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Physics 140 Sound Physics 140 Sound Chapter 12 Sound waves Sound is composed of longitudinal pressure waves. wave propagaon Compression è when Compression Compression parBcles come together RarefacBon RarefacBon RarefacBon è when parBcles are farthest apart 2 Physics 140, Prof. M. Nikolic Sound waves Described by the gauge pressure – difference between the pressure at a given point and average pressure 3 Physics 140, Prof. M. Nikolic The speed of sound waves T Let’s remember Chapter 11 and the speed of string waves v = µ The speed of sound B waves v = B is the bulk modulus of the medium ρ and ρ its density. The speed of sound Y v = waves in thin rods Y is the Young’s modulus of the ρ medium and ρ its density. All other equaons from Chapter 11 sBll apply. 2π v = f λ k = 4 Physics 140, Prof. M. Nikolic λ Conceptual quesBon – Speed of sound Q1 Given the informaon in the table below, which substance would have the highest speed of sound? Substance Bulk Modulus Density Speed (GPa) (kg/m3) (m/s) Aluminum 70 2700 Brass 61 8400 Copper 14 8900 Mercury 27 13600 A. aluminum B B. brass v = C. copper ρ D. mercury 5 Physics 140, Prof. M. Nikolic Conceptual quesBon – Speed of sound Q1 Given the informaon in the table below, which substance would have the highest speed of sound? Substance Bulk Modulus Density Speed (GPa) (kg/m3) (m/s) Aluminum 70 2700 5.1 x 103 Brass 61 8400 2.7 x 103 Copper 14 8900 1.25 x 103 Mercury 27 13600 1.4 x 103 A. aluminum B B. brass v = C. copper ρ D. mercury 6 Physics 140, Prof. M. Nikolic B Exercise: Speed of sound in gold v = ρ What’s the speed of sound in a gold ingot? Density of gold is ρ = 19,360 kg/m3 and its bulk modulus is 220 GPa. What’s given: B ρ = 19,360 kg/m3 v = B = 220 GPa = 220x109 Pa ρ 9 220×10 Pa 3 v = = 3.38 ×10 m/s 19300 kg/m 3 7 Physics 140, Prof. M. Nikolic CharacterisBcs of sound Sound can travel through any kind of maer, but not through a vacuum. The speed of sound is different in different materials; in general, it is slowest in gases, faster in liquids, and fastest in solids. The speed depends somewhat on temperature, especially for gases. T v = v0 T V0 is the absolute speed at 0 absolute temperature T0. e.g. v (air) = 343 m/s at T = 200 C. 0 0 8 Physics 140, Prof. M. Nikolic The speed of sound waves T Speed of sound in ideal gases: v = v0 T0 SI units of temperature: [K] To convert temperature from 0C to Kelvins (SI unit of temperature) you just need to add 273 K. T(in K) = TC (in C)+ 273 for the speed of sound in air (and only in air) we can use an approximate formula: v = (331+0.606TC ) m/s 0 where Tc is the air temperature in C. 9 Physics 140, Prof. M. Nikolic Exercise: A lightning flash A lightning flash is seen in the sky and 8.2 seconds later the boom of thunder is heard. The temperature of the air is 12.0 0C. (a) What is the speed of sound in air at that temperature? What is given: Since we are looking for the speed of sound in air, we can use the t = 8.2 s approximate formula: 0 TC = 12 C v = (331+ 0.606TC ) m/s v = (331+0.606⋅12 0C) m/s = 338 m/s (b) How far away is the lightning strike? d = vt d = (338 m/s)(8.2 s)= 2800 m = 2.8 km (c) The speed of light is 3.00x108 m/s. How long does it take the light signal to reach the observer? d 2800 m −6 t = t = 8 = 9.3×10 s v 3.0×10 km/s 10 Physics 140, Prof. M. Nikolic The amplitude and intensity The intensity of a wave is the energy transported per unit Bme across a unit area. It can be given as a funcBon Or by looking at the amplitude of a pressure amplitude (P0) of the displacement of parBcles P2 I = 0 P = ωvρs 2ρv 0 max ω is the angular frequency of sound v is the speed of sound ρ is the density of the medium smax is the maximum displacement Units of intensity: W/m2 11 Physics 140, Prof. M. Nikolic Loudness and decibels Loudness of a sound is measured by the logarithm of the intensity. Sound level (intensity level) is measured in decibels (dB) and is defined: I β = (10 dB)log I0 I0 is taken to be the threshold of hearing: −12 2 I0 = 1.0 × 10 W/m Sound level is not the same as the intensity of the sound wave! 12 Physics 140, Prof. M. Nikolic Exercise: Decibels from a speaker The sound level 25 m from a loudspeaker is 71 dB. a) What is the rate at which sound energy is being produced by the loudspeaker, assuming it to be an isotropic source? What is given: I r = 25 m β = (10 dB)log β = 71 dB -12 2 I0 I0 = 10 W/m We can solve for the intensity of the sound wave: I I β I β β = (10dB)log log = =1010dB I I 10dB 0 0 I0 71 dB I =10−12 W/m 2 ⋅1010 dB =1.3×10−5 W/m 2 13 Physics 140, Prof. M. Nikolic Exercise: Decibels from a speaker The sound level 25 m from a loudspeaker is 71 dB. a) What is the rate at which sound energy is being produced by the loudspeaker, assuming it to be an isotropic source? What is given: r = 25 m β = 71 dB And we calculated: I = 1.3x10-5 W/m2 We need to find the rate at which sound energy is being produced – that’s power, but we know intensity of sound and distance. OK, let’s go back to Chapter 11 and use the equaon for intensity of the isotropic source: P 2 I = 2 P I4 r 4πr = π P = (1.3×10−5 W/m2 )4π (25 m)2 = 0.10 W 14 Physics 140, Prof. M. Nikolic Exercise: Decibels from a speaker The sound level 25 m from a loudspeaker is 71 dB. b) What is the pressure amplitude of a sound wave? Density of air is 1.2 kg/m3 and the speed of sound is about 343 m/s. What is given: r = 25 m P2 β = 71 dB I = 0 3 P I 2 v ρ = 1.2 kg/m 2ρv 0 = ⋅ ρ v = 340 m/s And we calculated: -5 2 −5 2 3 I = 1.3x10 W/m P0 = 1.3×10 W/m ⋅2⋅1.2 kg/m ⋅343 m/s = 0.1 Pa 15 Physics 140, Prof. M. Nikolic Standing waves of sound All of these musical instruments are effecBvely sealed off on one end and open on the other. The air molecules right next to the closed end are essenBally not moving è displacement nodes The molecules at the open end are exposed to the atmosphere. This gives them the greatest freedom of movement è displacement annodes 16 Physics 140, Prof. M. Nikolic Standing waves of sound - Tube closed at one end - L first harmonic - fundamental Second harmonic Third harmonic λ1 5 4 L 4L 3λ1 4 λ1 = ⇒ λ1 = L = ⇒ λ = L L = ⇒ λ1 = L 4 4 1 3 4 5 17 Physics 140, Prof. M. Nikolic Standing waves of sound - Tube closed at one end - The general result for standing waves in a tube open at one end and closed at the other is 4L λn = n where n = 1, 3, 5,… (odd values only!!) v nv fn = = = nf1 λn 4L f1 is the fundamental frequency. Or 4L λn' = 2n'−1 where n’ = 1, 2, 3,… frequencies have only odd mulBples of the fundamental one 18 Physics 140, Prof. M. Nikolic è Even harmonics are not present Exercise: A bugle What are the first 3 harmonic frequencies emi`ed by a bugle that has a tube 50 cm long aer it was ler out in the cold, cold night (assume the temperature is -12oC)? What is given: (2n −1)v L = 50 cm = 0.5 m fn = = (2n −1) f1 0 TC = -12 C 4L Since we are looking for the speed of sound in air, we can use the approximate formula: v 331 0.606T m/s 0 = ( + C ) v = (331+0.606⋅(−12 C)) m/s = 323.7 m/s (2n −1)v v n = 1 - fundamental frequency: f 323.7 m/s 1 = = f1 = =161.8 Hz 4L 4L 4⋅0.5 m f = (2 ⋅ 2 −1) f = 3 f n = 2 2 1 1 f2 = 485.6 Hz n = 3 f3 = (2⋅3−1) f1 = 5 f1 f3 = 809.3 Hz 19 Physics 140, Prof. M. Nikolic Pressure variaons - Tube closed at one end - Displacement node at the closed end corresponds to the pressure anBnode è molecules at that locaon are the ones geng maximally compressed Displacement anBnode at the open end corresponds to the pressure node è molecules at that locaon are the ones geng maximally stretched out 20 Physics 140, Prof. M. Nikolic Standing waves of sound - Tube open at both ends - L first harmonic - fundamental Second harmonic Third harmonic λ1 3 2 L 2L 2λ1 λ1 = ⇒ λ1 = L = ⇒ λ = L L = ⇒ λ1 = L 2 2 1 2 3 21 Physics 140, Prof.
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