Transcript
Page 1: Presentation Chapter 2: Wave equation

Derivation of Wave Equation(Ta3520)

Derivation of Wave Equation – p. 1/11

Page 2: Presentation Chapter 2: Wave equation

Derivation wave equationConsider small cube of mass with volume ∆V :

∆z

∆x

∆y

p+∆p

p+∆p

z

p+∆px

y

Desired: equations in terms of pressure p and particlevelocity v

Derivation of Wave Equation – p. 2/11

Page 3: Presentation Chapter 2: Wave equation

Deformation EquationConservation of mass:

ρ(t0)∆V (t0) = ρ(t0 + dt)∆V (t0 + dt)

ρ0∆V ' (ρ0 + dρ)(∆V + dV )

or:

ρ0

' −

dV

∆V,

neglecting term dρdV .

Derivation of Wave Equation – p. 3/11

Page 4: Presentation Chapter 2: Wave equation

Deformation EquationAssume linear relation between density ρ and pressure p:

ρ0

=dp

K

where K is bulk modulus.

This is one. Now dV/∆V .

Derivation of Wave Equation – p. 4/11

Page 5: Presentation Chapter 2: Wave equation

1-Dimensional motion (in x-direction):

dV

∆V'

dx

∆x

dx is difference in change of particle displacement dux:

dx = (dux)x+∆x − (dux)x

=∂(dux)

∂x∆x =

∂vx

∂xdt ∆x

where vx is x–component of particle velocity.Derivation of Wave Equation – p. 5/11

Page 6: Presentation Chapter 2: Wave equation

Substitute dρ/ρ0 and dV/∆V :

dp

K= −

∂vx

∂xdt

Assuming low-velocity approximation:

dp

dt≈

∂p

∂t

Gives deformation equation:1

K

∂p

∂t= −

∂vx

∂x

Derivation of Wave Equation – p. 6/11

Page 7: Presentation Chapter 2: Wave equation

Equation of MotionNewton’s law applied to volume ∆V :

∆Fx = ρ∆Vdvx

dt

where Fx is force in x-direction (1-dimensional motion)

∆Fx = −∆px∆Sx

= −

(

∂p

∂x∆x +

∂p

∂tdt

)

∆Sx

' −

∂p

∂x∆V (for dt → 0)

Derivation of Wave Equation – p. 7/11

Page 8: Presentation Chapter 2: Wave equation

Assuming low-velocity approximation:

dvx

dt≈

∂vx

∂t

Then Newton’s Law becomes:

∂p

∂x= ρ0

∂vx

∂t

This is the equation of motion.

Derivation of Wave Equation – p. 8/11

Page 9: Presentation Chapter 2: Wave equation

Wave EquationCombine deformation equation and equation of motion.

Let operator (∂/∂x) work on equation of motion andassume ρ constant:

∂x

(

∂p

∂x

)

= ρ∂

∂t

(

∂vx

∂x

)

Substitute deformation equation:

Derivation of Wave Equation – p. 9/11

Page 10: Presentation Chapter 2: Wave equation

Wave Equation

∂2p

∂x2−

ρ0

K

∂2p

∂t2= 0

or:

∂2p

∂x2−

1

c2

∂2p

∂t2= 0

This is the wave equation.

c is velocity of sound: c =√

K/ρ

Derivation of Wave Equation – p. 10/11

Page 11: Presentation Chapter 2: Wave equation

Need to know for examConservation of mass, with some approximations, leadsto deformation equation:

1

K

∂p

∂t= −

∂vx

∂x

Conservation of forces, with some approximations, leadsto equation of motion:

∂p

∂x= ρ0

∂vx

∂t

From then on, everything.

Derivation of Wave Equation – p. 11/11


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