Class 12 Physics Acoustic phenomena Notes

Class 12 Physics Acoustic phenomena Notes
UNIT 3
CLASS 12 PHYSICS • WAVE AND OPTICS

Acoustic Phenomena

Chapter 9
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Doppler Effect

The opening line in the scanned note is written as: f = v/λ; velocity changes when observer is in motion and wavelength changes when source is in motion.
The apparent change in frequency due to relative motion between source and observer is known as Doppler effect.

For example, the pitch of horn of car appears to increase as it approaches a stationary observer and pitch appears to decrease as car passes the observer.

us = velocity of source
uo = velocity of observer
v = velocity of sound
f = frequency of sound
f′ = apparent frequency
λ′ = apparent wavelength

1. Source in Motion and Observer at Rest

Doppler effect for a moving source and stationary observer Source Observer uₛ Source moving towards observer Source Observer uₛ Source moving away
Moving source with stationary observer

a) Source Moving Towards Stationary Observer

For a moving source, the apparent wavelength is:

f′ = v/λ′
λ′ = (v − us)/f
f′ = v / [(v − us)/f]
f′ = [v/(v − us)]f
Therefore, f′ > f.

Hence apparent frequency increases when source is moving towards stationary observer.

b) Source Moving Away from Stationary Observer

f′ = v/λ′
λ′ = (v + us)/f
f′ = v / [(v + us)/f]
f′ = [v/(v + us)]f
Therefore, f′ < f.

Hence apparent frequency decreases when source is moving away from stationary observer.

2. Observer in Motion and Source at Rest

Doppler effect for moving observer and stationary source Source Observer uₒ Observer moving towards source Source Observer uₒ Observer moving away
Moving observer with stationary source

a) Observer Moving Towards Stationary Source

f′ = v′/λ
λ = v/f
v′ = v + uo
f′ = (v + uo)/(v/f)
f′ = [(v + uo)/v]f
Therefore, f′ > f.

Hence apparent frequency increases when observer is moving toward stationary source.

b) Observer Moving Away from Stationary Source

f′ = v′/λ
v′ = v − uo
λ = v/f
f′ = (v − uo)/(v/f)
f′ = [(v − uo)/v]f
Therefore, f > f′.

Hence apparent frequency decreases when observer is moving away from stationary source.

3. Both Source and Observer are in Motion

Four relative-motion cases for source and observer in Doppler effect Source Observer Moving away from each other Source Observer Approaching towards each other Source Observer Source toward, observer away Source Observer Source away, observer toward
Four source-observer motion cases

a) Source and Observer Moving Away from Each Other

f′ = [(v − uo)/(v + us)]f

Hence apparent frequency decreases when source and observer are moving away from each other.

b) Both Source and Observer Approaching Towards Each Other

f′ = [(v + uo)/(v − us)]f

Hence apparent frequency increases when source and observer approach towards each other.

c) Observer Moving Away from Source and Source Moving Towards Observer

f′ = [(v − uo)/(v − us)]f
If us > uo, apparent frequency increases.
If uo > us, apparent frequency decreases.

d) Source Moving Away from Observer and Observer Moving Towards Source

f′ = [(v + uo)/(v + us)]f
If us > uo, apparent frequency decreases.
If uo > us, apparent frequency increases.

Numericals — Doppler Effect

Q.1 — Source and Observer Cases

A source of sound generates sound waves which travel with a speed of 340 ms−1. The frequency of the source is 500 Hz. Find the frequency of the sound heard if: (i) the source is moving towards the stationary observer with a speed of 30 ms−1; (ii) the observer is moving towards the stationary source with a speed of 30 ms−1; (iii) both source and observer move with a speed of 20 ms−1 and approach one another.

v = 340 m/s,   f = 500 Hz

(i) us = 30 m/s

f′ = [v/(v − us)]f
= [340/(340 − 30)] × 500
f′ = 548.38 Hz

(ii) uo = 30 m/s

f′ = [(v + uo)/v]f
= [(340 + 30)/340] × 500
f′ = 544.11 Hz

(iii) The source line first writes uo = us = 20 m/s, but the substitution uses 30 m/s.

f′ = [(v + uo)/(v − us)]f
= [(340 + 30)/(340 − 30)] × 500
f′ = 596 Hz   [as written in the scan]
The handwritten statement and substitution in part (iii) use different speeds. This version preserves that inconsistency rather than silently correcting it.

Q.2 — Echo Heard by a Car Driver

A car is approaching towards a cliff at a speed of 20 m/s. The driver sounds a whistle of frequency 800 Hz. What will be the frequency of the echo heard by the car driver? Velocity of sound in air = 350 m/s.

v = 350 m/s
f = 800 Hz
us = 20 m/s

Case I: Frequency received at the cliff

f′ = [v/(v − us)]f
= [350/(350 − 20)] × 800
f′ = 848.48 Hz

Case II: Reflected sound heard by moving driver

f″ = [(v + uo)/v]f′
= [(350 + 20)/350] × 848.48
f″ = 896.96 Hz

Q.3 — Car Horn Passing a Stationary Observer

A car sounding a horn and producing a note of 500 Hz approaches and then passes a stationary observer at a steady speed of 20 ms−1. Calculate the change in frequency heard by the observer. Velocity of sound is 340 m/s.

v = 340 m/s,   f = 500 Hz,   us = 20 m/s

Approaching:

f′ = [v/(v − us)]f
= [340/(340 − 20)] × 500
f′ = 531 Hz

Moving away:

f″ = [v/(v + us)]f
= [340/(340 + 20)] × 500
f″ = 472 Hz

Change in frequency:

f′ − f″ = 531 − 472
= 59 Hz

Q.4 — Moving Observer Passing a Stationary Source

An observer travelling with constant velocity of 20 m/s passes close to a stationary source of sound and notices that there is a change of frequency of 50 Hz as he passes the source. What is the frequency of source? Speed of sound in air is 340 m/s.

uo = 20 m/s,   v = 340 m/s

Observer approaching source:

f′ = [(v + uo)/v]f
= [(340 + 20)/340]f
= 36f/34

Observer moving away:

f″ = [(v − uo)/v]f
= [(340 − 20)/340]f
= 32f/34
f′ − f″ = 50
36f/34 − 32f/34 = 50
4f = 1700
f = 425 Hz

Q.5 — Motion Detector and Approaching Truck

A stationary motion detector sends sound waves of 150 kHz towards a truck approaching at a speed of 100 km/hr. What is the frequency of wave reflected back to detector?

In the handwritten working, 100 km/hr is used as about 33.33 m/s. The following steps reproduce the source values.
v = 340 m/s,   f = 150000 Hz

Case I:

f′ = [v/(v − us)]f
= [340/(340 − 33.33)] × 150000
f′ = 166.3 kHz

Case II:

f″ = [(v + uo)/v]f′
= [(340 + 33.33)/340] × 166.3
f″ = 182.6 kHz

Threshold of Hearing

The minimum intensity of sound that can be perceived by human ear is called threshold of hearing. The threshold of hearing is taken as 10−12 watt/m2 for a pure tone of frequency 1000 Hz for a normal ear.

Unit of Intensity

The intensity is defined as the sound energy emitted or received per unit area per second.
I = Sound energy / (Area × time)
= J/(m2s)
I = watt/m2

Inverse Square Law

It states that the intensity of sound at a point is inversely proportional to the square of distance of the point from the source of sound.
Inverse square law showing sound spreading from a source to two distances Source r₁ r₂ I ∝ 1/r² I₁/I₂ = r₂²/r₁²
Inverse square relation of sound intensity
I ∝ 1/d2
If I1 and I2 are intensities at distances d1 and d2 respectively:
I1/I2 = d22/d12

Relation Between Intensity and Loudness

Since the loudness ‘L’ of a sound is directly proportional to the logarithm of its intensity:

L ∝ log10I
L = k log10I   — (i), where k is proportionality constant

Let I0 be the intensity of sound at threshold of hearing. Then loudness L0 for the corresponding threshold of hearing is:

L0 = k log10I0   — (ii)

Difference in loudness:

L − L0 = k log10I − k log10I0
L − L0 = k log10(I/I0)

Since loudness at threshold of hearing is taken as zero:

L0 = 0
L = k log10(I/I0)

Taking k = 1:

L = log10(I/I0)

Intensity Level

L = log10(I/I0)
where I = intensity of wave
I0 = intensity at threshold of hearing

Unit of Loudness

Bel (B)

The unit of loudness is called Bel ‘B’.

L = log10(I/I0)
When I = 10I0:
L = log10(10I0/I0)
L = 1 Bel

Thus the loudness of sound is said to be 1 Bel if its intensity is 10 times more than that of threshold of hearing.

Decibel (dB)

A small unit of loudness is decibel (dB). One decibel loudness is 10 times smaller than 1 Bel.

L = 10 log10(I/I0)

Intensity Level and Distance

I ∝ 1/r2
I1/I2 = r22/r12
L1 = 10 log I1
L2 = 10 log I2
L1 − L2 = 10 log(I1/I2)
L1 − L2 = 10 log(r22/r12)

Numericals — Loudness, Beats and Temperature

Q.6 — Loudspeaker Intensity Level at Different Distance

The intensity level from a loud speaker is 100 dB at a distance of 10 m. What is its intensity level at a distance of 100 m?

L1 = 100 dB
r1 = 10 m
r2 = 100 m
L1 − L2 = 10 log(r22/r12)
100 − L2 = 10 log(10000/100)
100 − L2 = 10 × 2
L2 = 80 dB

Q.7 — Frequency of a Note from Beats

A note produces 2 beats with a tuning fork of frequency 480 Hz and 6 beats with a tuning fork of 472 Hz. Find the frequency of the note.

With 480 Hz fork and 2 beats/s:
Frequency of note = 480 ± 2
= 482 Hz or 478 Hz

With 472 Hz fork and 6 beats/s:

Frequency of note = 472 ± 6
= 478 Hz or 466 Hz
Therefore, f = 478 Hz

Q.8 — Air Column and Loaded Tuning Fork

A column of air is set into vibration and the note emitted gives 10 beats per second when a tuning fork of frequency 440 Hz is sounded, the temperature being 20°C. The frequency of beats decreases when the tuning fork is loaded with a small piece of wax. At what temperature will the unloaded fork and the air column be in unison?

At 20°C:

f = 440 − 10
f = 430 Hz
v20 = fλ = 430λ   — (i)

Let the required temperature be T:

vT = 440λ   — (ii)
vT/v20 = 440λ/430λ   — (iii)

Also:

vT/v20 = √(T/293)   — (iv)

From equations (iii) and (iv):

√(T/293) = 440/430
T/293 = (44/43)2
T = 306.33 K

Acoustic Phenomenon

The branch of physics which deals with the producing, transmission and controlling factor relating to the design and construction of sound is called Acoustic Phenomenon.

Music

The sound which gives pleasant sensation to the ear is called music, e.g. sound produced by guitar, flute etc.

Noise

The sound which gives unpleasant sensation to ear is called noise, e.g. sound produced by airplane.

Characteristics of Musical Sound

There are three fundamental characteristics of musical sound:

1. Pitch

The sharpness or flatness of musical sound is known as pitch. Higher the frequency, higher will be pitch and vice-versa. For example, pitch of voice of girl is higher as compared to pitch of boys.

Pitch depends upon following factors:

  1. The frequency of sound source.
  2. The relative motion between source of sound and observer. This is called Doppler effect.

2. Loudness or Intensity

The loudness is magnitude of hearing sensation produced by musical sound. It depends upon sensitivity of the ear as well as intensity of sound.

3. Quality or Timbre

The quality of musical sound is the characteristic which enables us to distinguish between two sounds of same pitch and loudness.

Intensity of Sound

The intensity of sound wave at a place is defined as the rate of flow of sound energy per unit area held normally to the direction of wave propagation.

The displacement (y) of a vibrating layer of air due to the propagation of wave is given by:

y = a sinωt   — (i)
where a = amplitude
ω = angular velocity

Let v be the velocity of sound at any instant:

v = dy/dt
= d(a sinωt)/dt
v = aω cosωt   — (ii)

But kinetic energy is:

E = ½mv2
= ½ma2ω2cos2ωt

For maximum K.E., cosωt = 1:

Maximum K.E. (E0) = ½mω2a2   — (iii)

If v be the velocity of sound measured in one second, then length (l) of layer of air distributed in one second is:

l = v

Volume of air in one second:

V = A × l
V = v   [if A = 1 m2]

The mass of air distributed in one second:

m = Vρ
m = vρ

Maximum K.E. per unit area per second is intensity:

I = ½vρ(2πf)2a2
I ∝ a2

Hence, intensity of sound is directly proportional to square of amplitude of vibration.

Pressure Amplitude

Sound wave is longitudinal wave. This wave travels out in all direction from a source of sound. Suppose sound wave is travelling in x direction, then its displacement is given by:

y = a sin(ωt − kx)   — (i)
where a = amplitude
ω = angular velocity
k = 2π/λ = wave number
t = time

Here x and y are parallel because in longitudinal wave the displacement y is along the direction of wave travel.

Imaginary air cylinder showing displacement of two cross sections in a longitudinal wave Area A y₁ y₂ Δx Imaginary cylinder of air
Displacement of two cross-sections of an air cylinder

Let us consider an imaginary cylinder of cross-sectional area ‘A’ and length ‘Δx’. Then volume of cylinder is given by:

V = AΔx, when there is no wave.

When wave is produced, the size of cylinder is disturbed. Let left cross-sectional surface be displaced by y1 and right cross-sectional surface be displaced by y2. Therefore change in volume:

ΔV = A(y2 − y1)
ΔV = AΔy

If Δy is positive, volume increases and pressure in the cylinder decreases. If Δy is negative, volume decreases and pressure in the cylinder increases.

Hence fractional change in volume:

ΔV/V = AΔy/(AΔx)
ΔV/V = Δy/Δx
ΔV/V = dy/dx   — (ii)

Again pressure variation in cylinder is:

ΔP = −B(ΔV/V)   — (iii)
where B = bulk modulus

From equations (ii) and (iii):

ΔP = −B dy/dx
= −B d[a sin(ωt − kx)]/dx
= −Ba · d[sin(ωt − kx)]/d(ωt − kx) · d(ωt − kx)/dx
ΔP = −Bak cos(ωt − kx)   [as written in the source]
ΔP = ΔPm cos(ωt − kx)
ΔPm = Bak

ΔPm is pressure amplitude which is the maximum increase or decrease in pressure.

Also we have:

v = √(B/ρ)
v2 = B/ρ
B = v2ρ

Therefore:

ΔPm = v2ρak
ΔPm ∝ a

From above equation, pressure amplitude is directly proportional to amplitude.

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