PHYSICS 220. Lecture 21. Textbook Sections Lecture 21 Purdue University, Physics 220 1

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1 PHYSICS 220 Lecture 21 Sound Textbook Sections Lecture 21 Purdue University, Physics 220 1

2 Overview Last Lecture Interference and Diffraction Constructive, destructive Diffraction: bending of waves around obstacles Coherence Same frequency and constant phase difference Reflection and Refraction fixed point inverts wave sinθ 1 /sinθ 2 = v 1 /v 2 Standing Waves (fixed ends) Today λ n = 2L/n f n = n v / 2L Sound Waves Lecture 21 Purdue University, Physics 220 2

3 Quiz 1) A violin string of length L is fixed at both ends. Which one of these is not a wavelength of a standing wave on the string? A) 5L/3 B) L/2 C) L D) 2L λ = 2 L n n 2) Two strings, one thick and the other thin, are connected to form one long string. A wave travels along the string and passes the point where the two strings are connected. Which of the following does not change after crossing that point: A) frequency B) wavelength C) propagation speed Lecture 21 Purdue University, Physics 220 3

4 Sound Waves Lecture 21 Purdue University, Physics 220 4

5 Speed of Sound Recall for pulse on string: v = sqrt(f / μ) For fluids: v = sqrt(b/ρ) B = bulk modulus: ΔP = B ΔV V For ideal gas: the speed of sound is proportional to the square root of the absolute temperature, re but independent of pressure and density Medium Speed (m/s) B ρ T v = v 0 T T0 T in Kelvin Room T = 20 0 C or 293 K Air 343 Helium 972 Water 1500 Steel 5600 Lecture 21 Purdue University, Physics 220 5

6 Speed of Sound Lecture 21 Purdue University, Physics 220 6

7 Speed of Sound The general relationship for a wave, v = ƒ λ holds The speed of the sound wave is a function of the medium and is independent of the frequency and amplitude The frequency and amplitude depend on how the wave is generated Frequency range of normal human hearing is 20 Hz to 20 khz Sound waves below 20 Hz are called infrasonic Sound waves above 20 khz are called ultrasonic Lecture 21 Purdue University, Physics 220 7

8 Intensity and Loudness Intensity is the power per unit area I = P / A Units: Watts/m 2 For Sound Waves I = p 0 2 / (2 ρ v) (p o is the pressure amplitude) Proportional to p 0 2 (note: Energy goes as A 2 ) Loudness (Decibels) Loudness perception is logarithmic i Threshold for hearing I 0 = W/m 2 β = (10 db) log 10 (I/I I 0 ) β 2 β 1 = (10 db) log 10 (I 2 /I 1 ) Pitch Perception of frequency Lecture 21 Purdue University, Physics 220 8

9 Log 10 Review Log 10 (1) = 0 Log 10 (10) = 1 Log 10 (100) = 2 Log 10 (1,000) = 3 log 10 (10 x )= x logx 2 = 2 log x Log 10 (10,000,000,000) ) = 10 Lecture 21 Purdue University, Physics 220 9

10 Decibels The sensitivity of the ear depends on the frequency of the sound In the most favorable frequency range, the ear can detect intensities as small as about W/m 2 A more convenient unit, the decibel (db), is often used to look at sound intensity The sound intensity level, β, is measured in decibels Lecture 21 Purdue University, Physics

11 Decibels Sound intensity level β β = (10 db) log 10 ( I / I 0 ) Units: Bels 5 I = 10 5 W/m 2 I W/m = A ratio of 10 7 indicates a sound intensity of 7 bels or 70 decibels (db) An intensity of 0 db corresponds to the hearing threshold Intensity increase by factor 10 intensity level adds 10 db Adding 3 db to the intensity level doubles the intensity ty (log 10 2=0.3) 03) Alexander Graham Bell ( ) Lecture 21 Purdue University, Physics

12 Intensity Level of Some Sounds Lecture 21 Purdue University, Physics

13 Question If 1 person can shout with loudness 50 db. How loud will it be when 100 people shout? A) 52 db B) 70 db C) 150 db β 100 β 1 = (10 db) log 10 (I 100 /I 1 ) β 100 = 50 + (10 db) log 10 (100/1) β 100 = Lecture 21 Purdue University, Physics

14 Standing Sound Waves Pipe open at both ends 2 λ = L n n Lecture 21 Purdue University, Physics

15 Standing Sound Waves Pipe open at one end 4 λ = L n n Lecture 21 Purdue University, Physics

16 Standing Waves in Pipes Open at both ends: Pressure node at end λ = 2 L / n n=1,2,3,,,... Open at one end: Pressure antinode at closed end: λ = 4 L / n n=1,3,5,... Lecture 21 Purdue University, Physics

17 Standing Wave Summary The closed end of a pipe is a pressure antinode for the standing wave The open end of a pipe is a pressure node for the standing wave A pressure node is always a displacement antinode, and a pressure antinode is always a displacement node These three points will allow you to predict the allowed frequencies of standing sound waves in pipes Lecture 21 Purdue University, Physics

18 Organ Pipe Example A 0.9 m organ pipe (open at both ends) is measured to have it s n=2 harmonic at a frequency of 382 Hz. What is the speed of sound in the pipe? Pressure node at each end. λ = 2 L / n n=1,2,3..,, λ = L for n=2 harmonic f=v/λ λ v = f λ = (382 s -1 ) (0.9 m) = 343 m/s Lecture 21 Purdue University, Physics

19 Resonance ILQ What happens to the fundamental frequency of a closed pipe, if the air (v=343 m/s) is replaced by helium (v=972 m/s)? A) Increases B) Same C) Decreases λ 1 =4L f = v/λ f = v/(4l) Lecture 21 Purdue University, Physics

20 Musical Tones A sound wave described by a single frequency is called a pure tone Most sounds are combinations of many pure tones Many sound are a combination of frequencies that are harmonically related One way to characterize a combination tone is by a property called pitch Generally, the pitch of a pure tone is its frequency With more complex sounds, the pitch is associated with the fundamental frequency Lecture 21 Purdue University, Physics

21 Real Musical Tones Pipes are the basis for many musical instruments The standing wave in the instrument is a combination of standing waves with different frequencies These combination tones are often said to have a property called timbre Also known as tone color The timbre depends on the mix of frequencies The timbre is different for different instruments and how the instrument is played Lecture 21 Purdue University, Physics

22 Timbre Fourier decomposition Fundamental + Overtones Lecture 21 Purdue University, Physics

23 Superposition & Interference Consider two harmonic waves A and B meeting at x=0. Same amplitudes, but ω 2 = 1.15 x ω 1. The displacement versus time for each is shown below: A(ω 1 t) B(ω 2 t) What does C(t) = A(t) + B(t) look like? Lecture 21 Purdue University, Physics

24 Superposition & Interference Consider two harmonic waves A and B meeting at x=0. Same amplitudes, but ω 2 = 1.15 x ω 1. The displacement versus time for each is shown below: A(ω 1 t) B(ω 2 t) C(t) = A(t) + B(t) DESTRUCTIVE INTERFERENCE CONSTRUCTIVE INTERFERENCE Lecture 21 Purdue University, Physics

25 Beats If two tones are played after one another, you will be able to detect a minimum frequency difference, ƒ For most people, ƒ 0.01 ƒ This is the smallest detectable frequency difference for tones that are played consecutively If two tones are played simultaneously, you can detect much smaller frequency differences When two tones are played together, the superposition principle indicates the sound pressures will add Lecture 21 Purdue University, Physics

26 Beat Frequency The amplitude of this combination pressure wave oscillates These amplitude oscillations are called beats The frequency of the oscillations is the beat frequency, ƒbeat ƒ beat = ƒ 1 ƒ 2 Beats can be used to tune musical instruments Lecture 21 Purdue University, Physics

27 Beats Add two cosines and remember the identity: where 1 2 and ( ) Acos( ωt) + Acos( ω t) = 2Acos ω t cos( ω t) 1 ωl ω1 ω2 2 = ( ) ( ) L H 1 ωh = ω1 + ω2 ω 2 Beat Frequency: cos(ω L t) ω = 2ω =Δ ω or f =Δf beat L beat Lecture 21 Purdue University, Physics

28 Doppler Effect Moving Source Lecture 21 Purdue University, Physics

29 Doppler Effect Moving Source When source is coming toward you (v s > 0) Distance between waves decreases Frequency increases When source is going g away from you (v s < 0) Distance between waves increases Frequency decreases f o =f s /(1 (1- v s /v) Lecture 21 Purdue University, Physics

30 iclicker A police car passes you with its siren on. The frequency of the sound you hear from its siren after the car passes A) Increases B) Decreases C) Same Lecture 21 Purdue University, Physics

31 Doppler Effect Moving Observer Lecture 21 Purdue University, Physics

32 Doppler Effect Moving Observer When moving toward source (v o < 0) Distance between waves decreases Frequency increases When away from source (v o > 0) Distance between waves increases Frequency decreases f o =f s (1- v o /v) Combine: f o =f s (1-v o /v) / (1-v s /v) Lecture 21 Purdue University, Physics

33 Moving Observer and Source Combine: f o = f s (1-v o /v) / (1-v s /v) A: You are driving along the highway at 65 mph, and behind you a police car, also traveling at 65 mph, has its siren turned on. B: You and the police car have both pulled over to the side of the road, but the siren is still turned on. In which case does the frequency of the siren seem higher to you? A) Case A B) Case B C) same f f v v s v o Lecture 21 Purdue University, Physics

34 Velocity ILQ A sound wave having frequency f 0, speed v 0 and wavelength λ 0, is traveling through air when it encounters a large heliumfilled balloon. Inside the balloon the frequency of the wave is f 1, its speed is v 1, and its wavelength is λ 1. Compare the speed of the sound wave inside and outside the balloon A) v 1 < v 0 B) v 1 = v 0 C) v 1 > v 0 V 0 =343m/s V 1 =965m/s Lecture 21 Purdue University, Physics

35 Frequency ILQ A sound wave having frequency f 0, speed v 0 and wavelength λ 0, is traveling through air when it encounters a large heliumfilled balloon. Inside the balloon the frequency of the wave is f 1, its speed is v 1, and its wavelength is λ 1. Compare the frequency of the sound wave inside and outside the balloon A) f 1 < f 0 C) f 1 > f 0 f 0 f 1 B) f 1 = f 0 Time between wave peaks does not change! Lecture 21 Purdue University, Physics

36 Wavelength ILQ A sound wave having frequency f 0, speed v 0 and wavelength λ 0, is traveling through air when it encounters a large heliumfilled balloon. Inside the balloon the frequency of the wave is f 1 1, its speed is v 1 1,, and its wavelength is λ 1. Compare the wavelength of the sound wave inside and outside the balloon A) λ 1 < λ 0 λ 0 λ 1 B) λ 1 = λ 0 C) λ 1 > λ 0 λ = v / f Lecture 21 Purdue University, Physics

37 Human Ear Audible range: 20 Hz - 20 khz Your ear is sensitive to an amazing range! 1dB 100 db Watts/m 2 1 Watt/m 2 Like a laptop that can run using all power of Battery Entire Nuclear Power Plant Lecture 21 Purdue University, Physics

38 Hearing When a sound wave reaches your ear, the pressure on the outside of your eardrum is the atmospheric pressure plus the oscillating sound pressure On the inside of your eardrum, the pressure is equal to the atmospheric pressure Generally, your eardrum is insensitive to atmospheric pressure, but detects the sound pressure Lecture 21 Purdue University, Physics

39 Ear as a Pressure Detector Intensity is proportional to the square of the pressure amplitude There is a relationship between the threshold intensity for human hearing I o and the smallest pressure oscillation that can be detected by the ear A sound intensity of I o = 1.00 x W/m 2 corresponds to a pressure amplitude of p o = 2 x 10-5 Pa This is about 2 x times smaller than atmospheric pressure Lecture 21 Purdue University, Physics

40 Human Perception of Sound The lowest line plots the intensity of the minimum detectable sound as a function of the frequency Doubling the perceived loudness corresponds to moving from one contour line to the next The highest contour curve has a loudness at which the ear is damaged Occurs at β 120 db The loudness contours depend on frequency Lecture 21 Purdue University, Physics

41 Ultrasound Images Ultrasonic imaging uses sound waves to obtain images inside a material The most familiar use is to produce views inside the human body The depth of objects inside the body is determined by the time it takes the reflected ray to return to the detector Some medical ultrasounds also use the Doppler effect Lecture 21 Purdue University, Physics

42 Summary of Concepts Speed of sound v = sqrt(b/ρ) Intensity β = (10 db) log 10 ( I / I 0 ) Standing Waves f n = n v/(2l) Open at both ends n=1,2,3, f n = n v/(4l) Open at one end n=1,3,5, Doppler Effect f o = f s (v-v o ) / (v-v s ) Beats ω L = 1 ( 2 ω ω 1 2) Lecture 21 Purdue University, Physics