3.11 Non-linear models
(telephone wires, rocket
motion, pulled rope)
Table of Contents
Nonlinear Springs
Hard and Soft Springs
Nonlinear Pendulum
Telephone Wires
Rocket Motion
Variable Mass
Nonlinear Springs
for a linear function where F (x) = kx
d2
m 2 + F (x) = 0
dy
for a nonlinear function F (x) = kx + k1 x3
d2 x 2 d2 x
m 2
+ kx = 0 or m 2
+ ks + k1 x3 = 0
dt dt
Hard and Soft Springs
The restoring force is given by
F (x) = kx + k1 x3 , k > 0
The spring is hard if k1 >0
The spring is soft if k1 <0
Nonlinear Pendulum
3.11 Non-linear models (telephone wires, rocket motion, pulled rope) 1
, a simple pendulum is a special case of the physical pendulum and consists
of a rod of length l to which a mass m is attached at one end\
the angular acceleration is
d2 s d2 θ
a= 2 =l 2
dt dt
For Newton's second law
d2 θ
F = ma = ml 2
dt
Equate the the two different versions of the tangential force to obtain
d2 θ g
+ sin θ = 0
dt2 l
If we assume that the displacements θ are small enough to justify using the
replacement sin θ ≈ θ then
d2 θ g
+ θ=0
dt2 d
Telephone Wires
the firs order differential equation gives
dy W
=
dx T1
Since W = ρs
dy ρs
=
dx T1
Since the arc length between points P1 and P2 is given by
3.11 Non-linear models (telephone wires, rocket motion, pulled rope) 2
(telephone wires, rocket
motion, pulled rope)
Table of Contents
Nonlinear Springs
Hard and Soft Springs
Nonlinear Pendulum
Telephone Wires
Rocket Motion
Variable Mass
Nonlinear Springs
for a linear function where F (x) = kx
d2
m 2 + F (x) = 0
dy
for a nonlinear function F (x) = kx + k1 x3
d2 x 2 d2 x
m 2
+ kx = 0 or m 2
+ ks + k1 x3 = 0
dt dt
Hard and Soft Springs
The restoring force is given by
F (x) = kx + k1 x3 , k > 0
The spring is hard if k1 >0
The spring is soft if k1 <0
Nonlinear Pendulum
3.11 Non-linear models (telephone wires, rocket motion, pulled rope) 1
, a simple pendulum is a special case of the physical pendulum and consists
of a rod of length l to which a mass m is attached at one end\
the angular acceleration is
d2 s d2 θ
a= 2 =l 2
dt dt
For Newton's second law
d2 θ
F = ma = ml 2
dt
Equate the the two different versions of the tangential force to obtain
d2 θ g
+ sin θ = 0
dt2 l
If we assume that the displacements θ are small enough to justify using the
replacement sin θ ≈ θ then
d2 θ g
+ θ=0
dt2 d
Telephone Wires
the firs order differential equation gives
dy W
=
dx T1
Since W = ρs
dy ρs
=
dx T1
Since the arc length between points P1 and P2 is given by
3.11 Non-linear models (telephone wires, rocket motion, pulled rope) 2
