Free vibration (Introduction)
A system is said to undergo free vibration when it oscillates only under an initial disturbance with no external forces after the initial disturbance.
Examples
• The oscillations of the pendulum of a clock
• The vertical oscillatory motion felt by a bicyclist after hitting a road bump
• The motion of a child on a swing under an initial push
Free vibration (EOM)
) ( )
( )
( )
( t c x t kx t F t x
m && + & + =
EOM of a single degree of freedom system
System’s characteristics External forces Free vibration: External forces = 0
0 )
( )
( )
( t + c x t + kx t = x
m && &
The response of the system x(t) can be known from the solution of the EOM
Review differential equation (1)
3
0
2 2 2
1
+ + a y =
dx a dy
dx y a d
Consider the differential equation shown below
Auxiliary equation: a1r2 + a2r + a3 = 0 Roots of the auxiliary equation: r1,r2
Review differential equation (2)
Case 2: r1 and r2 are real numbers and r1 = r2
Solution of differential equation:
x
e
rx C C
y = (
1+
2)
1Case 1: r1 and r2 are real numbers and r1 ≠ r2
Solution of differential equation:
x r x
r
C e
e C
y =
1 1+
2 2Review differential equation (3)
Case 3: r1 and r2 are complex numbers r1 = a+bi, r2 = a−bi
Solution of differential equation:
x bi a x
bi
a
C e
e C
y =
1 ( + )+
2 ( − )θ θ
θ cos isin
ei = + From Euler identity:
or
)) sin(
) cos(
( A
1bx A
2bx
e
y =
ax+
) sin( + φ
= Ae bx
y
axFree vibration system (undamped)
0 )
( )
( t + kx t = x
m &&
EOM of the undamped free vibration system
2 + k = 0 Auxiliary equation: mr
Roots of the auxiliary equation: ± i
(
k m)
(m and k are always positive, the roots of the aux. eq. are complex numbers)
Solution of differential equation:
)) sin(
)
cos( 2
1 t
m A k
mt A k
x = +
) sin( +φ
= t
m A k
or x
(A1 and A2 or A and φ are obtained from initial conditions)
Response of the free vibration system
) sin( +φ
= t
m A k
x
Consider the response in the term
m k τ = 2π
m k
φ
is call “natural frequency”
m k
m
n
= k
ω
• Free vibration only occurs at a certain frequency ωn.
• Response is sinusoidal and not decaying (undamped).
τ π π
ω
n= 2 f
n= 2
and
Initial condition (1)
Given EOM: mx&&(t) + kx(t) = 0
Initial condition: x(0) = x0 , x&(0) = v0
0 )
( )
( + x t =
m t k
x&&
From given EOM x&&(t) +ωn2x(t) = 0 Response of the system is x(t) = Asin(ωnt +φ)
) cos(
)
(t = Aω ω t +φ
x& n n
) 0
0
( x
x =
Initial condition: x0 = Asin(φ)
)
0 Aωn cos(φ v =
) 0
0
( v
x& =
(1) (2)
Initial condition (2)
Solving equations (1) and (2) yield,
n
n x v
A ω
ω2 02 + 02
=
Given EOM: mx&&(t) + kx(t) = 0
Initial condition: x(0) = x0 , x&(0) = v0
and
0 1 0
tan v
nx φ = − ω
Therefore the response is
) tan
sin(
) (
0 1 0
2 0 2
0 2
v t x
v t x
x n n
n
n ω ω
ω
ω −
+ +
=
Relationship between x, v, and a.
) sin(
)
(t = A ω t +φ
x n
Displacement
) cos(
)
(t = ω A ω t +φ
x& n n
Velocity
) sin(
)
(t = −ω2A ω t +φ
x&& n n
Acceleration
Example: (m, k, ω n )
A vehicle wheel, tire, and suspension assembly can be
modeled crudely as a single-degree-of-freedom spring-mass system. The mass of the assembly is measured to be about 300 kilograms (kg). Its frequency of oscillation is observed to be 10 rad/s. What is the approximate stiffness of the tire,
wheel, and suspension assembly? [Inman ex1.1.2]
Example: Young’s Modulus from f
nA simply supported beam of square cross section 5mm×5mm and length 1m, carrying a mass of 2.3kg at the middle, is
found to have a natural frequency of transverse vibration of 30 rad/s. Determine the Young’s modulus of elasticity of the
beam.
