AcousticsMATERIAL TO READ

Web:1. http://www.physics.uoguelph.ca/tutorials/L

OG/ on Logarithms2. http://www.physics.uoguelph.ca/biophysics/

Text:1. chapter 2, sections 1 and 2, pp. 1-24

Handbook:1. study guide 22. lab 2 - Acoustics

For quiz on study guide 2, you should know:

1. Power and intensity of sound, logs and antilogs2. Standing wave resonance and beats3. Echolocation4. Anatomy and function of ear and its components

For Lab quiz #9, you should know:

1. Oscilloscope pattern and sketch of apparatus formeasuring speed of sound

2. Standing wave resonances3. v = λf

SOUND WAVES

• are waves of molecular displacements• require a medium (gas, liquid or solid)• produce regions of compression and rarefaction

high low density density

Analogous to waves on a "slinky".

x

Direction of wave motion displacement

Longitudinal wave:Displacement of molecules from their equilibriumpositions is in the same direction as wave velocity.

Before displacement (denoted by "y") wasperpendicular to velocity: transverse wave.

Travelling sound wave

y = y0sin(2πt/T - 2πx/λ)

y = y0sin(2πf (t - x/v))

y = y0sin(ωt - kx)

sound velocity: v = λλf DEPENDS ON MEDIUM

Medium Speed (ms-1)Air 340

Seawater 1550Iron 5000

Rubber 54Wood 3300

Standing Wave ResonanceResonance: Establishment of a standing wave

(not always possible).e.g. string fixed at both ends. String is pluckedand disturbance is reflected at both ends.Resonant wavelengths reinforced, others damped -stable standing wave results. Ends are fixed,therefore nodes at both ends.

Harmonicsx = 0 x = L

1st harmonic, or (primary resonance) (fundamental harm.)

2nd harmonic

3rd harmonicx =0 x = L

standing wave: y = 2y0cos(2πt/T)sin(2πx/λ)(from previous SG)

nodes occur at positions, x, where y = 0sin(2πx/λ) = 0 ⇒ 2πx/λ = nπ, n = 0, 1, 2, …

Nodes at positions: x = nλλ/2, n = 0, 1, 2, …

MUST be a node at x = L (end of string, n ≠ 0)

∴∴ L = nλλ/2 ⇒⇒ λλ = 2L/n (n = 1, 2, 3, …)

Only standing waves with wavelengths satisfyingλ=2L/n can occur

In terms of frequency, f = v/λ

⇒n

Lf

2v

=, n = 1, 2, 3, …

harmonic Index frequency wavelength1st harmonic 1 1v/(2L)=v/2L 2L/12nd harmonic 2 2v/(2L) = v/L 2L/2 = L3rd harmonic 3 3v/(2L) 2L/3nth harmonic n nv/2L 2L/n

Pitch of sound is proportional to frequency

Pressing strings against the neck of instrumentdecreases length⇒ increases f (pitch)

NB: Waves on string are transverse waves, notsound waves. These waves transfer vibrations(through the bridge) to walls of instrument’s cavitywhich cause air inside to resonate as a sound wave.Most sound is due to “fundamental” or 1st harmonicfrequency.

Therefore to understand acoustics we must studyresonance in air cavities.

Standing sound waves in pipes (Expt. 2)

OPEN CLOSEDEND END

must be antinode must be node

Allowed standing waves are:λ/4 λ/4

L λ/2λ/4 λ/4

λ/2 λ/2Distance between adjacent nodes = λλ/2

Distance from antinode to nearest node = λλ/4

Therefore to set up a standing wave,L = λλ/4 + nλλ/2 , n = 0, 1, 2, … = (1 + 2n)λλ/4 = m/4 , m = 1, 3, 5, … (odd)

Exercise II of lab 2

speaker L

piston

• Speaker buzzes with preset f : (500–2500 Hz)• Wave speed, v = 340 m/s (sound in air)• Hence wavelength λ = v/ f is fixed

Hence standing wave resonance is only heard when:

L = mλλ/4, m = 1, 3, 5, …e.g. shortest resonance

length = L1

longest resonance length = L2

L2 – L1 = k (λλ/2) = 2λ/2 (k = 1 + # of resonances between L1 and L2)

Pipe open at both ends

Antinode required at both ends

1st harmonic 2nd harmonic

In general;

Length L = n(λλ/2), n = 1, 2, 3, …

Resonance due to standing sound waves in pipesdetermines different notes (i.e. frequencies orpitches) of organs, brass and woodwindinstruments (and kazoos…).

Later we will see how resonance in auditory canalof ear amplifies sounds for easier detection.

BeatsDue to interference between sound waves of nearequal intensity and slightly different frequency.

Superposition principle: Resultant (red) is thesum two individual waves: wave 1 and wave 2.

Beat cycle: time between minima in resultant ortime for both waves to return to initial phases.

Beat frequency: fB = f2 - f 1 NB. fB > 0 (abs. val.)

e.g.: Two tuning forks are set side by side. Onehas f = 350 Hz. Five beats per second are heard.What are the possible frequencies of the 2nd fork?

1 2 3 4 5

1 2 3 4 5 6

BEAT CYCLE time

ENERGY, POWER AND INTENSITYOF SOUND

Sound wave energy is carried by the kinetic energyof the oscillations between the molecules.

Units and dimensions of energy:

Energy (E) = work = W = Force ×× distance

Force (F) = Mass (M) ×× acceleration (a)

Acceleration = rate of change of velocitya = v/t

velocity = rate of change of position, Lv = L/t ⇒ a = L/t2

Therefore the dimensions of Energy are: E = FL = MaL = ML2/t2

In SI units L = meter (m), t = second (s) and M =kilogram (kg)

⇒ SI unit for E is kgm2/s2 ≡ Joule (J)

Power (P) = rate of change of energy or energyper unit time.

P = E/t SI unit: J/s ≡ Watt (W)

Intensity (I) = power per unit area

I = P/A SI unit W/m2

Consider a source of sound:[A]

Assume sound isradiated equallyin all directions

r1 [B]Assume sound energyis not absorbed

r2 by surroundingmedium

⇓⇓ENERGY IS CONSERVED

∴Energy (or power) passing thru imaginary sphereat radius r = energy (power) emitted by source.e.g.:power thru sphere at r1 = power thru sphere at r2

Power is evenly distributed over entire surface ofeach sphere from assumption [A].

Area of sphere (radius r) = A = 4πr2

∴∴ Intensity of sound at distance r from source:

I = P/A = P/(4ππr2)

The intensity of sound decreases as the inversesquare of the distance from the source.

A sound’s loudness depends upon its intensity.

Sample problem

A foghorn is used to warn ships that rocky shoalsare nearby. The intensity of the foghorn at 1.0 mdistance is 10-3 W/m2. At sea, a sound intensity of10-8 W/m2 is required for a person to hear it. Howfar away can the horn be heard?

Given: Intensity at r1 = 1.0 m is I1 = 10-3 W/m2

I1 = P/(4πr12)

P = I1 (4πr12) = 4π(1.0 m)2 × 10-3 W/m2

= 4π × 10-3 W

Because energy is conserved, power through ANYsphere enclosing the source is 4π × 10-3 W. Wewant sphere of radius r2 such that I2 = 10-8 W/m2.

I2 = P/(4πr22)

r2 = [P/(4πI2)] ½

r2 = [4π × 10-3/(4π × 10-8)] ½

r2 = 3.2 × 102 m

Sound Intensity Level: LogarithmsEar’s sensitivity to sound varies with logarithm ofintensity.

Notation: x = zy (z raised to power y)⇒ y = logz(x) (log of x to base z)⇒ x = antilogz(y)

For sound we only use log’s to base 10. If: x = 10y

Then: y = log(x)

Notation base 10 often omitted

e.g.: 102 = 100 log(100) = 2103 =1000 log(1000) = 3

Rules: 1) log(xa) = alog(x)

2) log(x1x2) = log(x1) + log(x2)

3) log(x1/x2) = log(x1) - log(x2)

antilog(y) = 10y (function 10y on calculator)

Solved examples:Evaluate:1) the antilog of 0.85 = antilog(0.85) = 10.85 = 7.08

2) 6.000.420 = 2.12 (use “yx” function on calculator)OR

Let x = 6.000.420

Then log(x)=log(6.000.420) = 0.420×log(6.00)= 0.327

x = antilog(0.327) = 100.327 = 2.12

3) z where 5.0 × 6.0z = 4.0(z-2)

take log of both sides:

⇒ log(5.0 × 6.0z) = log(4.0(z-2))

⇒ log(5.0) + zlog(6.0) = (z-2)log(4.0)

⇒ z[log(6.0) - log(4.0)] = -2log(4.0) –log(5.0)

⇒ z log(6/4) = -log(16×5)

⇒ z = -log(80)/log(1.5) CAN’T SIMPLIFY MORE

⇒ z = -1.90/0.176 = -10.8

Sound Intensity Level

Intensity Level = L = 10 log(I/I0)

Although L is dimensionless, a unit, the decibel (db)is defined in conjunction with above equation.

I0 is a “standard” or “reference” intensity.For humans, I0 = 10–12 W/m2 (~minimum

audible intensity)Sample Problems1) A bat produces a pulse of intensity 10-2 W/m2

at a distance of a few centimeters. What isthe intensity level?

L = 10log(10-2/10-12) = 10log(10) = 10 × 10 = 100 db

2) A bat increases the intensity level of its soundpulse by 7.8 db. By what factor has theintensity increased? (S.G. page 2-4)

Must find ratio (Inew/Iold) given Lnew – Lold = 7.8 db

Lnew = 10log(Inew/I0) Lold = 10log(Iold/I0)

Lnew - Lold = 10[log(Inew/I0) - 10log(Iold/I0)] = 10log(Inew/Iold)

7.8 = 10log(Inew/Iold)0.78 = log(Inew/Iold)

107.8 = Inew/Iold = 6.0

3) The threshold of hearing for dogs is lower(10-15 W/m2) than for humans. If a sound isjudged by a human to be 40 db, what is the dbrating judged by the dog?

Lhuman = 10log(I/10-12) Ldog = 10log(I/10-15)

Ldog - Lhuman = 10log[(I/10-12)/(I/10-15)] = 10log(103) = 30 dB

Ldog = Lhuman + 30 dB = 70 dB

4) What is the intensity level produced of fouridentical loudspeakers if each produces anintensity of 60 db at a given distance?

Sound wave intensities are additive (superpositionprinciple))

If I1 = intensity from one loudspeaker, I = 4I1 isthe intensity for all four.

⇒ L = 10log(I/10-12) = 10log(4I1/10-12) = 10log(4) + 10log(I1/10-12)

⇒ L = 10log(4) + L1 This is just L1,intensity level

⇒ L = 6.0 + 60.0 from 1 speaker

⇒ L = 66 db NB: Increasing intensity byfactor of 4 has only changedintensity level by 10%

EcholocationPulses of sound are emitted and reflected fromtarget. Echo is detected which can indicatedistance, speed, size and texture of object. Usedby bats, porpoises, whales and submarines).Consider a sound source with power, P:

Imaginary sphereof radius r,area A = 4πr2

some smallsegment onsphere, area = a

Intensity of sound at distance r from source,

I = P/A = P/4πr2

Acoustical power passing through area, a:

Pa = Ia NB: PA = IA for entire sphere

Area, a, now becomes a ”source“ with power ∝ Pa.

Sample Problem (Modified from text prob. #2-13)

In searching for flying insects, a bat emits highfrequency pulses, about 3 ms long, having anintensity level of about 120 db at 1m from the batsmouth. (Use I0 = 10-12 W/m2 : as for humans.)

a) What is the acoustical power associatedwith each pulse?

L = 120 db at r = 1 mL = 10log(I/I0)10L/10 = I/I0

10120/10 = I/10-12

10-121012 = 1 W/m2 = I

P = IA = 1(4π(1)2) = 4ππ W

b) How much acoustical energy is associatedwith each pulse?

E = Pt = 4π × (3 × 10-3 s) =12ππ ××10-3 J

Duration of pulse

c) 0.01 s after emission, a pulse strikes aninsect. How far is the insect from the bat?

d = velocity × time (travel time for pulse)d = vt = 340 m/s × 0.01 s = 3.4 m

d) The effective area of the insect is 10 mm2.How much of the bat’s acoustical power isintercepted by the insect?

First we find the intensity at distance d

Id = P/(4πd2) Power emitted by bat = 4π W

Id = 4 π/(4π(3.4)2) = (3.4)-2 = 0.087 W/m2

Power intercepted by insect is: PI = IdaI

aI = 10 mm2 = 10 (10-3 m)2 = 10-5 m2 = Area of insect

PI = 0.087 × 10-5 = 8.7 ×× 10-7 W

e) The sound pulse is echoed back to the bat.What is the intensity of the echo at theposition of the bat?

Assume all power strikingthe insect, PI is reflected.

http://www.batworld.org/bats.html

PI is now the “source” power emitted in alldirections. Intensity due to this power at theposition of the bat, Ib is:

Ib = PI/(4πd2) Assuming the bat and insect movevery little during pulse flight

Therefore: Ib = (8.7 x 10-7 W)/(4π(3.4 m)2)

= 6.0 ×× 10-9 W/m2

f) What intensity level does the bat hear forthe echoed pulse?

L = 10log(I/I0) = 10log[(6.0 × 10-9)/(10-12)]

= 10log[6 × 103]

= 10[log(6) + 3]

= 37.8 db

Displacement Amplitude of Sound WaveEnergy transported by wave is related to thekinetic energy, K of the molecules each with mass,m. For molecule i,

Ei = Ki + Pi = ½ mvi2 + ½ kyi

2

where Pi = potential energy of molecule i, k is anelastic constant dependent upon the medium and vi

= velocity of molecule i. Energy is conserved,

[A] [B] [C]Pi = max Pi = 0 Pi = maxKi = 0 Ki = max Ki = 0Energy Energy EnergyMolecules = Molecules = Molecules rarefied equilibrated compressed

x+y 0 -y

yi ≅ average distance of molecule from equilibriumchanges as a travelling wave.

yi = y0sin(2πt/T - 2πx/λ)= y0sin[2πf (t – x/v)]

Velocity vi = dyi/dt = 2πνy0cos[2πf(t – x/v)]

vi has a maximum value: vi,max = 2πfy0

In case [B] above all the energy of the wave, willbe kinetic and vi = vi,max. Therefore for eachmolecule:

Ei = ½m(vi,max)2 = ½m(2πfy0)2 = 2π2mf2y02

TOTAL ENERGY OF WAVE of N molecules:E = NEi = N(2ππ2mf 2y0

2) = 2ππ2Mf 2y02

where M = Nm = total mass of medium

TOTAL ENERGY PER UNIT VOLUME:E/V = 2ππ2ρρf 2y0

2 where ρ = M/V = mass density

E/V relates A =areato the sound intensity:

distance moved in time ∆t = v∆∆t sound speed

Total energy transported = Energy in dashed box

Ebox = (E/V) × (volume of box)= (E/V) Av∆t

Intensity = P/unit area = (E/unit time)/unit areaI = (E/V)v = 2π2ρf 2y0

2v

Know how to use these 2 formulae (provided onformula sheets). Derivation only to motivateunderstanding

I = 2ππ2ρρf 2y02v

y0 = [I/(2ππ2ρρf 2 v)]½

y0 =molecular displacement amplitude in sound wave

Sample Problem The threshold of hearing at 4kHz for humans is 10-12 W/m2. What displacementamplitude is associated with the air molecules?(ρair = 1.3 kg/m3 : on formula sheets)

1222

-12

220 107.2(340))4000)(3.1(2

10

v2

Iy −×===

πρπ f m

(typical molecule ≈ 10-9 m)

Anatomy and Function of Ear and its Components

http://www.earaces.com/anatomy.htm

Section Special function Structures involvedOuter Amplification of

important frequenciesAuditory canal

Middle Amplification andtransduction of sound.(Air vibrations convertedto mechanical)

Tympanic membrane(Eardrum)Ossicles (small bones)

Inner Pitch resolution andperception of sound

Cochlea

Auditorycanal

Outer ear

Tympanicmembrane

Middle ear Inner ear

Ossicles

Amplification of sound in auditory canal

Mainly due to standing wave resonance. Canal islike a pipe with one open end and one closed end.

λ/4

eardrum

L ≈≈ 2.5 cmShown above: Fundamental Resonance when L =λ/4. ∴∴ λ = 4L = 10. cm = 0.10 m

∴∴ f = v/λ = 340/0.10 = 3400 Hzis the resonance frequency

Ear is most sensitive between 3000 and 4000 Hz.

Frequency /Hz

0 5000 10000 15000

Inte

nsity

Lev

el /

db

-20

0

20

40

60

80

Seetext:figure2-11forlog/logversion

Threshold of audibility in humans

Reference intensity = I0

NegativeLoudnesspossible

I0 ≈thresh-hold

at1000Hz

Transduction in middle ear

1. Tympanic membrane converts pressure variation(sound) into mechanical vibration

2. Ossicles (mallus,incus, stapes) transmit vibrationto inner ear with 20 times greater peak pressure

3. Vibration converted back into a pressure wave–entering cochlea at oval window.

Pitch resolution and perception in inner ear

Basilar membrane has hair-like vibration sensitiveneurons. Position of max. displacement depends onfrequency – therefore frequency can be resolved.

Position of maximumdisplacement for this frequency

Sounds with f >17000 Hz “spillenergy off” end

of membranethru

helicotrema

Basilar membrane