Chapter 17

Sound Waves

Particularly interesting waves are sound waves. When an object vibrates, it transmits its vibrations to the air surrounding it: as an example, hitting a drum causes its membrane to vibrate back and forth, and, in doing so, it alternatively compresses and "decompresses" the air around it. These pressure waves propagate through the air and, if they reach an ear, they induce the same vibration in the eardrum. The eardrum vibrations are then interpreted by the brain as sound, and the pitch of the sound is determined by the vibration frequency. Human ears are sensitive to vibrations in the approximate range 20 (very low pitch) to 20,000 (very high pitch) Hz. Any given sound in general consists of the superposition of many different waves of different frequencies but, in some particularly simple case, one can just have a single frequency (called a pure tone). Frequencies below the audible range correspond to infrasounds, above it to ultrasounds (notice the similarity with visible light, infrared and ultraviolet, but be aware that em waves and sound waves are completely different phenomena, even though they are both waves).

To generate sound, vibrating objects transmit their vibrations to a medium (it doesn't have to be air, sound propagates equally well through liquids and solids), but sound will not exist in a vacuum : Star War movies portraying battles in outer space should save money and not hire a sound effect specialist, since such battles would be accompanied by eery silence...

Most of the results we had obtained for traveling waves on a string also apply to sound waves. In particular

• speed of sound: in a fluid, the speed is $v = \sqrt{B/\rho}$, with

  1.

  B : "Bulk Module" is the elasticity factor (the bulk module is a measure of the resistance a fluid opposes to being compressed : a large value of B means that a large pressure will induce a small change of volume)

  2.

  $\rho$: density, is the inertia factor ($\rho = m/V$)

  Sound speed is the smallest in gases, followed by liquids and then solids. In air (at 20⁰C) sound speed is about 340 m/s More precisely, one has $v = 331\sqrt{1+T/273}~m/s$,where T the temperature measured in Celsius.
  
  Question: what happens to $\lambda$ and fwhen a sound wave goes from one medium to another of different sound velocity?
• Wave Function Under the influence of a sound source vibrating with frequency f=1/T, the air molecules will undergo (longitudinal) displacements described by
  $s(x,t) = s_{max}\cos [2\pi(x/\lambda -t/T)]$
  where, for any given medium ov sound velocity v, the wavelength is $\lambda = vT$. One could also show that the pressure variations within the fluid are described by
  $\Delta P(x,t) = \Delta P_{max}\sin [2\pi(x/\lambda -t/T)]$
  with $\Delta P_{max} = 2\pi\rho vs_{max}/T$
• Sound Energy: with procedures similar to those followed for waves on a string, one could obtain the expressions for (1)the energy carried by a sound wave in a cylinder of cross section A and length equal to one full wavelength, (2)the corresponding Power and, more importantly, (3) the Intensity I, i.e. the Power per Unit Area. The results are
  $E = 2\pi^{2}\rho As_{max}^{2}v/T$
  
  $\cal P$ $= 2\pi^{2}\rho As_{max}^{2}v/T^{2}$
  
  $I = 2\pi^{2}\rho s_{max}^{2}v/T^{2}$

From the definition of intensity, $I = \cal P$ /A, it follows that intensities will be measured in W/m². One can immediately see that the Intensity of sound decreases like the inverse square of the distance from the source: suppose that a sound source emits a certain amount of power $\cal P$, and the emission is uniform in all directions. At a distance r from the source, the sound energy will be spread equally over the surface of a sphere centered at the source (this is called a spherical wave), and its intensity will be :

$I = \cal P$$/A = \cal P$$/(4\pi r^2$)

If we move a large distance away from the sound source, even though the wave is still a spherical wave, the sphere curvature will be very small and, if one considers a small enough region, it can be neglected. In this approximation we are dealing with a plane wave.

Often it is preferable to measure not the absolute intensity but its relative value with respect to a reference level (in the case of sound, this is called the sound level).

The human ear can detect sound intensities as low as 10^-12 W/m² (animals with big ears can detect even lower intensities), and it can tolerate (with pain) intensities that are more than 1 million million times stronger, therefore corresponding to a few W/m². Given this very wide range, one can guess that the ear response will not be linear : a sound of 1000 times the intensity will not be perceived as 1000 times louder. Because of the non-linear response and the very wide range, it is customary to compare sound levels on a logarithmic scale. Specifically, one expresses the loudness of a sound of intensity I by comparing it to the threshold of hearing I₀ by means of decibels, defined as

decibel = 10 log₁₀ (I/I₀)

the factor of 10 in front is there since the primitive unit, that hardly anyone uses, is the bel, with 1 decibel = 1/10 of a bel, therefore (value in dB) = 10 x (value in bels)). The reason for using decibels is that it has been shown that 1 dB is roughly the smallest change in sound intensity that a normal human ear can detect. To what change in intensity does this correspond ?

1 = 10 log₁₀ (I₂/I₁)

1/10 = log₁₀ (I₂/I₁)

I₂/I₁ = 10^1/10 = 1.25

showing that a change of one decibel correspond to an intensity increase of about 25%.

Doppler effect

We have seen earlier how the frequency (i.e. the perceived pitch) of a sound depends only on the frequency of the emitting source, and not on the sound velocity in the medium. This is not true any longer if the sound transmitter or the receiver are in motion. In the most general situation, one could have both sound transmitter and receiver in motion with respect to the sound transmitting medium, and they could move either towards or away from each other. We will just examine the case of a receiver at rest and a transmitter moving away from it with velocity v_t (and we will use the symbol v_s for the speed of sound), similar arguments apply to other situations. To find out what is the frequency perceived by the receiver, we will determine the period, i.e. the difference in time of arrival between two successive wave crests. Suppose that at t=0 transmitter and receiver are at a distance d and the transmitter emits a crest. This signal will reach the receiver after a time

t₁ = d/v_s

the next crest will be emitted after one period, i.e. at a time t=T, and it will reach a receiver at a time

t₂ = T+d'/v_s = T +(d+v_tT)/v_s = d/v_s+T(1+v_t/v_s)

The period T' perceived by the receiver is the difference in arrival time between two crests T'=t₂-t₁=T(1+v_t/v_s). Given that T=1/f, the final result is

f'=fv_s/(v_s+v_t)

See the book for the most general formula.

Shock wave : read and enjoy