v wave Here F is the tension and µ is the mass/length.

Transcription

1 Main points of today s lecture: Transverse and longitudinal waves traveling waves v wave = Wave speed for a string fλ v = F µ Here F is the tension Intensity of sound I = and µ is the mass/length. P = ΔE A Decibel intensity values I threshold =I W/m 2, I pain 1 W/m 2 β=10 log 10 (I/ I 0 ) Spherical waves I = P 4πr ( AΔT ) 2

2 Types of Waves -- Transverse In a transverse wave, each element that is disturbed moves perpendicularly to the wave motion

3 Types of Waves -- Longitudinal In a longitudinal wave, the elements of the medium undergo displacements parallel to the motion of the wave A longitudinal wave is also called a compression wave

4 Waves If you tie a spring to a wall, hold one end, and rapidly move your arm up and down, you will create a pulse that travels down the spring Transverse the displacement of the spring is perpendicular to the direction the pulse moves Rapidly move the spring back and forth, and compressed pulse travels down the spring Longitudinal the displacement is parallel to the direction of the pulse What do these pulses transport? Ma1er? NO! Energy is transported!

5 Snapshot Graphs of an asymmetic wave: space vs. time All points at different times. Slide 15-14

6 Constructing a History Graph one point at all times. Slide 15-15

7 Checking Understanding The graph below shows a snapshot graph of a wave on a string that is moving to the right. A point on the string is noted. Which of the choices is the history graph for the subsequent motion of this point? Slide 15-16

8 Checking Understanding The graph below shows a snapshot graph of a wave on a string that is moving to the left. A point on the string is noted. Which of the choices is the history graph for the subsequent motion of this point? WWhat is the motion at this point? Slide 15-16

9 Traveling sinusoidal wave The figure at the right shows a wave traveling on a string at different times, measured in units of a period. The round point shows how wave-front on the traveling wave moves. At a fixed time and as a function of distance, the wave reproduces itself after an x displacement of one wavelength λ. At each point, the wave returns to its original y displacement after a time T. At time T, the wave front has moved a distance λ. This makes the wave velocity v= λ/t= λf. Here we use the relationship f=1/t between wavelength and frequency. λ

10 λ Describing a Traveling Wave t=0 y each mass point goes up and down like mass+spring λ E = 1 2 mv 2 max x t=t/4 t=2t/4 t=3t/4 t=t λ: wavelength = length (m) of one oscilla=on. T: period = =me for one oscilla=on : frequency (Hz) While the wave has traveled one wavelength, each point on the wave has made one period of oscilla=on. v = Δx/Δt = λ/t = λf

11 Example A wave traveling in the positive x direction has a frequency of 25.0 Hz as in the figure below. Find the (a) amplitude, (b) wavelength, (c) period, and (d) speed of the wave. The amplitude is: a) 9 cm b) 18 cm The wavelength is: a) 10 cm b) 20 cm a) A = 18cm / 2 = 9cm b) λ = 10cm 2 = 20cm c) T = 1/f =1/25 s = 0.04 s d) v = λf = 20cm 25Hz = 500cm / s

12 Wave velocity on string Waves travel faster on strings with higher tension and lower mass/unit length. The exact relationship is v = F µ Here F is the string tension and µ isthe mass/length. Example: A 0.5 m string is stretched so the tension is 1.7 N. A transverse wave of frequency 120 Hz and wavelength 0.31 m travels on the string. What is the mass of the string? m= µ L; v = λf µ = L = 0.5m; v ( )( ) µ = F 2 v = 0.31m 120Hz = 37.3m / s F 1.7N 3 = = 1.22x10 kg / m 2 2 v 37.3m / s ( ) ( 3 )( ) m = µ L = 1.22x10 kg / m 0.5m = 0.61g = F µ L 0.5 m F 1.7N f 120Hz λ 0.31m m?

13 Example Problem A particular species of spider spins a web with silk threads of density 1300 kg/m 3 and diameter 3.0 µm. A typical tension in the radial threads of such a web is N. What is the speed of the waves traveling on the web. If a fly lands in this web, which will reach the spider first, the sound or the wave on the web silk? (sound velocity=343 m/s) a) 172 m/s, no b) 215 m/s, no c) 286 m/s. no d) 430 m/s, yes e) 873 m/s, yes 3 ρ = 1300kg / m µ = ρ i A silk = ( 1300kg / m 3 )i i (3x10 6 / 2m) 2 = 9.19x10 9 kg / m F v = = µ.007kg i m / s x10 9 kg / m = 873m / s This is faster than the speed of sound in air. Slide 15-21

14 Conceptual question A weight is hung over a pulley and attached to a string composed of two parts, each made of the same material but one having four times the diameter of the other. The string is plucked so that a pulse moves along it, moving at speed v 1 in the thick part and at speed v 2 in the thin part.what is v 1 /v 2? a) 1/4 b) 1/2 c) 2 d) 4 v = F µ = F M / L 1 2 R1 M F F v = µ = v1 = π ρ v = Vρ = πrlρ µ = M/L= πr ρ F πr 2 1 ρ / F πr 2 2 ρ / / / / / = 1 R R 2 2 R 1 2 R 2 2 = R 2 R R R = R 2 R 1 = 1/ 4

15 Using a Tuning Fork to Produce a Sound Wave A tuning fork will produce a pure musical note of frequency f As it s tine swings to the right, it forces the air molecules near it closer together This produces a high density area in the air This is an area of compression As the tine moves toward the left, the air molecules to the right of the tine spread out This produces an area of low density This area is called a rarefaction As the tuning fork continues to vibrate, a sinusoidal succession of compressions and rarefactions spread out from the fork Crests correspond to compressions and troughs to rarefactions λ v = λf

16 Sound Waves Audible range: between 20 Hz to 20,000 Hz Infrasonic waves: f < 20 Hz; Ultrasonic:f > 20,000 Hz Sound velocity : v elastic _ property inertial _ property Sound velocity in liquid: v B ; B isthe bulk modulus ρ Sound velocity in solid: v Y ; Y isyoung's modulus ρ Sound velocity in air: v ( 331m / s) T ; where T is in Kelvin or 273K v 331m / s + 0.6T; where T is in Celsius.

17 Example You are watching a pier being constructed on the far shore of a saltwater inlet when some blasting occurs. You hear the sound in the water 4.50 s before it reaches you through the air. How wide is the inlet? (Assume the air temperature is 20 C and sound velocity in water is 1530 m/s at its temperature (25 C). ) = v = ( 331m / s) T ( 331m / s) 293 vs,water 1530m / s; t d / v ; air s,air t d / v ; = s,air water s,water 273K = 273 = 343m / s = t ( ) air twater = d 1/ vs,air 1/ vs,water ( ) ( ) air water s,air s,water d= t t / 1/v 1/v 1 1 = 4.5s / = 1989m 343m / s 1530m / s

18 Quiz A dolphin located in sea water at a temperature of 25 C emits a sound directed toward the bottom of the ocean 150 m below. How much time passes before it hears an echo? (The sound velocity in sea water = 1530 m/s at 25 0 C.) a)0.2 s b)0.4 s c)0.6 s d)0.8 s e) 0.1 s Δ t = 2d / v = 300m / (1530m / s) = 0.196s

19 Sound amplitude and intensity v! The amplitude of the sound wave is proportional to the maximum velocity of the air as it moves from the high pressure to the low pressure domains. The energy and the power of the sound wave is proportional to the square of amplitude: E ρv Amplitude vmax p xmax 2 More useful than the energy of a sound wave is the intensity, I, which is the power P that the sound wave transmits per unit area. I = P = ΔE A The ear responds logarithmically to the intensity of sound waves striking the eardrum. I threshold =I W/m 2, I pain 1 W/m 2 This logarithmic behavior motivates the decibel measure of sound wave intensity. β=10 log 10 (I/ I 0 ) ( AΔT ) Fig 14.2, p. 426 Slide 4

20 Decibel is named after Alexander Graham Bell How the decibel scale works: if I=I 0, (I is equal to hearing threshold intensity) β=10log 10 (I 0 /I 0 )=10log 10 (1)=0 log 10 (10 n ) = n, i.e. log 10 (1000)=log 10 (10 3 ) = 3 if I=10 2 I 0 =100I 0, then β=10log 10 (100I 0 /I 0 )=10log 10 (100)=20 log 10 (A/B)=log 10 (A)-log 10 (B) log 10 (AB)=log 10 (A)+log 10 (B)

21 Machine noise A machine produces sound with a level of 80dB. How many machines can you add before exceeding 100dB? 1 machine 80 db=10log(i/i 0 ) 8=log(I/I 0 )=log(i/1e-12) 10 8 =I/1E-12 I 1 =10-4 W/m 2 N machines 100 db=10log(i N /I 0 ) 10 = log(i N /I 0 )=log(i N /1E-12) = I N /1E-12 I N =10-2 W/m 2 I 1 /I N = 10-4 /10-2 = 1/100 The intensity must increase by a factor of 100; one needs to add 99 machines. Shortcut: x = 20 so the increase in I is 10 (x/10) = 100.

22 Example The intensity level of sound A is 5.0 db greater than that of sound B and 3.0 db less than that of sound C. a) Determine the ratio (I C /I B ) of the intensity of sound C to the intensity of sound B. I a) β = 10log10 I 0 I β A β B = 5 = 10log A 10 I 0 10log I B 10 I 0 = 10 log I A 10 I 0 log 10 I A βa βb = 10log10 = 5 I B I similarly : β C β A = 10log C 10 I A = 3 I C βc βb = 10log10 I also, β β = β β C B C A I = C 10log10 8 I B I I B C 0.8 = 10 = 6.3 B I = C log I B 10 log 10 I B I 0 I C I B = 10log 10 ( ) + ( β A β B ) = 5+ 3 = 8 = I A / I 0 I B I 0 /