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Physics of Sound

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... velocity: v(x,t), m/s, (around ... Uniform Tube Model (lossless) ... Let the crosssection of the tube be A(x,t), then. Now consider the model. Wave eqns. ... – PowerPoint PPT presentation

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Title: Physics of Sound


1
Physics of Sound
  • Wave equation Part. diff. equation relating
    pressure and velocity as a function of time and
    space
  • Nonlinear contributions are not considered.
  • Valid under small partical Velocity.
  • Three steps
  • Lossless uniform tube model
  • Nonuniform, losses due to voval tract walls.
  • Boundary effects (lip radiation)

2
Sound
  • Sound is vibration of particles in a medium.
  • Particle velocity
  • Pressure
  • Sound wave is the propagation of disturbance of
    particles through a medium.
  • c ? f
  • ? 2? f
  • ?/c 2? / ? wave number
  • At sea level c 344 m/s (70º F)
  • At f 50 Hz ? c / f 6.88 m

3
  • Isothermal processes
  • Slow variation (of pressure), temp. stays
    constant (no time for heat transfer)
  • Adiabatic processes
  • Fast variation (of pressure), temp. changes (time
    for heat transfer)
  • Example Bicycle pump
  • Typical usage of the terms Isothermal/adiabatic
    compression of a gas
  • For most frequencies (except very low
    frequencies) sound is adiabatic.

4
Wave Equation
  • Atmospheric pressure P0 105 N/m2
  • Pressure P0 p(x,t)
  • p(x,t)
  • 0 dB, threshold of hearing 2(10-5) N/m2 at 1000
    Hz.
  • threshold of pain20 N/m2.
  • Particle velocity v(x,t), m/s, (around zero
    average)
  • Density of air particles ?(x,t), kg/ m3 (around
    an average of ?0 --gt, ?0 ?(x,t) )

5
Wave Equation
  • 3 laws of physics, to be applied on the cubic
    volume of air.
  • F ma
  • P V? Const P total pressure, V volume, ?
    1.4
  • Conservation of mass The cube may be deformed if
    pressure changes but the of particles inside
    remains the same.

  • F - (?p/ ?x) ?x (?y ?z)

  • net press. vol.
  • (no
    frictional pressure, zero viscosity)

  • m ? ?x ?y ?z

8
6
Wave Equation
  • F m a ? - (?p/?x) ?x (?y ?z) ? ?x ?y ?z
    (dv/dt)
  • dv (?v/?x) dx v (?v/?t) dt
  • dv/dt v (?v/?x) (?v/?t)
  • nonlinear can be
    neglected in speech production since particle
    velocity is small
  • ? - (?p/ ?x) ? (?v/?t)
  • Gas law and cons. of mass yields
    coupled wave equation pair
  • ? - (?p/ ?t) ?c2 (?v/?x)
  • The two can be combined as
  • (?2p/?x2) (1/c2) (?2p/?t2) wave equation for
    p
  • or
  • (?2v/?x2) (1/c2) (?2v/?t2) wave equation v

7
Uniform Tube Model (lossless)
  • No air friction along the walls
  • For convenience volume velocity is defined
  • u (x,t) A v (x,t) m3/s
  • 2nd order wave equations are the same in this
    case except the replacement v(x,t) ? u(x,t)
  • Coupled pair becomes
  • The solutions are of the form

8
Uniform Tube Model (lossless)
  • To find the particular solution let,
  • at x 0, ug(t) u(0,t) Ug(O) ej O t
    (glottal flow)
  • at x l, p(l,t) 0 (no radiation at the
    lips)
  • The general solution is
  • To solve for unknown constants k and k-, apply
    the boudary conditions above.
  • and

9
Uniform Tube Model (lossless)
  • These are standing waves.
  • The envelopes are orthogonal in space and in time

l
0
x
10
Uniform Tube Model (lossless)
  • The frequency response for vol. Velocity, Va(O)
  • The resonances occur at
  • Ex Consider a uniform tube of length l 35 cm.
    For c 350 m/s, the roots, resonances, are at f
    O / (2?) 2000 k / 8 250, 750, 1250,...

As l decrease resonance frequencies increase
Volume velocity
O
11
Uniform Tube Model (lossless)
  • Acoustic impedance The ratio of presure to
    volume velocity.
  • The frequency response can be changed to transfer
    function O ? s / j
  • Under some restrictions it can be written as
  • The poles are the resonant frequencies of the tube

12
Energy Loss Due to Wall Vibration
  • Let the crosssection of the tube be A(x,t), then
  • Now consider the model

Wave eqns.
13
Energy Loss Due to Wall Vibration
  • Assuming A(x,t) A0(x,t) ?A(x,t), an equaton
    can be written for ?A(x,t)
  • Then, the three equations can be written (under
    some simplifications, AA0?A)

14
Energy Loss Due to Wall Vibration
  • Assuming again ug(t) u (0, t) Ug(O) ej O t
    yields solutions of the form
  • These forms eliminate time dependence and the
    equations become
  • They are solved by numerical techniques

15
Formants would be at 500, 1500, 2500,..., in the
lossless case.
l17.5 cm, A05cm2, mw 0.4gr/cm2, bw
6500dyne-sec/cm3 , kw0 Bandwidth is not zero!
Viscosity (friction of air with walls) and
thermal loss included.
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