Download Mathematical Modeling of Mechanical Vibrations

Lecture 12. Second Order DE: Mathematical
Modeling of Mechanical Vibrations (§3.7)
March 13, 2012 (Tue)
Lecture 12. Second Order DE: Mathematical Modeling of Mec
Vertically Vibrating Spring
A weight of mass m is attached to a spring suspended from a
beam. It is then streched and set to oscilling with an initial
velocity v0 .
We want to derive a mathematical model representing this motion.
I
First consider the spring with no mass
attached. It is assumed to be in
equilibrium, so there is no motion.
This is called the spring equilibrium.
Lecture 12. Second Order DE: Mathematical Modeling of Mec
Vertically Vibrating Spring
A weight of mass m is attached to a spring suspended from a
beam. It is then streched and set to oscilling with an initial
velocity v0 .
We want to derive a mathematical model representing this motion.
I
First consider the spring with no mass
attached. It is assumed to be in
equilibrium, so there is no motion.
This is called the spring equilibrium.
I
Now we attach a weight of mass m to
the spring. This weight stretches the
spring until it is once more in
equilibrium at L. This is called the
spring-mass equilibrium.
Lecture 12. Second Order DE: Mathematical Modeling of Mec
Vertically Vibrating Spring
I
The position of the bottom of the
mass in the spring-mass equilibrium is
the reference point from which we
measure displacement, so it
corresponds to u = 0, where u = u(t),
measured positive downward, denote
the desplacement of the mass from its
equilibrium position at time t.
Lecture 12. Second Order DE: Mathematical Modeling of Mec
Vertically Vibrating Spring
I
The position of the bottom of the
mass in the spring-mass equilibrium is
the reference point from which we
measure displacement, so it
corresponds to u = 0, where u = u(t),
measured positive downward, denote
the desplacement of the mass from its
equilibrium position at time t.
I
At this point there are two forces
acting on the mass: the force of gravity
m g and the restoring force of the
spring Fs that acts upward and
depends on the displacement u.
Lecture 12. Second Order DE: Mathematical Modeling of Mec
Vertically Vibrating Spring
I
The position of the bottom of the
mass in the spring-mass equilibrium is
the reference point from which we
measure displacement, so it
corresponds to u = 0, where u = u(t),
measured positive downward, denote
the desplacement of the mass from its
equilibrium position at time t.
I
At this point there are two forces
acting on the mass: the force of gravity
m g and the restoring force of the
spring Fs that acts upward and
depends on the displacement u.
I
The fact that we have equilibrium at
u = u0 means that the total force on
the weight is 0,
Lecture 12. Second Order DE: Mathematical Modeling of Mec
Vertically Vibrating Spring
I
We stretch the spring further beyond
the spring-mass equilibrium, which
makes the weight moving. Its velocity
is v = u 0 .
Lecture 12. Second Order DE: Mathematical Modeling of Mec
Vertically Vibrating Spring
I
We stretch the spring further beyond
the spring-mass equilibrium, which
makes the weight moving. Its velocity
is v = u 0 .
I
Now, in addition to the gravity and the
restoring force, there is a damping
force Fd , which is the resistance to the
motion of the weight due to the
medium (air?) through which the
weight is moving and perhaps to
something internal to the spring.
Lecture 12. Second Order DE: Mathematical Modeling of Mec
Vertically Vibrating Spring
I
We stretch the spring further beyond
the spring-mass equilibrium, which
makes the weight moving. Its velocity
is v = u 0 .
I
Now, in addition to the gravity and the
restoring force, there is a damping
force Fd , which is the resistance to the
motion of the weight due to the
medium (air?) through which the
weight is moving and perhaps to
something internal to the spring.
I
The damping force mostly dpends on
the velocity. So we will write it as
Fd (v ).
Lecture 12. Second Order DE: Mathematical Modeling of Mec
Vertically Vibrating Spring
I
We stretch the spring further beyond
the spring-mass equilibrium, which
makes the weight moving. Its velocity
is v = u 0 .
I
Now, in addition to the gravity and the
restoring force, there is a damping
force Fd , which is the resistance to the
motion of the weight due to the
medium (air?) through which the
weight is moving and perhaps to
something internal to the spring.
I
The damping force mostly dpends on
the velocity. So we will write it as
Fd (v ).
I
To be complete, we will allow for an
external force F (t) as well.
Lecture 12. Second Order DE: Mathematical Modeling of Mec
Vertically Vibrating Spring
I
Let a = v 0 = u 00 denote the acceleration of the weight. By
Newton’s second law,
ma = total force acting on the weight
= Fs (u) + mg + Fd (v ) + F (t)
mu
1
00
= Fs (u) + mg + Fd (u 0 ) + F (t).
Discovered by the English scientist Robert Hooke in 1660. It states that, for relatively small deformations of
an object, the displacement or size of the deformation is directly proportional to the deforming force or load. Under
these conditions the object returns to its original shape and size upon removal of the load.
Lecture 12. Second Order DE: Mathematical Modeling of Mec
Vertically Vibrating Spring
I
Let a = v 0 = u 00 denote the acceleration of the weight. By
Newton’s second law,
ma = total force acting on the weight
= Fs (u) + mg + Fd (v ) + F (t)
mu
I
00
= Fs (u) + mg + Fd (u 0 ) + F (t).
Experimental fact (Hook’s law
1
) says that
Fs (u) = −k(L + u),
where k > 0 is refered to a spring constant.
1
Discovered by the English scientist Robert Hooke in 1660. It states that, for relatively small deformations of
an object, the displacement or size of the deformation is directly proportional to the deforming force or load. Under
these conditions the object returns to its original shape and size upon removal of the load.
Lecture 12. Second Order DE: Mathematical Modeling of Mec
Vertically Vibrating Spring
I
Let a = v 0 = u 00 denote the acceleration of the weight. By
Newton’s second law,
ma = total force acting on the weight
= Fs (u) + mg + Fd (v ) + F (t)
mu
I
00
= Fs (u) + mg + Fd (u 0 ) + F (t).
Experimental fact (Hook’s law
1
) says that
Fs (u) = −k(L + u),
I
where k > 0 is refered to a spring constant.
Now the equation becomes
mu 00 = −k(L + u) + mg + Fd (u 0 ) + F (t).
1
Discovered by the English scientist Robert Hooke in 1660. It states that, for relatively small deformations of
an object, the displacement or size of the deformation is directly proportional to the deforming force or load. Under
these conditions the object returns to its original shape and size upon removal of the load.
Lecture 12. Second Order DE: Mathematical Modeling of Mec
Vertically Vibrating Spring
I
The damping force Fd (u 0 ) always acts against the velocity.
Hence we can write it as
Fd (u 0 ) = −γu 0 ,
where γ ≥ 0 is a nonnegative constant, called the damping
constant.
Lecture 12. Second Order DE: Mathematical Modeling of Mec
Vertically Vibrating Spring
I
The damping force Fd (u 0 ) always acts against the velocity.
Hence we can write it as
Fd (u 0 ) = −γu 0 ,
I
where γ ≥ 0 is a nonnegative constant, called the damping
constant.
Now, since mg − kL = 0 from Fs (u0 ) + mg = 0, our equation
becomes,
mu 00 = mg − k(L + u) − γu 0 + F (t) ⇔ mu 00 + γu 0 + ku = F (t)
Lecture 12. Second Order DE: Mathematical Modeling of Mec
Vertically Vibrating Spring
I
The damping force Fd (u 0 ) always acts against the velocity.
Hence we can write it as
Fd (u 0 ) = −γu 0 ,
I
I
where γ ≥ 0 is a nonnegative constant, called the damping
constant.
Now, since mg − kL = 0 from Fs (u0 ) + mg = 0, our equation
becomes,
mu 00 = mg − k(L + u) − γu 0 + F (t) ⇔ mu 00 + γu 0 + ku = F (t)
p
Let c = γ/2m, ω0 = k/m, f (t) = F (t)/m, and x = u.
Then the DE becomes
x 00 + 2cx 0 + ω02 x = f (t),
whice is refered as the equation for harmonic motion.
Lecture 12. Second Order DE: Mathematical Modeling of Mec
Harmonic Motion: Vertically Vibrating Spring
I
The equation for the motion of a vibrating spring:
x 00 + 2cx 0 + ω02 x = f (t), x(0) = u0 , x 0 (0) = v0 ,
where x(t) represents a motion
of vertical vibating spring at
p
time t, c = γ/2m, ω0 = k/m, and f (t) = F (t)/m.
Lecture 12. Second Order DE: Mathematical Modeling of Mec
Harmonic Motion: Vertically Vibrating Spring
I
The equation for the motion of a vibrating spring:
x 00 + 2cx 0 + ω02 x = f (t), x(0) = u0 , x 0 (0) = v0 ,
where x(t) represents a motion
of vertical vibating spring at
p
time t, c = γ/2m, ω0 = k/m, and f (t) = F (t)/m.
I
IS Units length = meter (m), time = second(s), mass =
kilogram (kg ), velocity = m/s, acceleration = m/s 2 ,
g = 9.8m/s 2 , force = kg · m/s 2 = a newton (N), k = N/m.
Lecture 12. Second Order DE: Mathematical Modeling of Mec
Harmonic Motion: Vertically Vibrating Spring
I
The equation for the motion of a vibrating spring:
x 00 + 2cx 0 + ω02 x = f (t), x(0) = u0 , x 0 (0) = v0 ,
where x(t) represents a motion
of vertical vibating spring at
p
time t, c = γ/2m, ω0 = k/m, and f (t) = F (t)/m.
I
IS Units length = meter (m), time = second(s), mass =
kilogram (kg ), velocity = m/s, acceleration = m/s 2 ,
g = 9.8m/s 2 , force = kg · m/s 2 = a newton (N), k = N/m.
Lecture 12. Second Order DE: Mathematical Modeling of Mec
Example
I
A mass weighing 4 lb stretches a spring 2 inches. The mass is
displaced an additional 6 inches and then released; and is in a
medium that exerts a viscous resistance of 6 lb when the mass
has a velocity of 3 ft/sec. Formulate the IVP that governs the
motion of this mass:
mu 00 + γu 0 + ku = F (t), u(0) = u0 , u 0 (0) = v0
Lecture 12. Second Order DE: Mathematical Modeling of Mec
Harmonic Motion
I
The equation for the motion of
a vibrating spring:
mu 00 + γu 0 + ku = F (t)
I
The equation for the flow of
electric current on a simple
Electric Circuit (see Lecture 2
slide):
L
d 2I
dI
1
dE
+R + I =
2
dt
dt
C
dt
x 00 + 2cx 0 + ω02 x = f (t)
Lecture 12. Second Order DE: Mathematical Modeling of Mec