Chapter 19 Sound Waves

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Chapter 19 Sound Waves

• Molecules obey SHM s(x,t)smcos(kx-?t)
• leads to variations in density and pressure in
  the air
• pressure is large where density is large but
  displacement s(x,t) is small
• hence pressure(or density) wave
• is 900 out of phase with displacement
• pp0 ?pm cos(kx-?t-?/2)
  p0 ?pm sin(kx-?t) (?p is change in
  pressure relative to equilibrium)
• ? ?0 ??m sin(kx-?t) rho
• ??m ?pm (?0/B)
• B is bulk modulus

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Speed of Sound

• B-?p/(?V/V)
• Water B2.2x109N/m2 Air B 105 N/m2
• gases are more compressible (B is smaller)
• speed of sound in a fluid depends on the density
  and the elasticity-gt i.e. the medium
• v (B/?0)1/2 wave speed in a fluid
• vair 343 m/s at room temperature
• vwater 1482 m/s ( B is larger!)
• vsteel 5941 m/s

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Intensity Level and Loudness

• IPav/area
• area of sphere 4?r2 gt I ? 1/r2
• sensation of loudness is not directly
  proportional to intensity
• varies as log(I) eg. log(100)2log(10)
• define sound level SL10 log(I/I0) in decibels
  (dB)
• I0 10-12 W/m2 is the hearing threshold
• hence for II0 , SL 10 log(1) 0
• pain threshold is I 1 W/m2 or
  SL10log(1012)120 dB

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Problem

• When a pin of mass 0.1 g is dropped from a
  height of 1 m, 0.05 of its energy is converted
  into a sound pulse with a duration of 0.1 s.
• (a) Estimate the range at which the dropped pin
  can be heard if the minimum audible intensity is
  10 -11 W/m2 .
• (b) Your result in (a) is much too large in
  practice because of background noise. If you
  assume that the intensity must be at least 10-8
  W/m2 for the sound to be heard, estimate the
  range at which the dropped pin can be heard.
  (In both parts, assume that the intensity is
  P/4?r2 .)
• (a) Sound energy is 5 x 10-4 (mgh) 5 x 10-4
  (1m)(10-4 kg)(9.8 m/s2 ) 4.9 x
  10-7 J
• Pav E/?t 4.9 x10-6 W 4?r2 x10-11 W
  gt r 200 m.
• (b) r 200/ (1000)1/2 6.24 m.

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Problem

• The equation I Pav / 4?r2 is predicated on the
  assumption that the transmitting medium does not
  absorb any energy.
• It is known that absorption of sound by dry air
  results in a decrease of intensity of
  approximately 8 dB/km.
• The intensity of sound at a distance of 120 m
  from a jet engine is 130 dB.
• Find the intensity at 2.4 km from the jet engine
  (a) assuming no absorption of sound by air, and
  (b) assuming a diminution of 8 dB/km. (Assume
  that the sound radiates uniformly in all
  directions.)
• (a) ?SL 10 log(I1/I0)-10 log (I2/I0) 10
  log(I1/I2) 10log(r22/r12)20log(r2/r1)
• 20 log (20) 26 SL at 2.4 km (130 - 26) dB
  104 dB
• (b) Subtract 2.28 x 8 dB from result of (a)
• SL (104 - 18.2) dB 85.8 dB

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Interference of Sound Waves
ripple
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Interference andPhase Difference

• Consider shifting one of the waves by a distance
  ? relative to the other
• they add up constructively
• corresponds to a phase shift by ?? 2?
• constructive interference for ?? m2?
• or path difference ?L m ?
• a shift by ?? ? corresponds to a shift of one
  of the waves by ? /2
• destructive interference for ?? (2m1)?
• or path difference ?L (2m1) ? /2

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Musical Sounds

• oscillating strings
• membranes
• air columns
• standing wave patterns correspond to resonances
• large amplitude oscillations push surrounding air
  and generate sound waves at the same frequency

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Pressure waves in open pipe
Pressure waves in pipe closed at one end