Energy flux at your eardrums

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 The intensity of a sound is related to the amount of energy flowing in the sound waves. It depends on the amplitude of the vibrations producing the.
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Presentation transcript:

Energy flux at your eardrums Sound Intensity Energy flux at your eardrums § 14.5 1

Sound Intensity I = (Power output)/(Area) Units: W/m2 2

Intensity and Distance Inverse-square law: Source transmits power P At distance R, energy is spread over area A = 4pR2 Intensity at distance R is P/A IR = P 4pR2 3

Sound Intensity Level Decibel scale I b = (10 dB) log10 I0 Reference intensity I0 = 10–12 W/m2 Audibility threshold at 1000 Hz A 10-dB increase in intensity level represents a factor-of-10 increase in sound intensity 4

Decibel Differences I b = (10 dB) log10 I0 I2 b2 – b1 = (10 dB) log10 5

Example Problem 30 m from an outdoor concert stage, the sound intensity level is 80 dB. What is the sound intensity level 40 m from the stage? b2 – b1 = (20 dB) log10(R1/R2) b2 = b1 + (20 dB) log10(R1/R2) = 80 dB + (20 dB) log10(30/40) = 80 dB + (20 dB) (–0.125) = 80 dB – 2.50 dB = 77.5 dB 6

Doppler Effect Distance and time § 14.6

Moving Source or Detector Source: successive wave fronts do not emanate from the same place Detector: successive wave fronts are not detected at the same place Simulation: http://www.astro.ubc.ca/~scharein/a311/Sim/doppler/Doppler.html

Moving Source or Detector uS v uD vO 1 2 3 D Equations of Motion: Source xS = uSt Observer xO = D + uDt nth wavefront xn = uS(nT) + v (t – nT)

Differing Formulas u – uD Mine: fD = fS u – uS 1 + vD/v Book’s: fD = fS 1 – vS/v 1 + vD/v Difference: Book uses |v|, propagation direction is –