This section will help you to better comprehend the nature of sound waves. An important fact for understanding how our ears works is that pressure variations control what we hear.

One can produce a one- dimensional periodic sound wave in a long, narrow tube containing a gas by means of an oscillating piston at one end as shown in figure.

The darker parts of the dark areas in this figure represent regions where the gas is compressed and thus the density and pressure are above their equilibrium values.

A compressed region is fanned whenever the piston is pushed into the tube. This compressed region, called a compression, moves through the tuber as a pulse, continuously compressing the region just in front of itself.

When the piston is pulled back, the gas in front of it expands, and the pressure and density in this region fall below their equilibrium values (represented by the lighter parts in figure). These low-pressure regions, called rare factions which also propagate along the tube, following the compressions. Both regions move with a speed equal to the speed of sound in the medium.

As the piston oscillates sinusoidally, regions of compression and rarefaction are continuously set up.

The distance between the centres of two successive compressions (or two successive rarefactions) equa ls the wavelength A.

As these regions travel through the tube, any small element of the medium moves with simple harmonic motion parallel to the direction of the wave. lf y(x, t) is the position of a small element relative to its equilibrium position, we can express this harmonic position function as

`y (x, t) =A cos (omegat - kx)`

Where A is the maximum position of the element relative to equilibrium called the displacement amplitude of the wave. The parameter k is the wave number and w is the angular frequency of the piston.

Note that the displacement of the element is along y, in the direction of propagation of the sound wave, which means we are describing a longitudinal wave.

Consider a thin disk-shaped element of gas whose circular cross section is parallel to the piston in figure.

This element will undergo changes in position, pressure, and density as a sound wave propagates through the gas. From the definition of bulk modulus , the pressure variation in the gas is

`DeltaP=-B(DeltaV)/(V_i`

The element has a thickness llx in the horizontal direction and a cross-sectional area A, so its volume is `V_i=ADeltax`.

The change in volume `DeltaV` accompanying the pressure change is equal to `ADeltay`, where `Deltay` is the difference between the value of y at `x +Deltax` and the value of y at x respectively. Hence, we can express to P as

`DeltaP=-B(DeltaV)/(V_i)=-(B[y^'-y])/(A(Deltax))\(ADeltay)/(ADeltax)=-B(Deltay)/(Deltax)`

As `Deltax` approaches zero,

`DeltaP=-B(dely)/(delx)`

If `y=Asin(omegat-kx).....(i)`

`=BAKcos(omegat-kx)`

`p=p_0sin{omega(t-x//v)+pi//2}....(ii)`

Equation (i) and (ii) represents that same sound wave where, p is excess pressure (at position x), i.e. pressure over and above the average atmospheric pressure and the pressure amplitude `p_0` is given by

`p_0=(BAomega)/v=BAK`

(B = Bulk modulus ofthe medium, K = angular wave number)
Note from equation (i) and (ii) we can observe that the displacement of a particle and excess pressure at any position are out of phase by `pi/2`. Hence a displacement maxima corresponds to a pressure minima and vice-versa.

Let the pressure of the undisturbed air be P and the pressure inside the pulse be `P + DeltaP`, where `DeltaP` is positive due to the compression. Consider an element of air of thickness l'ix and face area A, moving toward the pulse at speed v. The time interval that is taken by the pulse to cross the element

`Deltat=(Deltax)/v`

The forces acting on the leading and trailing faces (due to air pressure) as shown.

`F=PA-(P+DeltaP)A=-DeltaPA` (net force)

The minus sign indicates that the net force on the air element is directed to the left in figure.

The volume of the element is A (L'ix) so with the aid of equation, we can write its mass as

`Deltam=rhoDeltav=rhoADeltax=rhoAv(Deltat)`

The average acceleration of the element during `Deltat` is

`a=(Deltav)/(Deltat)`

Thus, from Newton 's second law (F = ma), we have,

`-DeltaPA=(rhoAvDeltat)(Deltav)/(Deltat)`

which we can write as

`rhov^2=-(DeltaP)/(Deltav//v)`

The air that occupies a volume `V (= AvDeltat)` outside the pulse is compressed by an amount `DeltaV(= ADeltavDeltat)` as it enters the pulse. Thus,

`(DeltaV)/V=(ADeltavDeltat)/(AvDeltat)=(Deltav)/v`

Substituting equation and then equation into equation leads to

`rhov^2=-(DeltaP)/(Deltav//v)=-(DeltaP)/(DeltaV//V)=B`

`:.v=sqrt(B/rho)`

`text(Newton's formula :)`
Newton assumed propagation of sound through a gaseous medium to be an isothermal process.

PV = constant

`=>(dP)/(dV)=-P/V`

and hence `B = P`

and thus he obtained for velocity of sound in a gas.

`v=sqrt(P/rho)=sqrt((RT)/M)` where `M=` molar mass

`text(Laplace's correction :)`
Later Laplace established that propagation of sound in a gas is not an isothennal but an adiabatic process
and hence PV' = constant

`=>(dP)/(dV)=-gammaP/V`

where, `B=-V(dP)/(dV)=gammaP`

and hence speed of sound in a gas,

`v=sqrt((gammaP)/rho)=sqrt((gammaRT)/M`

Velocity of sound in solids is given by `v=sqrt(Y/rho)`

where Y = young's modulus.

`text(Factors affecting speed of sound in atmosphere :)`

`text(Effect of temperature)`
As temperature (T) increases velocity (v) increases. For small change in temperature above room temperature v increases linearly by 0.6 m/s for every `1^@C` rise in temp.

`text(Effect of humidity)`
With increase in hunidity density decreases. This is because the molar mass of water vapour is less than the molar mass of air.

`text(Effect of pressure)`
The speed of sound in a gas is given by `v=sqrt((gammaP)/rho)=sqrt((gammaRT)/M`
So at constant temperature, if P changes then `rho` also changes in such a way that `P//rho` remains constant.
Hence pressure does not have any effect on velocity of sound as long as temperature is constant.

Consider a harmonic sound wave propagating along a tube of cross-sectional area S, as shown in figure.

The quantity pis the excess pressure caused by the wave, and oy/ot is the velocity of an element of the fluid. The instantaneous power (P) supplied by the wave to the element is

`P=Fv=pA(deltay)/(deltat)`

Using equation

`P=[-p_0cos(kx-omegat)]S[-omegaAcos(kx-omegat)]`

or `P=p_0AomegaScos^2(kx-omegat)`

At any position say x = 0, the average of `cos^2 omegat` over one period is ; hence the average power transmitted by the wave is

`P_(av)=1/2rhoS(omegaA)^2v`

Note that this has the same form as equation for the power transported by a wave on a string.

The intensity `I` of a wave is defined as the energy incident per second per unit area normal to the direction
of propagation:

`I=text(Power)/text(Area)=P/S` The SI unit of intensity is `W //m^2`

From equation we know `p_0=BAK`

Since `K=omega/v` and `B=rhov^2`, therefore `p_0=rho omega vA`

Thus `I_(av)=p_0^2/(4pir^2)`

That is, `Iprop1/r^2`, the intensity decreases as the inverse square of the distance from a point source.

For cylindrical waves `Iprop1/r`

`text(Loudness and Intensity)`
The sound intensities that the human ear can hear range from `10^(-12) W//m^2`.

The intensity of a sound is perceived by the ear as the subjective sensation of loudness. However, if the intensity doubles, the loudness does not increase by a factor of 2.

Experiments first carried out by A.G. Bell showed that to produce an apparent doubling in loudness, the intensity of sound must be increased by a factor of about 10.

Therefore, it is convenient to specify the intensity level `beta` in terms of the decibel (dB). which is defined as

`beta=10log(I/I_0)`.........(i)

Where `I` is the measured intensity and `I_0` is some reference value. If one takes `I_0` to be `10^(-12)` `W//m^2`, then the threshold of hearing corresponds to `beta=10` `I=0 dB`. At the threshold of pain , `1W/m^2` the intensity level is

At the threshold of pain , `1W//m^2` the intensity level is

`beta=10los(1/10^(-12))=120dB`

A list of the intensity level of various sources is given in Table

`text(Sources Intensity Level)`
Threshold of hearing`text( )`0
Leaves rustling`text( )`10
Quiet hall`text( )`25
Office`text( )`60
Conversation`text( )`60
Heavy traffic (3m)`text( )`80
Loud classical music`text( )`95
Loud rock music`text( )`120
Jet engine (20m)`text( )`130

`text(Coherent sources)`
If `p_1=p_(0_1)sin(omega_1t-k_1x_1+phi_1)`

and `p_2=p_(0_2)sin(omega_2t-k_2x_2+phi_2)`

Where `p_1` and `p_2` are pressures at point O due to `S_1` and `S_2` respectively

The phase difference of the or arriving wave is

`Delta theta=Delta theta_2 -Delta theta_1=(k_1x_1-k_2x_2)+(omega_2-omega_1)t+(phi_2-phi_1)`

The sources are called coherent if the phase difference between them does not change with time.

`Delta theta` is independent of time if
(a) `omega_2-omega_1=0` `=>` `omega_2=omega_1`
Thus for sources to be coherent, their frequencies must be equal.

(b) `k_1=k_2`
So `Delta theta=k(x_1-x_2)+(phi_2-phi_1)=k(Deltax)+Deltaphi`
where `Deltax` is called path difference of the waves.

If source are coherent the resultant intensity at any point is constant.

Sources are called incoherent, if their frequencies are not equal. For incoherent sources, phase difference of the waves continuously changes with rime and interference effects can not be observed at a point.

`text(Interference of Sound waves)`
If `p_1=p_(0_1)sin(omegat-k_1x_1+phi_1)`
`p_2=p_(0_2)sin(omegat-k_2x_2+phi_2)`

Phase difference `alpha=k(Deltax)+phi`

If initial phases fo the two waves are same i.e. the sources vibrate in same source

i.e. `Deltaphi=0`

then phase difference is due to the path difference only i.e. `alpha=k(Deltax)=(2pi)/lamda(Deltax)`

Resultant amplitude is

`p_0=sqrt(p_(0_1)+p_(0_2)+2p_(0_1)p_(0_2)cosalpha)`

Resultant intensity `I=I_1+I_2+2sqrt(I_1I_2)cosalpha`

`text(Condition for constructive Interference)`

when `cosalpha=+1`

i.e. `alpha=0,2pi,4pi...`
i.e. `2npi`

i.e. `(2pi)/lamda=Deltax=2npi`

`=> Deltax=nlamda\\(n=0,1,2,3.....)\\ [ninI]`

`p_0=p_(0_1)+p_(0_2)`

`I=(sqrtI_1+sqrtI_2)^2`

`text(Condition for destructive Interference)`
`p_0=|p_(0_1)-p_(0_2)|`

`I=(sqrtI_1+sqrtI_2)^2`

when `cosalpha=-1`

i.e. `alpha=pi,3pi,5pi.....`

`alpha=(2n-1)pi`

and `Deltax=(2n-1)lamda/2=(n-1/2)lamda` (where `n=1,2,3,.....`) [`ninI`]

If `I_1=I_2=I_0` then

`I=I_0 + I_0 +2 sqrt(I_0^2)cosalpha`

`=2I_0(I+cosphi)`

`I=4I_0cos^2(phi/2)`

`=> I_(max)=4I_0` [where `alpha=2npi`]

and `I_(m i n)=0` [when `alpha=(2pi-1)pi`]