PHYSICS 231 Sound 1
Sound: longitudinal waves A sound wave consist o longitudinal oscillations in the pressure o the medium that carries the sound wave. Thereore, in vacuum: there is no sound. 2
Relation between amplitude and intensity A x -A time (s) For sound, the intensity I goes linear with the amplitude o the longitudinal wave squared I~A 2 3
Intensity Intensity: rate o energy low through an area Power (P) J/s A (m 2 ) Intensity: I=P/A (J/m 2 s=w/m 2 ) Even i you have a powerul sound source (say a speaker), the intensity will be small when ar away. 4
Intensity and distance rom the source Sound rom a point source produces a spherical wave. Why does the sound get ainter urther away rom the source? 5
Intensity and distance r=1 I=P/(4r 2 )=P/(4) 1 r=2 I=P/(4r 2 )=P/(16) 4 r=3 I=P/(4r 2 )=P/(36) 9 The amount o energy passing through a spherical surace at distance r rom the source is constant, but the surace becomes larger. I 1 /I 2 =r 22 /r 1 2 I=Power/Surace=P/A=P/(4r 2 ) 6
Wave ronts sound emitted rom a point source are spherical. Far away rom that source, the wave are nearly plane. plane waves spherical waves 7
The speed o sound Depends on the how easy the material is compressed (elastic property) and how much the material resists acceleration (inertial property) v=(elastic property/inertial property) v=(b/) B: bulk modulus : density The velocity also depends on temperature. In air: v=331(t/273 K) so v=343 m/s at room temperature material Air (20 o C) Helium Water Aluminum Diamond speed o sound 343 m/s 972 m/s 1493 m/s 5100 m/s 12000 m/s 8
Quick question The speed o sound in air is aected in changes in: a) wavelength b) requency c) temperature d) amplitude e) none o the above 9
Intensity Faintest sound we can hear: I~1x10-12 W/m 2 (1000 Hz) Loudest sound we can stand: I~1 W/m 2 (1000 Hz) sound wave vibrating ear drum Factor o 10 12? Loudness works logarithmic 10
sound/decibel level =10log(I/I 0 ) I 0 =10-12 W/m 2 y=log 10 x inverse o x=10 y (y=ln(x) x=e y ) log(ab) =log(a)+log(b) log(a/b) =log(a)-log(b) log(a n ) =nlog(a) 11
decibels =10log(I/I 0 ) I 0 =10-12 W/m 2 An increase o 10 db: intensity o the sound is multiplied by a actor o 10. 2-1 =10 10=10log(I 2 /I 0 )-10log(I 1 /I 0 ) 10=10log(I 2 /I 1 ) 1=log(I 2 /I 1 ) 10=I 2 /I 1 I 2 =10I 1 12
sound levels Table o sound levels L and corresponding sound pressure and sound intensity Sound Sources Examples with distance Sound Pressure Level L p dbspl Sound Pressure p N/m 2 = Pa Sound Intensity I W/m 2 Jet aircrat, 50 m away 140 200 100 Threshold o pain 130 63.2 10 Threshold o discomort 120 20 1 Chainsaw, 1 m distance 110 6.3 0.1 Disco, 1 m rom speaker 100 2 0.01 Diesel truck, 10 m away 90 0.63 0.001 Kerbside o busy road, 5 m 80 0.2 0.0001 Vacuum cleaner, distance 1 m 70 0.063 0.00001 Conversational speech, 1 m 60 0.02 0.000001 Average home 50 0.0063 0.0000001 Quiet library 40 0.002 0.00000001 Quiet bedroom at night 30 0.00063 0.000000001 Background in TV studio 20 0.0002 0.0000000001 Rustling leaves in the distance 10 0.000063 0.00000000001 Threshold o hearing 0 0.00002 0.000000000001
Frequency vs intensity 1000 Hz 14
Example A person living at Cherry Lane (300 m rom the rail track) is tired o the noise o the passing trains and decides to move to Abbott (3.5 km rom the rail track). I the sound level o the trains was originally 70dB (vacuum cleaner), what is the sound level at Abbott? 15
example A machine produces sound with a level o 80dB. How many machines can you add beore exceeding 100dB? 16
PHYSICS 231 Doppler eect 17
Doppler eect: a non-moving source v sound =v sound / source you 18
doppler eect: a source moving towards you v source source you the distance between the wave ront is shortened v source v v sound vsound vsource prime : heard observable v sound v source The requency becomes larger: higher tone 19
doppler eect: a source moving away rom you the distance between the wave ront becomes longer you v source source v source v source v v v vsound : negative!!! sound sound v v source source The requency becomes lower: lower tone 20
doppler eect: you moving towards the source v sound additional waveronts detected per second : source you v observer v observer v observer v v sound sound 21
doppler eect: you moving away rom the source v sound source you additional waveronts detected per second : v observer v observer v : negative observer v observer v v sound sound 22
doppler eect: general source you v v v v observer source v observer : positive i moving towards to source v source : positive i moving towards the observer 23
question An ambulance is moving towards you with its sirens on. The requency o the sound you hear is than the requency you would hear i the ambulance were not moving at all. a) higher b) the same c) lower v v v v observer source 24
example A police car using its siren (requency 1200Hz) is driving west towards you over Grand River with a velocity o 25m/s. You are driving east over grand river, also with 25m/s. a)what is the requency o the sound rom the siren that you hear? b) What would happen i you were also driving west (behind the ambulance)? v sound =343 m/s a) b) 25
applications o doppler eect: weather radar Both humidity (relected intensity) and speed o clouds (doppler eect) are measured. 26
shock waves v v v v observer source what happens i v source v sound? vt sin=v sound /v source v source > v sound waveront: high pressure wave that carries a lot o energy 27
shock waves: passing jet 28
applications o the doppler eect: speed radar v v v v observer source v v v approachingcar 29
example 300m you 1000m jet A low-lying jet passes by at 300m away rom you. When it is 1 km urther, you hear the sonic shock-wave. What was the speed o the jet in mach? 30
PHYSICS 231 String instruments 31
standing wave v 1 v 2 i two waves travel in opposite directions and v 1 =v 2, the superposition o the two waves produces a standing wave: maxima and minima always appear at the same location 32
standing waves in a guitar string waves in the string travel back and orth and create standing waves. a wave bouncing back rom a ixed point, returns inverted 33
we can produce dierent wave lengths L L L 1 =2L 2 =L 3 =2L/3 L L 4 =2L/4 5 =2L/5 both ends ixed n =2L/n or L=n n /2 34
standing waves both ends ixed n =2L/n or L=n n /2 n n v 1 2 n nv 2L v 2L 2v 2L nv 2L n 1 n 2L F 1 : undamental requency nth harmonics F: tension in rope : mass per unit length 35
example: the guitar n th harmonics: depends where and how the string is struck note that several harmonics can be present and that non-harmonics are washed out n n 2L F tension can be varied by stretching the wire changes rom string to string: bass string is very heavy length can be chosen by placing ingers 36
beats Superposition o 2 waves with slightly dierent requency The amplitude changes as a unction o time, so the intensity o sound changes as a unction o time. The beat requency (number o intensity maxima/minima per second): beat = a - b 37
example A guitar string is struck. Assume that the irst harmonic is only excited. What happens to the requency i: a) The player put a inger at hal the length o the string? b) The player makes the tension 10% larger (by turning the tuning screw)? c) A string is struck in the same way, but its mass is 3 times higher? 38
example Someone is trying to tune a guitar. One o the strings is supposed to have a requency o 500 Hz. The person is using a tuning ork which produces a sound o exactly this requency, but while sounding the ork and the playing the guitar, hears a beat in the sound with a requency o 3 Hz (3 beat per second). a) What is the real requency o the guitar string? b) By what raction does the person need to change the tension o the guitar string to tune it properly? 39