for this equivalent. The pressures in Pa of sound waves are equivalents in the same way.

Table 1.2 Sound frequencies for different wavelengths

*It is worth noting that this is roughly equal to 1 foot.

The range of frequencies covered in the table (16 Hz to 16 kHz) is approximately the range of frequencies which the normal adult human ear can detect.

Decibels

Correctly expressed, the number of decibels representing a ratio of two powers is

dB = 10 log(power ratio)

If we are dealing with pressures or voltages, the expression becomes

dB = 20 log(pressure ratio) or dB = 20 log(voltage ratio)

As an example, suppose that the power amplifier feeding a loudspeaker is delivering 50 W but is then replaced by one delivering 100 W:

the dB increase is 10 log(100/50)

= 10 log(2)

= 10 × 0.3010 (easily found from a scientific calculator)

= 3.010 or approximately 3 dB

It will be remembered from Part 1 of the chapter that this increase of 3 dB is only just about detectable! To get twice the loudness from the loudspeaker, assuming that it will take the increased power without being damaged, the power must be increased by 10 dB. This means that the 50 W amplifier will have to be replaced by a 500 W one:

Questions

Try to answer these as honestly as possible. You could cheat by looking at the answers first, but that doesn't really help. The answers are given at the end of the book.

1.   What range of sound wave frequencies can the normal adult detect?

2.   What is the speed of sound waves in air at normal temperatures?

3.   A hall being used for music recordings has its gallery supported by square-sectioned pillars. These are 0.5 m by 0.5 m cross-section. What frequency range of sound will be reflected from each pillar?