The Differential Equation for a Vibrating String · 2008-11-06 · The Differential Equation for a Vibrating String. logo1 Model Forces The Equation Modeling Assumptions 1. The string

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Model Forces The Equation

The Differential Equation for a VibratingString

Bernd Schroder

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Differential Equation for a Vibrating String

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Modeling Assumptions

1. The string is made up of individual particles that movevertically.

2. u(x, t) is the vertical displacement from equilibrium of theparticle at horizontal position x and at time t.

��

��u > 0

u < 0

u = 0

- x

��

�



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Modeling Assumptions1. The string is made up of individual particles that move

vertically.

2. u(x, t) is the vertical displacement from equilibrium of theparticle at horizontal position x and at time t.

��

��u > 0

u < 0

u = 0

- x

��

�



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vertically.2. u(x, t) is the vertical displacement from equilibrium of the

particle at horizontal position x and at time t.

��

��u > 0

u < 0

u = 0

- x

��

�



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vertically.2. u(x, t) is the vertical displacement from equilibrium of the

particle at horizontal position x and at time t.

��

��u > 0

u < 0

u = 0

- x

��

�



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Decomposing the Tensile Force

-



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-



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x

-



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x

-

+ ~Ft



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x

-

+

�

~Ft



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x

-

+

�

?

~Fv

~Ft



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x

-

+

�

?

α

~Fv

~Ft



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x x+∆x

-

+

�

?

α

~Fv

~Ft



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x x+∆x

-

+

�

?

:

α

~Fv

~Ft

~Ft



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x x+∆x

-

+

�

?

:6

α

~Fv

~Fv ~Ft

~Ft



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x x+∆x

-

+

�

?

:6

-

α

~Fv

~Fv ~Ft

~Ft



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x x+∆x

-

+

�

?

:6

-

α

α

~Fv

~Fv ~Ft

~Ft

F(x) ≈ Fv(x+∆x)−Fv(x)



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x x+∆x

-

+

�

?

:6

-

α

α

~Fv

~Fv ~Ft

~Ft

F(x) ≈ Fv(x+∆x)−Fv(x)



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The Vertical Force at a Point

F(x) ≈ Fv(x+∆x)−Fv(x)= Ft sin(α)−Ft sin(α)

0.25 ≈ 14.3◦



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The Vertical Force at a PointF(x) ≈ Fv(x+∆x)−Fv(x)

= Ft sin(α)−Ft sin(α)

0.25 ≈ 14.3◦



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0.25 ≈ 14.3◦



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0.25 ≈ 14.3◦



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0.25 ≈ 14.3◦



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= Ft sin(α)−Ft sin(α)(

sin(θ) ≈ tan(θ),θ small)

≈ Ft tan(α)−Ft tan(α)

-

6



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-

6



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-

6



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-

6

x



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-

6

x



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-

6

x



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-

6

x

1



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-

6

x

1



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-

6

x

f ′(x)

1



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-

6

x

f ′(x)

1θ



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≈ Ft tan(α)−Ft tan(α)(tan(θ) = f ′(x)

)

= Ft

(ddx

u(x+∆x)− ddx

u(x)) (

f (x+∆x) ≈ f (x)+ f ′(x)∆x)

≈ Ft

(ddx

u(x)+∆x · d2

dx2 u(x)− ddx

u(x))

= Ft∆xd2

dx2 u(x)



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)= Ft

(ddx

u(x+∆x)− ddx

u(x))

(f (x+∆x) ≈ f (x)+ f ′(x)∆x

)≈ Ft

(ddx

u(x)+∆x · d2

dx2 u(x)− ddx

u(x))

= Ft∆xd2

dx2 u(x)



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)= Ft

(ddx

u(x+∆x)− ddx

u(x)) (

f (x+∆x) ≈ f (x)+ f ′(x)∆x)

≈ Ft

(ddx

u(x)+∆x · d2

dx2 u(x)− ddx

u(x))

= Ft∆xd2

dx2 u(x)



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)= Ft

(ddx

u(x+∆x)− ddx

u(x)) (

f (x+∆x) ≈ f (x)+ f ′(x)∆x)

≈ Ft

(

ddx

u(x)+∆x · d2

dx2 u(x)− ddx

u(x))

= Ft∆xd2

dx2 u(x)



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)= Ft

(ddx

u(x+∆x)− ddx

u(x)) (

f (x+∆x) ≈ f (x)+ f ′(x)∆x)

≈ Ft

(ddx

u(x)+∆x · d2

dx2 u(x)

− ddx

u(x))

= Ft∆xd2

dx2 u(x)



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)= Ft

(ddx

u(x+∆x)− ddx

u(x)) (

f (x+∆x) ≈ f (x)+ f ′(x)∆x)

≈ Ft

(ddx

u(x)+∆x · d2

dx2 u(x)− ddx

u(x))

= Ft∆xd2

dx2 u(x)



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)= Ft

(ddx

u(x+∆x)− ddx

u(x)) (

f (x+∆x) ≈ f (x)+ f ′(x)∆x)

≈ Ft

(ddx

u(x)+∆x · d2

dx2 u(x)− ddx

u(x))

= Ft∆xd2

dx2 u(x)



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Using Newton’s Second Law

ma = F(x) = Ft∆x∂ 2

∂x2 u(x, t)

ρl∆x∂ 2

∂ t2u(x, t) = Ft∆x

∂ 2

∂x2 u(x, t)

ρl

Ft

∂ 2

∂ t2u(x, t) =

∂ 2

∂x2 u(x, t)



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ma

= F(x) = Ft∆x∂ 2

∂x2 u(x, t)

ρl∆x∂ 2

∂ t2u(x, t) = Ft∆x

∂ 2

∂x2 u(x, t)

ρl

Ft

∂ 2

∂ t2u(x, t) =

∂ 2

∂x2 u(x, t)



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ma = F(x)

= Ft∆x∂ 2

∂x2 u(x, t)

ρl∆x∂ 2

∂ t2u(x, t) = Ft∆x

∂ 2

∂x2 u(x, t)

ρl

Ft

∂ 2

∂ t2u(x, t) =

∂ 2

∂x2 u(x, t)



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ma = F(x) = Ft∆x∂ 2

∂x2 u(x, t)

ρl∆x∂ 2

∂ t2u(x, t) = Ft∆x

∂ 2

∂x2 u(x, t)

ρl

Ft

∂ 2

∂ t2u(x, t) =

∂ 2

∂x2 u(x, t)



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ma = F(x) = Ft∆x∂ 2

∂x2 u(x, t)

ρl∆x

∂ 2

∂ t2u(x, t) = Ft∆x

∂ 2

∂x2 u(x, t)

ρl

Ft

∂ 2

∂ t2u(x, t) =

∂ 2

∂x2 u(x, t)



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ma = F(x) = Ft∆x∂ 2

∂x2 u(x, t)

ρl∆x∂ 2

∂ t2u(x, t)

= Ft∆x∂ 2

∂x2 u(x, t)

ρl

Ft

∂ 2

∂ t2u(x, t) =

∂ 2

∂x2 u(x, t)



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ma = F(x) = Ft∆x∂ 2

∂x2 u(x, t)

ρl∆x∂ 2

∂ t2u(x, t) = Ft∆x

∂ 2

∂x2 u(x, t)

ρl

Ft

∂ 2

∂ t2u(x, t) =

∂ 2

∂x2 u(x, t)



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ma = F(x) = Ft∆x∂ 2

∂x2 u(x, t)

ρl∆x∂ 2

∂ t2u(x, t) = Ft∆x

∂ 2

∂x2 u(x, t)

ρl

Ft

∂ 2

∂ t2u(x, t) =

∂ 2

∂x2 u(x, t)



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The One-Dimensional Wave Equation

The equation of motion for small oscillations of a frictionlessstring is

∂ 2

∂x2 u(x, t) = k∂ 2

∂ t2u(x, t),

where k =ρl

Ft> 0, with ρl being the linear density of the string

and Ft being the tensile force.This equation is also called the one-dimensional waveequation.Our derivation is valid for small oscillations and negligiblefriction.The cancellation of the ∆x was “clean”.



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The One-Dimensional Wave EquationThe equation of motion for small oscillations of a frictionlessstring is

∂ 2

∂x2 u(x, t) = k∂ 2

∂ t2u(x, t),

where k =ρl


and Ft being the tensile force.

This equation is also called the one-dimensional waveequation.Our derivation is valid for small oscillations and negligiblefriction.The cancellation of the ∆x was “clean”.



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∂ 2

∂x2 u(x, t) = k∂ 2

∂ t2u(x, t),

where k =ρl


and Ft being the tensile force.This equation is also called the one-dimensional waveequation.

Our derivation is valid for small oscillations and negligiblefriction.The cancellation of the ∆x was “clean”.



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∂ 2

∂x2 u(x, t) = k∂ 2

∂ t2u(x, t),

where k =ρl


and Ft being the tensile force.This equation is also called the one-dimensional waveequation.Our derivation is valid for small oscillations and negligiblefriction.

The cancellation of the ∆x was “clean”.



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∂ 2

∂x2 u(x, t) = k∂ 2

∂ t2u(x, t),

where k =ρl


and Ft being the tensile force.This equation is also called the one-dimensional waveequation.Our derivation is valid for small oscillations and negligiblefriction.The cancellation of the ∆x was “clean”.



The Differential Equation for a Vibrating String · 2008-11-06 · The Differential Equation for a Vibrating String. logo1 Model Forces The Equation Modeling Assumptions 1. The string

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