Download Free mechanical vibrations / Couple mass

Section 4.9; Section 5.6
Free Mechanical Vibrations/Couple Mass-Spring
System
June 30, 2009
Free Mechanical Vibrations/Couple Mass-Spring System
Today’s Session
Free Mechanical Vibrations/Couple Mass-Spring System
Today’s Session
A Summary of This Session:
Free Mechanical Vibrations/Couple Mass-Spring System
Today’s Session
A Summary of This Session:
(1) Free Mechanical Vibration (no forcing term).
Free Mechanical Vibrations/Couple Mass-Spring System
Today’s Session
A Summary of This Session:
(1) Free Mechanical Vibration (no forcing term).
(2) Coupled Mass-Spring systems
Free Mechanical Vibrations/Couple Mass-Spring System
Today’s Session
A Summary of This Session:
(1) Free Mechanical Vibration (no forcing term).
(2) Coupled Mass-Spring systems
(3) Our first exposure to systems of differential equations
Free Mechanical Vibrations/Couple Mass-Spring System
Today’s Session
A Summary of This Session:
(1) Free Mechanical Vibration (no forcing term).
(2) Coupled Mass-Spring systems
(3) Our first exposure to systems of differential equations
(4) Eigenvalues and Eigenvectors
Free Mechanical Vibrations/Couple Mass-Spring System
Today’s Session
A Summary of This Session:
(1) Free Mechanical Vibration (no forcing term).
(2) Coupled Mass-Spring systems
(3) Our first exposure to systems of differential equations
(4) Eigenvalues and Eigenvectors
Free Mechanical Vibrations/Couple Mass-Spring System
Free Mechanical Vibrations
Free mechanical vibration=no forcing function, so f (t) = 0. We
are dealing with
my ′′ + by ′ + ky = 0
where m = mass attached to a spring of stiffness k, subject to
friction (or damping) proportional to speed with damping constant
b.
Free Mechanical Vibrations/Couple Mass-Spring System
Free Mechanical Vibrations
Free mechanical vibration=no forcing function, so f (t) = 0. We
are dealing with
my ′′ + by ′ + ky = 0
where m = mass attached to a spring of stiffness k, subject to
friction (or damping) proportional to speed with damping constant
b.
Four cases: (1) undamped free case; (2) underdamped case; (3)
overdamped case; (4) critical damped case.
Free Mechanical Vibrations/Couple Mass-Spring System
Free Mechanical Vibrations
Free mechanical vibration=no forcing function, so f (t) = 0. We
are dealing with
my ′′ + by ′ + ky = 0
where m = mass attached to a spring of stiffness k, subject to
friction (or damping) proportional to speed with damping constant
b.
Four cases: (1) undamped free case; (2) underdamped case; (3)
overdamped case; (4) critical damped case.
Free Mechanical Vibrations/Couple Mass-Spring System
Case 1: Undamped mass-spring system: b = 0.
The equation is given by:
my ′′ + ky = 0
Free Mechanical Vibrations/Couple Mass-Spring System
Case 1: Undamped mass-spring system: b = 0.
The equation is given by:
my ′′ + ky = 0
or
y ′′ +
k
y =0
m
Free Mechanical Vibrations/Couple Mass-Spring System
Case 1: Undamped mass-spring system: b = 0.
The equation is given by:
my ′′ + ky = 0
or
k
y =0
m
q
k
k
2
Let ω = m . The quantity ω = m
is called the angular
frequency (measured in radians per second)
y ′′ +
Free Mechanical Vibrations/Couple Mass-Spring System
Case 1: Undamped mass-spring system: b = 0.
The equation is given by:
my ′′ + ky = 0
or
k
y =0
m
q
k
k
2
Let ω = m . The quantity ω = m
is called the angular
frequency (measured in radians per second)
y ′′ +
Period: T =
2π
ω
(measured in seconds)
Free Mechanical Vibrations/Couple Mass-Spring System
Case 1: Undamped mass-spring system: b = 0.
The equation is given by:
my ′′ + ky = 0
or
k
y =0
m
q
k
k
2
Let ω = m . The quantity ω = m
is called the angular
frequency (measured in radians per second)
y ′′ +
Period: T = 2π
ω (measured in seconds)
ω
Frequency: f = T1 = 2π
(measured in Hertz=1/seconds=# cycles
per second)
Free Mechanical Vibrations/Couple Mass-Spring System
Case 1: Undamped mass-spring system: b = 0.
The equation is given by:
my ′′ + ky = 0
or
k
y =0
m
q
k
k
2
Let ω = m . The quantity ω = m
is called the angular
frequency (measured in radians per second)
y ′′ +
Period: T = 2π
ω (measured in seconds)
ω
Frequency: f = T1 = 2π
(measured in Hertz=1/seconds=# cycles
per second)
The solution is given by: y = C1 cos ωt + C2 sin ωt = A sin(ωt + φ).
Free Mechanical Vibrations/Couple Mass-Spring System
Undamped case, cont’d
Here A is called the amplitude, and φ is called the phase. They are
given by
q
C1
A = C12 + C22
φ = arctan .
C2
Free Mechanical Vibrations/Couple Mass-Spring System
Undamped case, cont’d
Here A is called the amplitude, and φ is called the phase. They are
given by
q
C1
A = C12 + C22
φ = arctan .
C2
Example A 1/8 kg manss is attached to a spring with stiffness
k = 16N/m. The mass is displaed 0.5 m to the right of the
√
equilibrium point and given an outward initial velocity of 2 m/s.
(a) Neglecting damping, find a formula of the displacement as a
function of time. Display the values of ω, T , f , A and φ.
Free Mechanical Vibrations/Couple Mass-Spring System
Undamped case, cont’d
Here A is called the amplitude, and φ is called the phase. They are
given by
q
C1
A = C12 + C22
φ = arctan .
C2
Example A 1/8 kg manss is attached to a spring with stiffness
k = 16N/m. The mass is displaed 0.5 m to the right of the
√
equilibrium point and given an outward initial velocity of 2 m/s.
(a) Neglecting damping, find a formula of the displacement as a
function of time. Display the values of ω, T , f , A and φ.
(b) How long after release does the mass pass first through the
equilibrium position?
Free Mechanical Vibrations/Couple Mass-Spring System
Undamped case, cont’d
Answers:
√
(a) ω = 8 2 rad/sec; f = 1.8 cycles/sec (or Hz); T = 0.555 sec.
Free Mechanical Vibrations/Couple Mass-Spring System
Undamped case, cont’d
Answers:
√
(a) ω = 8 2 rad/sec; f = 1.8 cycles/sec (or Hz); T = 0.555 sec.
y (t) =
√
√
1
1
cos 8 2t + sin 8 2t
2
8
Free Mechanical Vibrations/Couple Mass-Spring System
Undamped case, cont’d
Answers:
√
(a) ω = 8 2 rad/sec; f = 1.8 cycles/sec (or Hz); T = 0.555 sec.
y (t) =
A=
q
√
√
1
1
cos 8 2t + sin 8 2t
2
8
√
1/2
and tan φ = 1/8
= 4 so φ = 1.326 rad.
√
√
17
sin(8 2t + 1.326)
y (t) =
8
( 12 )2 + ( 81 )2 =
17
8
Free Mechanical Vibrations/Couple Mass-Spring System
Undamped case, cont’d
Answers:
√
(a) ω = 8 2 rad/sec; f = 1.8 cycles/sec (or Hz); T = 0.555 sec.
y (t) =
A=
q
√
√
1
1
cos 8 2t + sin 8 2t
2
8
√
1/2
and tan φ = 1/8
= 4 so φ = 1.326 rad.
√
√
17
sin(8 2t + 1.326)
y (t) =
8
( 12 )2 + ( 81 )2 =
17
8
√
√
(b) y (t) = 0 means sin(8 2t + φ) = 0 so 8 2t + φ = kπ for k
= 0.16 sec. Every 1/2 period (or
integer. Solving gives: t = kπ−φ
ω
0.28 sec) the mass goes through the equilibrium point.
Free Mechanical Vibrations/Couple Mass-Spring System
Overdamped case: b 2 − 4k m > 0
The equation is
my ′′ + by ′ + ky = 0
Free Mechanical Vibrations/Couple Mass-Spring System
Overdamped case: b 2 − 4k m > 0
The equation is
my ′′ + by ′ + ky = 0
The characteristic equation is given by
mr 2 + br + k = 0
Free Mechanical Vibrations/Couple Mass-Spring System
Overdamped case: b 2 − 4k m > 0
The equation is
my ′′ + by ′ + ky = 0
The characteristic equation is given by
mr 2 + br + k = 0
so
b
±
r =−
2m
The solution is given by:
√
b 2 − 4k m
2m
y (t) = C1 e r1 t + C2 e r2 t
Free Mechanical Vibrations/Couple Mass-Spring System
Overdamped case: b 2 − 4k m > 0
The equation is
my ′′ + by ′ + ky = 0
The characteristic equation is given by
mr 2 + br + k = 0
so
b
±
r =−
2m
The solution is given by:
√
b 2 − 4k m
2m
y (t) = C1 e r1 t + C2 e r2 t
Note that both r1 and r2 are negative (why?), so as t → ∞,
y (t) → 0.
Free Mechanical Vibrations/Couple Mass-Spring System
Underdamped case: b 2 − 4k m < 0
With ω =
√
4k m−b2
,
2m
the roots of the characteristic equation are:
r =−
b
±iω
2m
So the solution is given by:
b
b
y (t) = C1 e − 2m t cos ωt + C2 e − 2m t sin ωt.
Free Mechanical Vibrations/Couple Mass-Spring System
Underdamped case: b 2 − 4k m < 0
With ω =
√
4k m−b2
,
2m
the roots of the characteristic equation are:
r =−
b
±iω
2m
So the solution is given by:
b
b
y (t) = C1 e − 2m t cos ωt + C2 e − 2m t sin ωt.
Over time y (t) dies in an oscillatory fashion. ω is a quasi-angular
frequency, T = 2π/ω is called the quasiperiod and f = 1/T is
called the quasi-frequency.
Free Mechanical Vibrations/Couple Mass-Spring System
Underdamped case: b 2 − 4k m < 0
With ω =
√
4k m−b2
,
2m
the roots of the characteristic equation are:
r =−
b
±iω
2m
So the solution is given by:
b
b
y (t) = C1 e − 2m t cos ωt + C2 e − 2m t sin ωt.
Over time y (t) dies in an oscillatory fashion. ω is a quasi-angular
frequency, T = 2π/ω is called the quasiperiod and f = 1/T is
called the quasi-frequency.
The solution can be put in the form
b
y (t) = A e − 2m t sin(ωt + φ)
where (as before):
A=
q
C12 + C22
φ = arctan
C1
.
C2
Free Mechanical Vibrations/Couple Mass-Spring System
Underdamped case: b 2 − 4k m < 0
With ω =
√
4k m−b2
,
2m
the roots of the characteristic equation are:
r =−
b
±iω
2m
So the solution is given by:
b
b
y (t) = C1 e − 2m t cos ωt + C2 e − 2m t sin ωt.
Over time y (t) dies in an oscillatory fashion. ω is a quasi-angular
frequency, T = 2π/ω is called the quasiperiod and f = 1/T is
called the quasi-frequency.
The solution can be put in the form
b
y (t) = A e − 2m t sin(ωt + φ)
where (as before):
A=
q
C12 + C22
φ = arctan
C1
.
C2
Free Mechanical Vibrations/Couple Mass-Spring System
Critically damped case: b 2 − 4k m = 0
In this case, the characteristic equation has a double root
b
, So the solution is
r = − 2m
b
b
y (t) = C1 e − 2m t + C2 t e − 2m t
Free Mechanical Vibrations/Couple Mass-Spring System
Critically damped case: b 2 − 4k m = 0
In this case, the characteristic equation has a double root
b
, So the solution is
r = − 2m
b
b
y (t) = C1 e − 2m t + C2 t e − 2m t
Example: (related to webassign question and example p. 234)
Free Mechanical Vibrations/Couple Mass-Spring System
Critically damped case: b 2 − 4k m = 0
In this case, the characteristic equation has a double root
b
, So the solution is
r = − 2m
b
b
y (t) = C1 e − 2m t + C2 t e − 2m t
Example: (related to webassign question and example p. 234)
Find the value of b for which
y ′′ + b y ′ + 25y = 0 y (0) = 1,
y ′ (0) = 0
is critically damped. Solve for y (t) in this case and sketch it.
Free Mechanical Vibrations/Couple Mass-Spring System
Critically damped case: b 2 − 4k m = 0
In this case, the characteristic equation has a double root
b
, So the solution is
r = − 2m
b
b
y (t) = C1 e − 2m t + C2 t e − 2m t
Example: (related to webassign question and example p. 234)
Find the value of b for which
y ′′ + b y ′ + 25y = 0 y (0) = 1,
y ′ (0) = 0
is critically damped. Solve for y (t) in this case and sketch it.
Answer: b = 10; y (t) = (1 − t)e −5t
Free Mechanical Vibrations/Couple Mass-Spring System
5.6: Coupled Mass-Spring System: Example 1, p. 308
See the graph and description of the couple mass-spring system on
p. 308.
Free Mechanical Vibrations/Couple Mass-Spring System
5.6: Coupled Mass-Spring System: Example 1, p. 308
See the graph and description of the couple mass-spring system on
p. 308.
I will show the derivation of the following equations in class:
Free Mechanical Vibrations/Couple Mass-Spring System
5.6: Coupled Mass-Spring System: Example 1, p. 308
See the graph and description of the couple mass-spring system on
p. 308.
I will show the derivation of the following equations in class:
We wish to solve:
n m x ′′ = −k x + k (y − x)
1
1
2
m2 y ′′ =
−k2 (y − x)
Free Mechanical Vibrations/Couple Mass-Spring System
5.6: Coupled Mass-Spring System: Example 1, p. 308
See the graph and description of the couple mass-spring system on
p. 308.
I will show the derivation of the following equations in class:
We wish to solve:
n m x ′′ = −k x + k (y − x)
1
1
2
m2 y ′′ =
−k2 (y − x)
In our example: m1 = 2kg ; k1 = 4N/m; m2 = 1kg ; and
k2 = 2N/m. So:
n 2x ′′ = −6x + 2y
y ′′ = 2x − 2y
Free Mechanical Vibrations/Couple Mass-Spring System
5.6: Coupled Mass-Spring System: Example 1, p. 308
See the graph and description of the couple mass-spring system on
p. 308.
I will show the derivation of the following equations in class:
We wish to solve:
n m x ′′ = −k x + k (y − x)
1
1
2
m2 y ′′ =
−k2 (y − x)
In our example: m1 = 2kg ; k1 = 4N/m; m2 = 1kg ; and
k2 = 2N/m. So:
n 2x ′′ = −6x + 2y
y ′′ = 2x − 2y
Free Mechanical Vibrations/Couple Mass-Spring System
Example 1, p. 308, cont’d
Let’s clean this up a little bit:
n x ′′ = −3x + y
y ′′ = 2x − 2y
Free Mechanical Vibrations/Couple Mass-Spring System
Example 1, p. 308, cont’d
Let’s clean this up a little bit:
n x ′′ = −3x + y
y ′′ = 2x − 2y
This is a linear system of differential equations of second order
(which is homogeneous; why?) We can put it in the form:
n (D 2 + 3)x − y
= 0
−2x + (D 2 + 2)y
= 0
Free Mechanical Vibrations/Couple Mass-Spring System
Example 1, p. 308, cont’d
Let’s clean this up a little bit:
n x ′′ = −3x + y
y ′′ = 2x − 2y
This is a linear system of differential equations of second order
(which is homogeneous; why?) We can put it in the form:
n (D 2 + 3)x − y
= 0
Therefore
−2x + (D 2 + 2)y
= 0
(D 2 + 2)(D 2 + 3)x − (D 2 + 2)y = 0
Free Mechanical Vibrations/Couple Mass-Spring System
Example 1, p. 308, cont’d
Let’s clean this up a little bit:
n x ′′ = −3x + y
y ′′ = 2x − 2y
This is a linear system of differential equations of second order
(which is homogeneous; why?) We can put it in the form:
n (D 2 + 3)x − y
= 0
Therefore
−2x + (D 2 + 2)y
= 0
(D 2 + 2)(D 2 + 3)x − (D 2 + 2)y = 0
We note that: (D 2 + 2)y = 2x.
Free Mechanical Vibrations/Couple Mass-Spring System
Example 1, p. 308, cont’d
Let’s clean this up a little bit:
n x ′′ = −3x + y
y ′′ = 2x − 2y
This is a linear system of differential equations of second order
(which is homogeneous; why?) We can put it in the form:
n (D 2 + 3)x − y
= 0
Therefore
−2x + (D 2 + 2)y
= 0
(D 2 + 2)(D 2 + 3)x − (D 2 + 2)y = 0
We note that: (D 2 + 2)y = 2x.
So:
(D 2 + 2)(D 2 + 3)x − 2x = 0.
Free Mechanical Vibrations/Couple Mass-Spring System
Example 1, p. 308, cont’d
Let’s clean this up a little bit:
n x ′′ = −3x + y
y ′′ = 2x − 2y
This is a linear system of differential equations of second order
(which is homogeneous; why?) We can put it in the form:
n (D 2 + 3)x − y
= 0
Therefore
−2x + (D 2 + 2)y
= 0
(D 2 + 2)(D 2 + 3)x − (D 2 + 2)y = 0
We note that: (D 2 + 2)y = 2x.
So:
(D 2 + 2)(D 2 + 3)x − 2x = 0.
We clean this up a little bit:
(D 4 + 5D 2 + 4)x = 0.
Free Mechanical Vibrations/Couple Mass-Spring System
Example 1, p. 308, cont’d
Let’s clean this up a little bit:
n x ′′ = −3x + y
y ′′ = 2x − 2y
This is a linear system of differential equations of second order
(which is homogeneous; why?) We can put it in the form:
n (D 2 + 3)x − y
= 0
Therefore
−2x + (D 2 + 2)y
= 0
(D 2 + 2)(D 2 + 3)x − (D 2 + 2)y = 0
We note that: (D 2 + 2)y = 2x.
So:
(D 2 + 2)(D 2 + 3)x − 2x = 0.
We clean this up a little bit:
(D 4 + 5D 2 + 4)x = 0.
Free Mechanical Vibrations/Couple Mass-Spring System
Example 1, p. 308, cont’d
Then factor:
(D 2 + 1)(D 2 + 4)x = 0
Free Mechanical Vibrations/Couple Mass-Spring System
Example 1, p. 308, cont’d
Then factor:
(D 2 + 1)(D 2 + 4)x = 0
Therefore:
x(t) = C1 cos t + C2 sin t + C3 cos 2t + C4 sin 2t
and
y = (D 2 + 3)x = 2C1 cos t + 2C2 sin t − C3 cos 2t − C4 sin 2t
Free Mechanical Vibrations/Couple Mass-Spring System
Example 2
Use the method of the previous example (the annihilator method)
to solve the first order (homogeneous) linear system:
n x ′ = 3x + 4y
y ′ = 4x + 3y
Free Mechanical Vibrations/Couple Mass-Spring System
Example 2
Use the method of the previous example (the annihilator method)
to solve the first order (homogeneous) linear system:
n x ′ = 3x + 4y
y ′ = 4x + 3y
Answer:
x(t) = C1 e 7t + C2 e −t
and
y (t) = C1 e −t − C2 e −t
Free Mechanical Vibrations/Couple Mass-Spring System