BENG 221: Mathematical Methods in Bioengineering
Lecture 17
Wave Equation in One Dimension for Vibrating Strings
References
Haberman APDE, Ch. 4.
http://en.wikipedia.org/wiki/Wave_equation
http://en.wikipedia.org/wiki/Vibrating_string
BENG 221
Lecture 17
M. Intaglietta
The one dimensional wave equation. The vibrating string as a boundary value
problem
Given a string stretched along the x axis, the vibrating string is a problem where forces
are exerted in the x and y directions, resulting in motion in the x-y plane, when the string
is displaced from its equilibrium position within the x-y plane, and then released.
The free body diagram of an element of string of length ∆s subjected to a tension T is
shown. The string material has density ρ . The equation of motion is obtained by
applying Newton’s second law of motion to the element of length ∆s in both directions.
For the x direction (and ignoring the torsional effects due to the applied torque):
(T + ∆T ) cos(θ + ∆θ ) − T cos θ = ρ A∆s
∂2 x
∂t 2
and in the y direction:
(T + ∆T ) sin(θ + ∆θ ) − T sin θ = ρ A∆s
∂2 y
∂t 2
Where the bar over the partial derivative signifies the average acceleration over the
element ∆s . A is the cross section of the string, assumed constant and equal to 1.
Dividing through by ∆s and taking the limit ∆s → 0 we obtain:
∂
∂2 x
(T cos θ ) = ρ 2
∂s
∂t
∂
∂2 y
(T sin θ ) = ρ 2
∂s
∂t
(15)
1
since cos θ =
∂x
∂y
and sin θ =
the above equations can be reduced to the form (for
∂s
∂s
constant T)
∂2 x
∂2 x
ρ
=
∂s 2
∂t 2
∂2 y
∂2 y
T 2 =ρ 2
∂s
∂t
T
(15a)
Since there is no motion in the x direction
Note that tan θ =
sin θ =
∂2 x
= 0.
∂t 2
∂y
, therefore
∂x
tan θ
1 + tan 2 θ
≈
∂y
∂x
for small θ therefore in (15a):
T
∂ ∂y
∂ ∂y
∂
∂ ∂y
∂2 y
=T
= T sin θ ≈ T
=T 2
∂s ∂x
∂x ∂s
∂x
∂x ∂x
∂x
and the system of equations reduces to:
∂2 x
=0
∂t 2
∂2 y T ∂2 y
=
∂t 2 ρ ∂x 2
(16)
The first equation has a trivial solution. The second equation can be solved by the
method of separation of variables by assuming that y ( x, t ) = X ( x) F (t ) which leads to:
X
d 2F T d 2 X
T
= F
and dividing both sides by XF we obtain:
2
2
ρ dx
ρ
dt
ρ 1 d 2F
T F dt 2
=
1 d2X
= −κ 2
2
X dx
according to previous reasoning indicating that an equality between functions of different
variables implies that the functions are equal to a constant. Therefore we can write:
2
d 2F
T
= −κ 2 F
2
ρ
dt
and
d2X
= −κ 2 X
2
dt
as previously, the solutions of these equations correspond to the case where the constant
is positive and therefore the characteristic equation of the second order differential
equation has imaginary coefficients, leading to the following solutions in terms of
trigonometric functions:
F = A1 cosκ
T
ρ
t + B1 sin κ
T
ρ
t
X = A2 cos κ x + B2 sin κ x
Therefore the system of equations given in (16) has product solutions of the form:
⎛
T
T ⎞
y (t , x) = ⎜⎜ A1 cosκ
t + B1 sin κ
t ⎟ ( A cos κ x + B2 sin κ x )
ρ
ρ ⎟⎠ 2
⎝
When κ = 0 then:
F0 = A0 + B0t
X 0 = C0 + D0 x
y0 = ( A0 + B0t )( C0 + D0 x )
Boundary conditions. Vibrating string clamped at both ends
If we impose the B.C.s that the string is clamped at the ends, namely y = 0 at x = 0, L then
there is no motion in the y direction and:
⎛
T
T ⎞
y (t , x) = ⎜⎜ A1 cosκ
t + B1 sin κ
t ⎟⎟ A2 = 0
ρ
ρ
⎝
⎠
y0 = ( A0 + B0t )C0 = 0
⎛
T
T ⎞
y (t , x) = ⎜⎜ A1 cosκ
t + B1 sin κ
t ⎟ ( A cos κ L + B2 sin κ L ) = 0
ρ
ρ ⎟⎠ 2
⎝
y0 = ( A0 + B0t )( C0 + D0 L ) = 0
These B.C.s are satisfied by A2 = C0 = D0 = 0 and the eigenvalues
κ L = nπ
n = 1, 2,3,...
leading to eigenfunctions:
3
∞
yn = ∑ sin
n =1
nπ x ⎛
nπ
⎜⎜ An cos
L ⎝
L
T
ρ
t + Bn sin
nπ
L
T ⎞
t⎟
ρ ⎟⎠
(17)
where the constant A1 is now include in An and Bn . Note that for any time t0 the string
has a configuration that depend son on n:
yn ( x, t0 ) = Cn (t0 ) sin
nπ x
L
which describes a family of modes of the string for the specific B.C.s of clamped ends.
These are called the normal modes of vibration. The intensity of sound depends on the
amplitude Cn = An2 + Bn2 which is derived from:
A cos θ + B sin θ = A2 + B 2 sin(θ + λ ) λ = tan −1
A
B
The number of oscillations per unit time or frequency in cycles per second is:
ω=
1 nπ
2π L
T
ρ
=
n T
2L ρ
Sound is produced by the superposition of natural frequencies n = 1,2,3,…. The normal
mode is the first harmonic or fundamental n = 1. The larger the natural frequency, the
higher the pitch. Tuning is accomplished by varying either L, ρ or T. For vibrating
strings the frequencies of the higher harmonics are all integral multiples of the
fundamental. Note that sound is produced by strings vibrating in a lateral direction,
however it is transmitted by waves of compression and rarefaction in the longitudinal
direction (the direction of propagation).
Standing waves and summation of traveling waves
Each standing wave is composed by the summation of two waves traveling in opposite
directions.
Recall that: sin x sin y =
1
1
cos( x − y ) − cos( x + y )
2
2
cos( x + y ) = cos x cos y − sin x sin y
cos( x − y ) = cos x cos y + sin x sin y
Subtracting we obtain:
4
sin x sin y =
1
1
cos( x − y ) − cos( x + y )
2
2
Therefore any single component y j ( x, t ) of the function yn ( x, t )
y j ( x, t ) = sin
jπ x
jπ
sin
L
L
T
ρ
t=
1
jπ ⎛
T ⎞ 1
jπ ⎛
T ⎞
cos
t ⎟⎟ − cos
t⎟
⎜⎜ x −
⎜⎜ x +
2
L ⎝
L ⎝
ρ ⎠ 2
ρ ⎟⎠
Wave traveling to Wave traveling to the left.
the right.
Where the wave velocity V is V =
T
ρ
.
Figure 2
Figure 2 shows how the equation in time is set up for advancing pulse. The upper
diagram shows the pulse at t = 0 given by the equation y = f(x). The lower diagram
shows the same pulse at time t = t having advanced a distance Vt without changing shape.
A new axis Y’ is constructed, displaced a distance x = Vt to the right, x’ being the new
coordinate of any point referred to the new origin. The equation f(x) at time t in terms of
x’ is the same as the equation at t = 0 in terms of x, or y = f(x’) for t = t. However:
x ' = x − Vt therefore
y = f ( x − Vt )
Initial conditions
Suppose the I.C.s are given by a function y(x) such that for t = 0:
5
2 xd
L
0≤ x≤
L
2
2d
L
y ( x, 0) =
( L − x) ≤ x ≤ L
L
2
dy ( x, 0)
=0
dt
y ( x, 0) =
Applying the velocity boundary conditions for t = 0 to (17) we obtain:
dy ∞
nπ
= ∑ Bn
dt n =1
L
T
ρ
cos
nπ x
=0
L
Which satisfies the I.C.s by setting Bn = 0 . Solution of the problem requires determining
the constants An so that that the initial conditions are satisfied:
∞
∑ A sin
n =1
n
nπ x 2d
x
=
L
L
2d
=
( L − x)
L
0≤ x≤
L
2
(18)
L
≤x≤L
2
multiplying both sides of (18) by sin
nπ x ⎞
mπ x
⎛ ∞
An sin
sin
dx =
⎟
∫0 ⎜⎝ ∑
L ⎠
L
n =1
L
L /2
∫
0
mπ x
and integrating between 0 and L we obtain:
L
2d
mπ x
2d
mπ x
sin
dx + ∫
dx
( L − x ) sin
L
L
L
L
L /2
L
(19)
Integrating (19) term by term:
Am
L /2
L
L 2d ⎡
mπ x
mπ x ⎤
x
sin
dx
( L − x) sin
dx ⎥
=
+
⎢∫
∫
2
L ⎣0
L
L
L /2
⎦
(20)
These integrals can be evaluated by considering that:
∫ x sin xdx = sin x − x cos x
and
1
1
2
2
∫ x sin axdx = a ∫ ax sin axdax = a ∫ θ sin θ dθ
(21)
where ax = θ . In (20), upon integration and evaluation at the limits the only non-zero
terms exist at L/2, of which there are two, with same sign. Therefore, in (21) we can set
a = mπ / L , then
6
Am =
8d
mπ x
sin
2 2
mπ
L
which vanishes when m is even and when it is odd the sine term oscillates between the
values ±1 .
The product solution therefore is:
8d
mπ x
mπ
sin
cos
2 2
L
L
m =1 m π
∞
y ( x, t ) = ∑
T
ρ
t m = 1,3,5,...
(22)
7