1
Linearized Vibrations in Moving Frames
1.1
Free vibrations of a string
We discuss here the shapes and frequencies of the free vibrating modes of a continuous string. Before actually starting with this problem, that will be solved
using the tools given in this course, we first look at a similar and simpler problem, using more classical methods.
1.1.1
Concentrated mass under traction
Consider a concentrated mass M , attached to two massless strings of length L
(see figure 1). The strings pull on both sides of the mass with a traction N0 , in
the equilibrium position. Suppose that the mass is taken out of its equilibrium
position, and moved by uez , where ez is a unit vector perpendicular to the axes
of the strings in the equilibrium position.
N
N
M
L
u
L
Figure 1: Mass M pulled by two strings in traction N
1. What are the forces applied on the mass, in the position out of equilibrium?
2. Write the equation of motion, and derive the value of u.
When in equilibrium, the traction forces applied by the strings cancel out.
However, when the mass is taken out its central position, the total force applied
on the mass is then
u
F = −2N0 ez
L
where we used the fact that, in small transformations, sin α = tan α = u/L. On
the other hand, the acceleration of the mass is üez , so that the equilibrium of
forces gives
2N0
u = 0.
M ü +
L
The system will therefore vibrate at frequency
r
2N0
ω0 =
.
ML
1.1.2
Free vibrations of a string
Let V be a cylindrical string along axis ez with a length L, a circular cross
section of radius a, and clamped at both sides.
1. describe the stress state σ ref of the string at rest;
1
2. assuming that the string has an isotropic linear elastic behavior around
this state, and that the free vibrating solutions are in the form
u = u(z) cos(ωt)ey ,
derive the values of u(z) and ω;
3. in the case that the string we are considering is a guitar steel string, tuned
at a traction of 60 N, compare the value of the shear modulus and the norm
of the stress tensor at rest.
The string at rest undergoes uniaxial stress, in the form
σ ref = σ0 ez ⊗ ez .
Using the proposed form for the free vibrating modes, we have D x (u) =
u0 (z) cos(ωt)ey ⊗ez , = u0 (z) cos(ωt)ey ⊗s ez , and σ = 2µu0 (z) cos(ωt)ey ⊗s ez .
We then have
Divx (σ + D x (u)σ ref ) = (σ0 + µ)u00 (z) cos(ωt)ey ,
and
ρü = −ω 2 ρu(z) cos(ωt)ey .
so that equation (1) yields
(σ0 + µ)u00 (z) = −ω 2 ρu(z)
Using the boundary conditions (the string is clamped at both ends, u(−L) =
u(L) = 0), this equation yields
r
z
σ0 + µ t
cos nπ
ey .
u = sin nπ
L
ρ L
The shear modulus of steel is µ = 80 GPa. With σ0 S = 60 N, and a section of
the string S ≈ 1 mm2 , we have σ0 = 60 MPa. Hence µ σ0 . This would mean
that the tension of the strings of a guitar has no influence on their frequency
of vibration, and on their tuning ! However, here, the hypothesis used for the
kinematics (the displacement is constant in each section: u depends only on
z) is not valid for µ σ0 . Indeed, it implies a stress state in the string that
does not verify the boundary conditions on the sides of the string σ · n = 0.
In the case of a guitar string, more refined hypotheses should be used for the
kinematics.
1.2
Drums
Consider a drumming instrument, modeled as a circular membrane of radius a
and thickness e. We suppose that the horizontal displacements of the membrane
on the outside, at r = R, vanish. The gravity forces will be neglected.
1. propose a general form for the stress state of such a membrane, when at
rest?
2
Figure 2: The Bessel function
2. assuming that the membrane has an isotropic linear elastic behavior around
this state, and possibly making additional suppositions on the value of the
shear modulus, derive the value of the vibrating frequencies for free vibrating modes in the form u = u(r) cos(ωt)ez .
Remark The non singular solution of the Bessel equation f 00 + f 0 /r + k 2 f = 0
is the Bessel function J0 (kr), whose first roots are x0 = 2.40, x1 = 5.52, and
x2 = 8.65 (see figure 2).
The stress state in the membrane is in the form σ ref = T (I d − ez ⊗ ez ), with
T ≥ 0 (traction).
Considering the displacement in the proposed form, one gets D x (u) =
u0 (r) cos(ωt)ez ⊗ er , = u0 (r) cos(ωt)ez ⊗s er , and σ = 2µu0 (r) cos(ωt)ez ⊗s er .
We then have
Divx (σ + D x (u)σ ref ) = (µ + T )(u00 +
u0
) cos(ωt)ez ,
r
and
ρü = −ω 2 ρu(r) cos(ωt)ez .
so that equation (1), with the additional hypothesis that µ T (see previous
exercice), yields
u0 (r)
ρ
u00 (r) +
= ω 2 u(r),
r
T
the solution of which is
r
ρ
r .
u(r) = J0 ω
T
The boundary condition, u(r = R) = 0, yields
s
T
,
ωn = xn
ρR2
where the xn are the roots of the Bessel function.
3
1.3
Important formulas
In a moving frame:
γ
= γo + Ω2 u + 2Ωu̇ + ü
v
= v o + Ωu + u̇
The linearized balance of momentum in a moving frame with follower forces and
applied pressure :
Divx (σ + D x (u)σ ref ) − ρ(Ω2 − D x (f ref ))u + f d = ρ (ü + 2Ωu̇)
(1)
The boundary condition with follower pressure on Γσ :
σn = − (pd + pref tr() + (gradpref · u)) n + 2pref n
The linear isotropic elastic constitutive behavior or the Hooke’s law for small
strains :
σ
= λtr()I d + 2µ
=
D x (u) + D Tx (u)
2
4