Maarten van Walstijn
PBASS Session 5: Vibrations of Strings and Membranes
1
Aims
• to explore the motion of strings and membranes, through the physics.
• to get familiar with the wave equation and its solutions.
• to explore the influence of physical parameters, such as damping, string length, on the string vibrations.
• Learn about modal synthesis as a simulation method.
PBASS Session 5: Vibrations of Strings and Membranes
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1

Content
• Transversal Vibration in Strings
• Wave Equation
• Travelling-Wave Solution
• Boundary Conditions
• String Normal Modes of Vibration
• Damping Effects
• Lossy Wave Equation
• Modal Synthesis
• Transversal Vibration in Membranes
PBASS Session 5: Vibrations of Strings and Membranes
Transversal Vibration in Ideal Strings (No Damping) y small string segment θ
2
Transversal components of the Tension:
T y , x
T y
= T
, x + dx sin( θ
1
)
= T sin( θ
2
) dy
θ
1 dx
PBASS x x + ∆ x x
Session 5: Vibrations of Strings and Membranes
Net force on the string segment:
F = T y , x + dx
= T
− T y sin( θ
2
) − T
, x sin( θ
1
)
4
3
2

Transversal Vibration in Ideal Strings (cont.)
Let’s write the net force on the segment as:
F =
=
T
T sin(
[ g (
θ x
2
+
) − dx
T
) sin(
− g (
θ
1 x )
)
] g ( x ) = sin
( ) x
For an infinitely small dx , we have:
∂
∂ g x
= g ( x + dx ) − dx g ( x )
⇔ g ( x + dx ) − g ( x ) =
∂ g
∂ x dx
Thus the net force then equals
F = T
∂ g
∂ x dx
PBASS Session 5: Vibrations of Strings and Membranes
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Transversal Vibration in Ideal Strings (cont.)
We have the function: g ( x ) = sin
( ) x
For very small angles, we may write sin( θ ) x
≈ tan( θ ) x
= dy dx
≈
∂ y
∂ x
θ dx dy
F = Tdx
∂ g
∂ x
= Tdx
∂
∂ y
∂ x
∂ x
= Tdx
∂
2 y
∂ x
2
PBASS Session 5: Vibrations of Strings and Membranes
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3

Transversal Vibration in Ideal Strings (cont.)
Newton’s second law:
F = m
∂
∂ t
2
2 y
= ρ dx
∂
2 y
∂ t
2
Hence the equation of motion is:
ρ dx
∂
2 y
∂ t
2
= Tdx
∂
2 y
∂ x
2 or:
∂
2 y
∂ t
2
=


T
ρ


∂
2 y
∂ x
2
( m = ρ dx
) mass per unit length (Kg/m)
PBASS Session 5: Vibrations of Strings and Membranes
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Transversal Vibration in Ideal Strings (cont.)
Dimensional analysis:
T
ρ
→
N
Kg m
− 1
=
Kg m s
2
Kg m
− 1
=

 m s


2 define the wave velocity c = T ρ that indicates how fast waves travel through the string. The wave equation can now be written:
∂ 2 y
∂ t
2
= c
2
∂ 2 y
∂ x
2
PBASS Session 5: Vibrations of Strings and Membranes
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4

Travelling-Wave Solution f g
The general solution to the second order PDE is of the form of a sum of two travelling waves: y ( x , t ) = f ( ct − x ) + g ( ct + x )
Where we may also write the travelling components as a function of time: y ( x , t ) = f ( t − x c ) + g ( t + x c )
PBASS Session 5: Vibrations of Strings and Membranes
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Fixed Boundary Conditions f f
If we “fix” (or “pin”) the string at x = 0 , the boundary boundary condition is: y ( 0 ,t ) = f ( t − 0 c ) + g ( t + 0 c ) = 0
-f
-f
Thus g ( t ) + f ( t )
⇔
= 0 g ( t ) = − f ( t )
In other words, the wave reflects with inverted sign at the boundary. The same reflection with sign-inversion happens at the other end of the string if that is fixed as well.
PBASS Session 5: Vibrations of Strings and Membranes
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5

Forced Vibration of an Infinite String f
For a string of infinite length and driven with a sinusoidal force from the left end, there is only a wave travelling in positive direction, hence the solution takes the form: y ( x , t ) = f ( t
42 t x
'
4
)
A string is a linear system, thus the resulting string vibrations will also be sinusoidal of the same frequency: f ( t ' ) =
=
=
Ae j ω t '
Ae j ω
( t − x
Ae c j
(
ω t − kx
)
) k =
ω c wave number
PBASS Session 5: Vibrations of Strings and Membranes
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Free Vibration of a Finite Length String
For a string of finite length, fixed at both ends, the general travellingform: y ( x , t ) = Ae j
(
ω t − kx
)
+ Be j
(
ω t + kx
)
If we then apply the boundary conditions:

 y y (
( 0 , t )
L , t )
= 0
= 0
One obtains the equations:


Ae
Ae j j
(
ω t
+
− kL
)
Be
+ j
Be j
=
(
ω t
0
+ kL
)
= 0
PBASS Session 5: Vibrations of Strings and Membranes
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Free Vibration of a Finite Length String (cont.)
Omitting the trivial solution , this results into:
A + B =
Ae
− jkL
0
+
⇔
Be
B jkL
= − A
= 0
Combining these equations gives:
Ae
− jkL
− Ae jkL
= 2 Aj sin( kL ) = 0
Which holds for all wave numbers for which: kL = n π
Where n is an integer. In other words, we have a series of harmonically related frequencies at which the string can freely vibrate: f n
=
ω n
2 π
= nc
2 L n = 1 , 2 , 3 , 4 ....
These are the frequencies of the normal modes of the string.
PBASS Session 5: Vibrations of Strings and Membranes
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Free Vibration of a Finite Length String (cont.)
The general solution for free vibration at one particular frequency takes the form: y n
( x , t ) = A n e j
(
ω n t − k n x
)
+ B n e j
(
ω n t + k n x
)
Which, if we substitute B = -A and use Euler’s equations, becomes: y n
( x , t ) = A n sin( k n x ) e j ω n t
In general, if a string is plucked, all modes tend to be excited.
Spectrum of possible frequencies for the string to oscillate at n = 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 . . .
frequency
PBASS Session 5: Vibrations of Strings and Membranes
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Normal Modes of the String
The vibration amplitude is position-dependent, and is scaled by the term: sin( k n x ) = sin
( n π x L
) n = 1 n = 2 n = 3 n = 4 x = 0 x = L f
1
= c
2 L f
2
= c
L
= 2 f
1 f
3
=
3 c
2 L
= 3 f
1 f
4
=
2 c
L
= 4 f
1
PBASS Session 5: Vibrations of Strings and Membranes
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Damping Effects: Air Damping
The lossless wave equation, in ‘Newtonian form’ is:
ρ dx
∂
2 y
∂ t
2
= Tdx
∂
2 y
∂ x
2
We may take into account the resistance that the surrounding air exerts on the vibration string by adding a damping force:
ρ dx
∂
2 y
∂ t
2
= Tdx
∂
2 y
∂ x
2
− r
1 dx
∂ y
∂ t
 =
 air per damping unit length


Now writing the wave equation in classical form, we obtain:
∂
∂ t
2 y
2
= c
2
∂
2 y
∂ x
2
− 2 b
1
∂ y
∂ t

 b
1
= r
1
2 ρ

 air damping constant
PBASS Session 5: Vibrations of Strings and Membranes
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8

Damping Effects : Internal Friction
Similarly, we can adapt the wave equation in order to take into account internal friction effects, which takes the form:
ρ dx
∂
2 y
∂ t
2
= Tdx
∂
2 y
∂ x
2
+ r
2 dx
∂
3 y
∂ t ∂ x
2 r
2
= internal damping per unit length
The classical wave equation with internal friction then becomes
∂
2
∂ t
2 y
= c
2
∂
2 y
∂ x
2
+ 2 b
2
∂
3 y
∂ t ∂ x
2

 b
2
= r
2
2 ρ


Internal friction damping constant
PBASS Session 5: Vibrations of Strings and Membranes
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Lossy Wave Equation (Air & Internal Damping)
Taking both effects into account, we get the lossy wave equation:
∂
∂ t
2 y
2
= c
2
∂
2 y
∂ x
2
− 2 b
1
∂ y
∂ t
+ 2 b
2
∂
3 y
∂ t ∂ x
2
Let’s ‘try’ a solution of the form: y = e st e
± jkx where we expect that, just as with the harmonic oscillator, s will contain a real part, representing exponential decay, i.e.
s = j ω − α
PBASS Session 5: Vibrations of Strings and Membranes
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9

Lossy String: Free Vibration Solutions
Given a solution for free vibration of the form: y = e st e
± jkx
The relevant derivatives are:
∂ y
∂ t
∂
2 y
=
=
∂ t
2 se st e
± jkx s
2 e st e
=
± jkx sy
= s
2 y
∂
3 y
∂ t ∂ x
2
=
∂


∂
2 y
∂ x
2
∂ t


=
∂ k
2 e st e
± jkx
∂ t
∂ y
∂ x
∂
2 y
∂ x
2
=
= jke st e
± jkx j
2 k
2 e st e
±
= ± jkx jky
= − k
2 y
= − k
2 s e st e
± jkx
= − k
2 sy
PBASS Session 5: Vibrations of Strings and Membranes
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Lossy String: Free Vibration Solutions (cont.)
Substitution of all the derivatives into the lossy wave equation gives: s
2 y = − k
2 c
2 y − 2 b
1 sy − 2 b
2 k
2 sy or, omitting the trivial solution y = 0 : s
2
+ 2 b
1
+ b
2 k
2 s + k
2 c
2
= 0 so that s k
=
=
=
−
− 2 α k
α k
− α k
±
± j j
± k
2
4 α
2 k
2 c
2
− 4
− α
2 k k
2 c
2
ω k
α k
ω k
=
= b
1
+ b
2 k
2 k
2 c
2
− α
2 k
PBASS Session 5: Vibrations of Strings and Membranes
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Lossy String: Free Vibration Solutions (cont.)
So we have the expected solutions of the form:
Where y k
= e s k t e
± jkx s k
= ± j ω k
− α k with
α k
ω k
= b
1
+ b
2 k
2
= k
2 c
2
− α
2 k
Need to know the wave number to compute solutions -> this follows from imposing boundary conditions.
PBASS Session 5: Vibrations of Strings and Membranes
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Lossy String: Free Vibration Solutions (cont.)
The general solution takes the form: y k
( x , t ) =


Ae s k t − jkx
+ Be s k t + jkx 
 e
− α k t
Imposing the (fixed, fixed) boundary conditions gives


 y y
( 0 , t
( L , t
) =
) =
0
0







Ae


Ae s k t
+ Be s k t − jkL
+ s k t
Be

 e
− α k t s k t + jkL
= 0

 e
− α k t
= 0
PBASS Session 5: Vibrations of Strings and Membranes
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11

Lossy String: Free Vibration Solutions (cont.)
Since the exponential cannot be zero for finite t , this becomes:


Ae s k t
Ae s k t
+ Be
− jkL
+ s k t
Be
= 0 s k t + jkL
= 0
Which leads again to sin( kL ) = 0
Hence we have again a series of wave numbers at which the string can freely vibrate: k n
= n π
L n = 1 , 2 , 3 , 4 , 5 ....
PBASS Session 5: Vibrations of Strings and Membranes
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Lossy String: Free Vibration Solutions (cont.)
Hence we can write the possible free-vibration solutions ( = modes of vibration) in terms of the mode number n : y n
( x , t ) = A n sin( k n x ) e
− α n t e j ω n t positiondependent amplitude scaling exponential damping sinusoidal oscillation
α
Where and
ω n n
=
= b
1 k
+ n b
2 c k
2 n
− α
2 n k n
= n π
L
PBASS Session 5: Vibrations of Strings and Membranes
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Modal Frequency and Damping
Example:
L = 0.6; c = 329 b1 = 20.1
b2 = 2.7e-2 purely harmonically related partials)
Damping
Inharmonicity
Damping introduces slight inharmonicity.
PBASS Session 5: Vibrations of Strings and Membranes
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External Force Excitation
The string can be excited by adding a force density term:
∂
∂ t
2 y
2
= c
2
∂ 2 y
∂ x 2
− 2 b
1
∂ y
∂ t
+ 2 b
2
∂ 3 y
∂ t ∂ x 2
+ F ( x , t )
It can be shown that if we excite the string at x=x e up at x = x p is:
, the resulting motion y ( x p
, t ) = n
N
= 1 sin n π x e
L sin

 n π x p
L


F ( x e
, t ) * h n
( t )
Where we have the individual ‘modal oscillators’ impulse responses: h n
( t ) = e
− α n t
M ω n sin( ω n t )
M
1 n
=
" modal
σ L
42
2 mass"
PBASS take ω n out for auralisation, scales better to ‘heard’ air pressure
Session 5: Vibrations of Strings and Membranes
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Modal Synthesis
This can be used for simulation on basis of summing modal contributions, i.e. “modal synthesis”:
F
F ( x e
) x e y x p f
1
α
1 f
2
α
2 f
3
α
3 f
4
α
4 f
N
α
N modal weights (excitation) sin n π x e
L modal oscillators h n
( t ) = modal weights (pick-up) e
− α n t m ω n sin( ω n t ) sin

 n π x p
L


PBASS y ( x p
)
N.B.: implement modal oscillators as discretised harmonic oscillators
Session 5: Vibrations of Strings and Membranes
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2D Vibration: Rectangular Membrane
The (transversal) forces on the membrane surface element dxdy are:
F x
F y
=
=
− Tdx
∂ 2 u
∂ x 2
− Tdy dx
∂
∂
2 u y 2 dx u y y + dy Now aplpying Newton’s 2 nd law yields:
F = F x
+ F y
= ( σ dx dy mass
) ⋅
∂
∂
2 u
2 accelarati on
PBASS Session 5: Vibrations of Strings and Membranes x x + dx
28
14

Rectangular Membrane: Equation of Motion
Equating force with mass times acceleration gave
T dx dy
 ∂
∂ x
2 u
2
+
∂ 2 u
∂ y
2

=
( σ dx dy
) ∂ 2 u
∂ t
2
From which we obtain the equation of motion:
∂ 2 u
∂ t 2
=


T
ρ


 

∂ 2 u
∂ x 2
+
∂ 2 u
∂ y 2


= c
2 ∇ u
Where c is the wave speed: c = T ρ
PBASS Session 5: Vibrations of Strings and Membranes
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Rectangular Membrane with Fixed Ends: Free Vibrations
Assume solution of a factorised form: u ( x , y , z ) = X ( x ) Y ( y ) e j ω t
Impose boundary conditions u = 0 at



 x y x y
=
=
=
=
0
0
L x
L y
L y
0
This leads to solutions of the form: u nm
( x , y , t ) = A sin

 m π x
L x


 sin n π y
L y e j ω mn t with modal frequencies:
ω mn
= π c m 2
L
2 x
+ n 2
L
2 y
L x
PBASS Session 5: Vibrations of Strings and Membranes
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Rectangular Membrane: Modal Frequencies
ω mn
= π c m
2
L 2 x
+ n
2
L 2 y
• spectrum is inharmonic
• modal density increases with frequency
PBASS Session 5: Vibrations of Strings and Membranes
Rectangular Membrane: Modes of Vibration
31
PBASS Session 5: Vibrations of Strings and Membranes
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Free, Damped Membrane Vibrations
Damping can be incorporated in a way similar to with the string, leading to so-called “proportional damping”: u nm
( x , y , t ) = A sin


 m π x
L x


 sin n π y
L y e
− α mn t e j ω mn t where
α mn
= b
1
+ b
2 k 2 mn
ω mn
= k
2 mn c
2 − α 2 mn
PBASS Session 5: Vibrations of Strings and Membranes
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Modal Synthesis
As with the string, we can show that the output is a sum of modal contributions: u ( x p
, t ) =
M m
N
= 1 n = 1
A ( x e
, y e
, x p
, y p
) F ( x e
, t ) * h mn
( t ) y p y e
F x e u x p
A ( x e
, y e
, x p
, y p
) = sin


 n π x e
L x

 sin n π y e
L y sin

 n π x
L x p

 sin n π y p
L y h mn
( t ) = e
− α mn t
M mn
ω mn sin( ω mn t ) M mn
=
σ L x
L y
4
PBASS take ω n out for auralisation, scales better to ‘heard’ air pressure
Session 5: Vibrations of Strings and Membranes
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Summary
• Learned how the wave equation is constructed from considering transversal tension forces strings and membranes.
• Understood how damping forces can be added to the wave equation in order to take into account air damped and internal friction.
• Learned how to apply simple boundary conditions
• Understood that the motion at any point along the string is the sum of two travelling waves.
• Learned how to calculate the modes of vibration (frequencies and damping) of string and membranes.
• Learned about modal synthesis.
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