Sound

Speed of Sound Recall for pulse on string: v = sqrt(F /  ) For fluids: v = sqrt(B/  ) 05 MediumSpeed (m/s) Air343 Helium972 Water1500 Steel5600

Velocity Question A sound wave having frequency f 0, speed v 0 and wavelength 0, is traveling through air when in encounters a large helium-filled balloon. Inside the balloon the frequency of the wave is f 1, its speed is v 1, and its wavelength is 1 Compare the speed of the sound wave inside and outside the balloon 1. v 1 < v 0 2. v 1 = v 0 3. v 1 > v 0 V 1 =965m/s V 0 =343m/s 10

Frequency Question A sound wave having frequency f 0, speed v 0 and wavelength 0, is traveling through air when in encounters a large helium-filled balloon. Inside the balloon the frequency of the wave is f 1, its speed is v 1, and its wavelength is 1 Compare the frequency of the sound wave inside and outside the balloon 1. f 1 < f 0 2. f 1 = f 0 3. f 1 > f 0 f1f1 f0f0 13 Time between wave peaks does not change!

Wavelength Question A sound wave having frequency f 0, speed v 0 and wavelength 0, is traveling through air when in encounters a large helium-filled balloon. Inside the balloon the frequency of the wave is f 1, its speed is v 1, and its wavelength is 1 Compare the wavelength of the sound wave inside and outside the balloon 1. 1 < = > = v / f 15

Intensity and Loudness l Intensity is the power per unit area. è I = P / A è Units: Watts/m 2 l For Sound Waves  I = p 0 2 / (2  v) (p o is the pressure amplitude) è Proportional to p 0 2 (note: Energy goes as A 2 ) l Loudness (Decibels) è Loudness perception is logarithmic è Threshold for hearing I 0 = W/m 2   = (10 dB) log 10 ( I / I 0 )   2 –  1 = (10 dB) log 10 (I 2 /I 1 ) 18

Log 10 Review l Log 10 (1) = l Log 10 (10) = l Log 10 (100) = l Log 10 (1,000) = l Log 10 (10,000,000,000) = 19

Decibels Question l If 1 person can shout with loudness 50 dB. How loud will it be when 100 people shout? 1) 52 dB2) 70 dB3) 150 dB 22

Intensity Question l Recall Intensity = P/A. If you are standing 6 meters from a speaker, and you walk towards it until you are 3 meters away, by what factor has the intensity of the sound increased? 1) 22) 43) 8 Area goes as d 2 so if you are ½ the distance the intensity will increase by a factor of 4 27 Speaker radiating power P I 1 = P/(4  D 1 2 ) D1D1 I 2 = P/(4  D 2 2 ) D2D2

Standing Waves in Pipes Open at both ends: Pressure Node at end  = 2 L / n n=1,2,3.. Open at one end: Pressure AntiNode at closed end : = 4 L / n n odd 29

Example: Organ Pipe A 0.9 m organ pipe (open at both ends) is measured to have it’s first harmonic at a frequency of 382 Hz. What is the speed of sound in the pipe? 32

Dopper shift l As a police car passes you with its siren on, the frequency of the sound you hear from its siren 1) Increases2) Decreases3) Same Doppler Example Audio Doppler Example Visual 36

Doppler Effect moving source v s l When source is coming toward you (v s > 0) è Distance between waves decreases è Frequency increases l When source is going away from you (v s < 0) è Distance between waves increases è Frequency decreases l f o = f s / (1- v s /v) 38

Doppler Effect moving observer (v o ) l When moving toward source (v o < 0) è Time between waves peaks decreases è Frequency increases l When away from source (v o > 0) è Time between waves peaks increases è Frequency decreases l f o = f s (1- v o /v) Combine: f o = f s (1-v o /v) / (1-v s /v) 40

Example: police car A police car is approaching you from behind so you pull over to the side of the road and stop. If the police car is moving at 25m/s and the siren is emitting a sound with frequency 200Hz. What frequency do you hear? 44 Speed of sound = 330m/s

Example: police car If the police car is moving at 25m/s and the siren is emitting a sound with frequency 200Hz. After the police car passes you what frequency do you hear? 44 Speed of sound = 330m/s

Superposition & Interference l Consider two harmonic waves A and B meeting at x=0.  Same amplitudes, but  2 = 1.15 x  1. l The displacement versus time for each is shown below: What does C(t) = A(t) + B(t) look like?? A(  1 t) B(  2 t) 46

Superposition & Interference l Consider two harmonic waves A and B meeting at x=0.  Same amplitudes, but  2 = 1.15 x  1. l The displacement versus time for each is shown below: A(  1 t) B(  2 t) CONSTRUCTIVE INTERFERENCE DESTRUCTIVE INTERFERENCE C(t) = A(t) + B(t) 47

Summary Speed of sound v = sqrt(B/  ) Intensity  = (10 dB) log 10 ( I / I 0 ) l Standing Waves è f n = n v/(2L) Open at both ends n=1,2,3… è f n = n v/(4L) Open at one end n=1,3,5… l Doppler Effect f o = f s (v-v o ) / (v-v s ) l Beats 50