Pressure Amplitude

Sound is a longitudinal wave which consists of alternate compressions and rarefactions. Due to this, the volume of the medium is altered and hence, there is a variation of pressure. Hence, the longitudinal waves are also called pressure waves.

Let us consider a sound wave travelling in a medium along +ve x-axis as,

y = a sin(ωt-kx) ….. (i)

where, a = displacement amplitude or simply amplitude, [latex]\omega[/latex] = angular frequency and k = propagation constant.

Consider a tube BC of air of cross-sectional area A and length [latex]\Delta x[/latex] and P be the instantaneous pressure variation at a point. So, its volume will be:

V = V₁ = [latex]A.\Delta x[/latex]

When the wave passes the end of the cylinder, it is displaced by y and the other end is displaced by [latex]y + \Delta y[/latex].

Hence, the change in volume of the cylinder is: ΔV = V₂ – V₁

                        = [latex]A.(y+\Delta y + \Delta x – y) – A.\Delta x[/latex]

            = [latex]A.\Delta y[/latex]

Now, [latex]\frac{\Delta V}{V} = \frac{A.\Delta y}{A.\Delta x} = \frac{\Delta y}{\Delta x}[/latex]

On the limit Δx → 0, the fractional change in volume is:

[latex]\frac{dV}{V} = \lim_{\Delta x\to 0} \frac{\Delta y}{\Delta x}[/latex]

            = [latex]\frac{dy}{dx}[/latex] …… (ii)

But, the Bulk modulus of the gas is:

B = [latex]-\frac{P}{\frac{dV}{V}} = -\frac{P}{\frac{dy}{dx}}[/latex]

∴ P = -B.[latex]\frac{dy}{dx}[/latex]  …….. (iii)

Now, differentiating eq^n. (i), w.r.t. x, we get,        

[latex]\frac{dy}{dx} = \frac{d[a sin(\omega t – kx)]}{dx} = a\frac{d sin(\omega t – kx)}{d(\omega t – kx)} .\frac{d(\omega t – kx)}{dx}[/latex]

            = – akCos(ωt-kx).

Now, eq^n. (iii) becomes,

P = Bka cos(ωt-kx) …….. (iv)

Here, P will be maximum if cos(ωt-kx) = 1, So,

Pressure Amplitude = Bka.

The equation (iv) is the wave equation in terms of pressure variation. The quantity Bka is called pressure amplitude.

Musical Sound and Noise

The audible sound is divided into two categories:

Noise
An unpleasant, discontinuous and non-uniform sound produced by irregular and periodic vibration is called noise. For example: thunder, firing of gun, etc.

Musical sound
A pleasant, continuous and uniform sound produced by regular and periodic vibration is called musical sound. For example, sound produced by various musical instruments.

Characteristics of musical sound

1. Pitch
   Pitch is the sensation received by our ear due to frequency of sound which measures the shrillness or flatness of sound. It depends on frequency of sound. Higher is the frequency, higher will be its pitch. Eg. Female voice has high pitch as compared to male voice.
2. Loudness
   It is defined as the sensation of sound in our ear due to intensity of sound wave. Experimentally, Weber – Fechner established the relation for loudness as,
   Loudness [latex]\propto[/latex] Log I.
   The greater the amplitude of vibration, greater will be the intensity (as [latex] I \propto a^2[/latex]) and louder will be the sound.
3. Quality or timber of sound
   Quality is the sensation received by our ear due to the waveform of a sound wave. The shape of resultant wave depends on the number of waves superimposed on each other. Since, waveform depends on number of harmonics present, quality of sound depends on the number of harmonics on the sound wave.

Intensity of sound
Sound waves carry energy when they travel from one point to another. The intensity of sound at a point is defined as the amount of sound energy per second through unit surface area kept normal to the direction along with the sound waves travel. If [latex]E[/latex] is the energy flowing in time [latex]t[/latex] through a surface area [latex]A[/latex] and perpendicular to the wave front, then, the intensity at that place is given by

[latex]I = \frac{E}{At}[/latex] [SI unit: [latex]Js^{-1}m^{-1}[/latex]]

Since, the power is energy per unit time, the intensity is: I = P/A

Hence, the intensity can be defined as the power passing through a unit area perpendicular to the wave direction. The SI unit of intensity is Wm^-2.

Relation between amplitude of vibration and intensity

When sound wave travel in a medium, the particles of the medium vibrate simple harmonically along the direction of wave velocity. Consider, the vibrating layer of air, whose displacement is given by:

y = aSinωt ……………. (i), where, a = amplitude of vibration.

            ω = 2πf = angular frequency of the particle.

For the given wave, velocity remains constant. Therefore, the velocity of the particle is,

v = [latex]\frac{dy}{dt} = \frac{d}{dt}(aSin\omega t)[/latex]

            = aCosωt.ω = aωCosωt

Hence, the kinetic energy of the vibrating air is:

KE = [latex]\frac{1}{2}mv^2 = \frac{1}{2}ma^2\omega^2Cos^2\omega t[/latex] ……………. (ii)

Where, m = mass of vibrating air. The vibrating layer of air has potential energy also but according to principle of conservation of energy, the sum of kinetic energy and potential energy is the maximum K.E. of the system. Hence, the total energy of vibrating air is:

E = [latex]\frac{1}{2}ma^2\omega^2[/latex] …………… (iii)

If ‘v’ be the velocity of sound in air, then, in 1 second, the air is disturbed by the wave over a distance of v meter. If we consider, the area of cross section is equal to 1m², the volume of air disturbed is v m³. Hence, the mass of air disturbed per second is v x ρ. where, [latex]\rho[/latex] = density of air. i.e., m = ρv.

Now, from eq^n. (3), we have,

Total energy flow (E) = [latex]\frac{1}{2}v\rho a^2\omega^2[/latex]

From the definition, this energy is the intensity of sound. Therefore, intensity (I) is equal to:

I = [latex]\frac{1}{2}v\rho a^2\omega^2[/latex]

For the given source, ‘f’ is constant and for the given medium, v and ρ are constant. Therefore,

I ∝ a²         [I =[latex]\frac{1}{2}v\rho(2\pi f)^2a^2][/latex]

i.e., intensity of sound is directly proportional to the square of the amplitude of vibration.

Relation between intensity and loudness

According to Weber-Fechner, the loudness of the sound perceived by the ear is directly proportional to logarithm of intensity. i.e., L ∝ logI

Or, L = k logI, where, k = constant.

Intensity Level

The lowest intensity of sound that can be just perceived by normal human ear is called threshold of hearing. It is taken as 10^-12 Wm^-2 for pure note of frequency 100 Hz, denoted by I_o. Therefore, I_o = 10^-12 Wm^-2.

Our ear responds to various levels of intensity than the intensity itself. Intensity level [latex](\beta)[/latex] is defined in logarithm scale and is given by: [latex]\beta[/latex] = log[latex]\frac{I}{I_o}[/latex] , where, I = Intensity of given sound.

The unit of intensity level of sound is bel.

If I = 10I_o, then,

[latex]\beta[/latex] = log[latex]\frac{10I_o}{I_o}[/latex]

[latex]\therefore \beta[/latex] = 1 bel.

Hence, intensity level of a sound is said to be 1 bel if its intensity is 10 times the threshold of hearing. The smaller unit of intensity level is decibel (dB) which is 1/10^th of a bel.

Therefore, intensity level in decibel is:

[latex]\beta[/latex] = 10log[latex]\frac{I}{I_o}[/latex]

Change in intensity level with intensities

Consider, two sound waves of intensities with I₁ and I₂ at distances r₁ and r₂ respectively. If I_o is the threshold of hearing, then,

[latex]\frac{I_2}{I_1}=\frac{I_2/I_o}{I_1/I_o}[/latex]

log[latex]\frac{I_2}{I_1}[/latex]  = log[latex]\frac{I_2}{I_o}[/latex]  – log[latex]\frac{I_1}{I_o}[/latex]

Multiplying both sides by 10, we get,

10log[latex]\frac{I_2}{I_1}  = 10 log\frac{I_2}{I_o}  – 10log\frac{I_1}{I_o}[/latex]

            = β₂ – β₁, where, β₂ = 10 log[latex]\frac{I_2}{I_o}[/latex]  and β₁ = 10log[latex]\frac{I_1}{I_o}[/latex] , are the intensities level due to I₂ and I₁ respectively.

[latex]\therefore[/latex] β₂ = β₁ + 10log[latex]\frac{I_2}{I_1}[/latex] ………. (i)

Now, from inverse square law, I ∝ [latex]\frac{1}{r^2}[/latex]

[latex]\therefore \frac{I_2}{I_1} = \frac{r_1^2}{r_2^2}[/latex]

So, [latex]\frac{I_2}{I_1}=(\frac{r_1}{r_2})^2[/latex]

So, eq^n. (i) becomes,

[latex]\beta_2 = \beta_1 + 10log(\frac{r_1}{r_2})^2[/latex]

Doppler Effect

When there is relative motion between a source of sound and an observer, the frequency of sound heard by the observer is different from the actual frequency of the source. This Apparent change in frequency due to relative motion of source and observer is called Doppler’s effect.

Calculation of Apparent frequency

a) When the source moves towards the stationary observer

Consider, a source of sound ‘S’ is moving towards a stationary observer ‘O’ with velocity v_s. Let, ‘v’ be the velocity of sound in the medium and f is the actual frequency of the source. If the source ‘S’ were stationary, the sound waves would distribute along a distance v so that the wavelength would be

[latex]\lambda = \frac{v}{f}[/latex]

But, when the source moves with velocity v_s, these sound waves are now confined to a shorter distance (v-v_s) so that the wavelength of the wave becomes,

[latex]\lambda’ = \frac{v-v_s}{f}[/latex]

Due to this apparent decrease in wavelength, the apparent frequency also changes. Hence, the apparent frequency is given by:

f’ = [latex]\frac{v’}{\lambda’} = \frac{v}{\frac{v-v_s}{f}}[/latex]    (velocity constant)

[latex]\therefore f’ = \frac{v}{v-v_s}\times f[/latex] Since, v > (v – v_s), f ^‘ > f. Hence, apparent frequency (pitch) increases when the source of sound moves towards a stationary observer.

b) When the source moves away from a stationary observer

Consider, a source of sound ‘S’ is moving away from a stationary observer ‘O’ with velocity v_s. Let, ‘v’ be the velocity of sound in the medium and f is the actual frequency of the source. If the source ‘S’ was stationary, the f-waves would distribute along a distance v so that the wavelength would be

[latex]\lambda = \frac{v}{f}[/latex]

But, when the source moves away with velocity v_s, these f- waves are now spread to a greater distance (v+v_s) so that the wavelength of the wave becomes,

[latex]\lambda’ = \frac{v+v_s}{f}[/latex]

Due to this apparent increase in wavelength, the apparent frequency also changes. Hence, the apparent frequency is given by:

[latex]f’ = \frac{v’}{\lambda’} = \frac{v}{v+v_s}[/latex]    (velocity constant)

[latex]\therefore f’ = \frac{v}{v+v_s}\times f.[/latex] Since, v < (v + v_s), f ^‘ < f. Hence, apparent frequency (pitch) decreases when the source of sound moves away from a stationary observer.

c) When an observer moves towards stationary source

Consider, an observer is moving towards the stationary source with velocity v_o. Let, v be the velocity of sound in the medium and f be the actual frequency of the source. When observer moves, there is no change in wavelength of the wave but there is relative velocity between the velocity of sound and velocity of observer.

Since, they are moving in opposite direction, the relative velocity v’ is:

v’ = v + v_o

Hence, the apparent frequency is given by:

[latex]f’ = \frac{v’}{\lambda’} = \frac{v+v_o}{\lambda} [\because \lambda’ = \lambda][/latex]

[latex]f’ = \frac{v+v_o}{\frac{v}{f}}[/latex]

[latex]f’ = \frac{v+v_o}{v}\times f[/latex]

Since, (v + v_o) > v, f ^‘ > f. Hence, apparent frequency (pitch) increases when the observer moves towards a stationary source.

d) When an observer moves away from a stationary source

Consider, an observer is moving away from the stationary source with velocity v_o. Let, v be the velocity of sound in the medium and ‘f’ be the actual frequency of the source. When observer moves, there is no change in the wavelength of the wave but there is relative velocity between the velocity of sound and velocity of observer.

Since, they are moving in same direction, the relative velocity is given by:

v^‘ = v – v_o.

Hence, the apparent frequency is given by:

[latex]f’ = \frac{v’}{\lambda’} = \frac{v-v_o}{\lambda} [\because \lambda = \lambda’][/latex]

f’ = [latex]\frac{v-v_o}{\frac{v}{f}}[/latex]

[latex]\therefore f’ = \frac{v-v_o}{v}\times f[/latex] Since, (v – v_o) < v, f ^‘ < f. Hence, apparent frequency (pitch) decreases when the observer moves away from a stationary source.

e) When an observer and a source move towards each other

Consider, an observer O and a source S are moving towards each other with velocity v_o and v_s respectively. Let, v be the velocity of sound in the medium and ‘f’ be the actual frequency of the source. Since, both observer and source are moving, relative velocity and wavelength of the sound wave gets changed. Hence, the apparent frequency is given by:

f^‘ = [latex]\frac{v’}{\lambda’}[/latex]
 = [latex]\frac{v+v_o}{\frac{v-v_s}{f}}[/latex]

[latex]\therefore f’ = \frac{v+v_o}{v-v_o} \times f[/latex]

Since, v + v_o > v – v_s, f ^‘ > f. Hence, the apparent frequency increases when source and observer are moving towards each other.

f) When an observer and a source move away from each other

Consider, an observer O and a source S are moving away from each other with velocity v_o and v_s respectively. Let, v be the velocity of sound in the medium and ‘f’ be the frequency of the source. Since, both observer and source are moving away from each other, relative velocity and wavelength of the sound waves gets changed. Hence, the apparent frequency

f^‘ = [latex]\frac{v’}{\lambda’}[/latex]
 = [latex]\frac{v-v)o}{\frac{v+v_s}{f}}[/latex]

[latex]\therefore f’ = \frac{v-v_o}{v+v_s} \times f[/latex]

Since, v – v_o < v + v_s, f ^‘ < f. Hence, the apparent frequency decreases when source and observer are moving away from each other.

g) When source and observer are moving in same direction (Observer is ahead of the source)

Consider, an observer O and a source S are moving in the same direction. Let, v be the velocity of sound in the medium and f be the actual frequency of the source. Since, both observer and source are moving in same direction, relative velocity and wavelength changes. So,

f^‘ = [latex]\frac{v’}{\lambda’}[/latex]
 = [latex]\frac{v-v_o}{\frac{v-v_s}{f}}[/latex]

[latex]\therefore f’ = \frac{v-v_o}{v-v_s} \times f[/latex]

If v_o > v_s, [latex]v-v_o < v-v_s[/latex], so, f ^‘ < f.

If v_o < v_s, [latex]v-v_o > v-v_s[/latex], so, f ^‘ > f.

h) When source and observer are in same direction (source is ahead of the observer)

If the source and observer are moving in the same direction, v and v_o are in the opposite directions, so that the velocity of sound waves relative to O is: v^‘ = v + v_o.

Since, the source is moving away from the observer, wavelength of the wave reaching O is: [latex]\lambda’ = \frac{v+v_s}{f}[/latex]

[latex]\therefore f’ = \frac{v’}{\lambda’}[/latex]

 = [latex]\frac{v+v_o}{\frac{v+v_s}{f}}[/latex]

[latex]\therefore f’ = \frac{v+v_o}{v+v_s} \times f[/latex]

Hence, if v_o = v_s, then f ‘ = f.

Limitation of Doppler’s Effect

The Doppler’s effect is not applicable in following conditions:

1. If the velocity of the source is greater than that of the sound, because the wave gets distorted due to which no change in frequency will be observed. i.e., source moves with supersonic velocity (velocity more than velocity of sound).
2. If the velocity of the observer is greater than that of the sound.

Doppler’s effect in light

Just like sound, there is change in frequency and wavelength of light when a star moves towards the earth or away from the earth. When the star moves towards the earth, the wavelength of the source decreases which means increase in the frequency of the light wave. The star is said to be blue shifted. While, when the star moves away from the earth, the wavelength of the source increases which means decrease in the frequency of the light wave. The star is said to be red shifted.

Let, [latex]\lambda[/latex] be the wavelength of the light emitted by the star when it is in the stationary position and [latex]\lambda'[/latex] be that when the star is in motion. If u_s is the velocity of the star and c be the velocity of light, then,
[latex]\lambda’ = \frac{c-u_s}{c}\lambda[/latex]

            = [latex](1 – \frac{u_s}{c})\times \lambda[/latex]

            = [latex]\lambda – \frac{u_s}{c}\lambda[/latex]

Or, [latex]\frac{u_s}{c}\lambda = \lambda – \lambda'[/latex]

Or, u_s = [latex]\frac{c(\lambda – \lambda’)}{\lambda}[/latex]

With the help of above equation, the velocity of the star can be calculated provided [latex]\lambda[/latex] and the shift ([latex]\lambda – \lambda'[/latex]) are known.

Application of Doppler’s Effect

1. The velocity of the star approaching the earth or moving away from the earth can be calculated.
2. Doppler’s effect in light gives us enough information to predict that the universe is expanding as the spectral lines emitted by stars appears to shift towards the red end.
3. This concept can be useful for military army in radar in order to locate the position, velocity and altitude of airplanes, fighter planes, etc. with the help of microwaves.

Infrasonic and Ultrasonic Sound

The sound produced below the audible frequency (frequency of sound as heard by normal human i.e., between 20 Hz to 20 KHz) is called infrasonic. Its range is below 20 Hz. It is produced by large objects like earthquake waves.

The sound of frequency above the audible (i.e., above 20 KHz) are called ultrasonic. It has very short wavelength.