VIBRATION This is a branch of science of engineering that deals with repetitive motion of mechanical system from machine parts to large structures. Sometimes these vibrations are desirable as in the case of a string vibrating on an instrument. Sometimes vibrations are unwanted as in the case of vibrating ground motion produced during an earthquake. The physical explanation of the phenomena of vibration concerns the interplay between potential (PE) and kinetic Energy (KE). A vibrating element system must have a component that stores potential energy and releases it as kinetic energy in the form of motion (Vibrating) of a mass.
Note that within the elastic limit of the spring the mass held at one end causes a proportional elongation on the spring Thus: f s α e f s = Ks e Where: Ks = Spring Constant (N/m) fs = Force from the mass (N) e = Elongation (m)
General Assumption on the loading of the spring 1. When at rest, the system held in equilibrium because the force of the spring is equal to the weight or load. 2. When in motion, the unbalanced force on the spring systems causes that motion. (a)
(b)
(c)
xs -x +x
Situation (b) illustrates the deflection (x s) of a spring having a load (m)
Situation (c) illustrates an up and down direction resulting to a simple harmonic motion.
From: FBD of (c) f s Where: w = mg …………………...…………. …………………...………….eq. 1
f s’
f s ………………………………eq.2 s = kx s s ………………………………eq.2
w
f s ’ = kx s ± kx = k (x s xs ’ ’ …… ……eq. 3 s ’ s ± kx s ± x) = k x s f s ’ = ma (unbalanced force) …………eq.4 s ’ =
Again from the illustration -f s s ’ = -w + f s s = -w + [kx s s + kx]
where: w = kx s
ma = -kx
Thus x = -ma/k where: x – is the instantaneous deflection (elongation positive; compression negative) which is in the reverse direction of the unbalanced force. a – is the acceleration due to the unbalanced force.
Situation of spring-mass system in SHM 1. At rest (ƩF = 0) fs f s = w = mg w 2. Moving in downward direction
w f s f s ' kx s ( kx s kx ) ma x
ma k
3. Moving in an upward direction fs w fs' ( kx s
kx ) kx s ma
x
ma k
Where: X s’ = Xs + X (total instantaneous deformation)
Review of Sine & Cosine Function Sine Function
A)
0
sin (ϴ + ϕ)
π/2
π
3π/2
2π
3π/2
2π
where: ϕ = 0
B)
0
sin (ϴ + ϕ)
π/2
π
where: ϕ = π/2
C)
-π/2
0
sin (ϴ + ϕ)
π/2
π
where: ϕ = -π/2
3π/2
5π/2
Cosine Function
A)
0
cos (ϴ + ϕ)
π/2
π
3π/2
2π
π
3π/2
2π
π
3π/2
where: ϕ = 0
B)
0
cos (ϴ + ϕ)
π/2
where: ϕ = π/2
C)
-π/2
cos (ϴ + ϕ)
0
π/2
where: ϕ = -π/2
To determine (Ø) phase angle from: x A cos t let t 0 when the motion is about to start and see if x A if so : 0
5π/2
EXAMPLE: A block weighing 96.5 lbf is dropped from a height of 4ft upon a spring whose modulus is 100 lb / in. What velocity will the block have at the instant if the spring is deformed by 4 in. Given: W = 96.5 lbf
4 ft
k = 100 lb / in
Situation 1
PE mgh KE Wh
1
mv 2
2 1
mgh
2
mv 2
v 2gh
2
2( 32.2 ft / sec )( 4 ft )
.. . 16.05 ft / sec
Situation 2
PEs
1 2
kx 2
1
m( v f v o ) mgh where : h x 4 in 2
2 1 2
2
m Vf 2 Vo 2 mgx
from…… W mg
kx 2 2mgx mVf 2 Vo 2 ( kx 2 2Wx )g V o 2 V f W 2
Vf
16.05
ft 2 s
2 ft in ( 100 lbin 12 ft 4 / 12 ft ) 2 96.5lb 4 / 12 ft ) 32.2 s 96.5lb
Vf 15.32 ft / sec
2
EXAMPLE NO.2 At the instant shown, an external load has pushed the 500lb f block against the spring thereby compressing the spring 6in. if the spring constant is 100 lb /in. How far will the block be f projected along the level plane ( =0.2) when the external force is released. What will be its maximum velocity? Figure: W= 500lbf 6 in k = 100 lb f / in F = 0.2
Solution:
fs kx fs (100lbf / in) x6in fs 600lbf
N 0.2(500lbf ) 100lbf
f f f f f f
Force Diagram:
PE s f f ( x) 1 2
kx2
100( s)
100(6)
s
s
18in
2
2(100)
k X
-6 in
-1
0
12 in
Maximum Velocity V ma x happens at a=0
PE KE 1 2
kx 2
1 2
m(Vf V 0 ) 2
2
2
Vf
kx g
Vf
100lb / in(5in) (32.2 f t / sec)(1 f t / 12in)
w 2
Vf 3.6629 f t / sec
500lbf
SPRING CONNECTIONS 1. Springs attached in parallel -we usually determine first the constant ‘k T ’ of a single spring equivalent to the two springs by finding the magnitude of the force required to cause a given deflection ( X T ). Since for a deflection (X T ), the magnitude of the forces exerted by the spring are k 1 X T and k 2 X T respectively, then: Force ( F ) = k 1 X T + k 2 X T = (k 1 k 2 ) X T
k 1 k 2
thus:
k T
where:
k T = spring constant for the springs connected in parallel
2. Spring attached in series -we first determine the constant ‘ k T ’ of a single spring equivalent to the two spring by finding the total elongation, ( X T ) of the spring under a given static load P. Given: X T X 1 X 2 P kX X T
Comparison
Equivalent spring constant Compressed distance Energy stored
P k 1
P k 2
In Parallel
In Series
SIMPLE HARMONIC MOTION A. Equilibrium Position Ʃ Fh = 0 ma = 0 since the block is at rest
-xs’ B. Ʃ Fh = fs’ fs = fs’ k(-xs’ ) = ma k a x m
f s’
(+)
Following a sign convention, situation (B) presents that the initial displacement of the block (x s’) gets a negative sign because it goes to the direction of the negative x-axis. The force gets a positive sign because it goes to the positive side of the axis. Situation (C) presents the reverse of its precedent situation.
C.)
xs’ Ʃ Fh = fs’ fs = - fs’ kx = -ma k a x m
(-)
0.1 inch
4
f s’
From calculus Velocity (v)
=
dx dt 2
Acceleration (a)
d x 2
dt
thus: d 2 x 2
dt
k m
x ; let : 2
k m
then: d 2 x 2
dt
2 x
where: ω - angular velocity in radians/s
transposing the right hand side gives us a Differential Equation d 2 x 2
dt
2 x 0
using auxiliary equation: m 2 2 0
and the roots are: m = ± i ω
thus the general solution is: x ( t ) c1 cos t c2 sin t
applying possible initial conditions when t = 0 1 0 x ( 0 ) c1 cos ( 0 ) c 2 sin ( 0 ) x ( 0 ) c 1
also
0 1 v ( 0 ) c1 sin ( 0 ) c 2 cos ( 0 ) v ( 0 )
c2
the graphical representation of the motion is: 0.6
C1 cos ωt C2 sin ωt
0.4
x(t)
ϕ C2
0.2
x(0)
0
x(t) C1
0 A
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
-0.2 -0.4 -0.6
the simplified form of SHM: x A sin
where
x A sin ( t )
t
k
and
m
then: Velocity ( v )
dx dt
A cos t
Acceleration ( a )
d 2 x 2
dt
2 A sin t
where: A – Amplitude ( m or ft ) ω – Angular velocity ( rads/s ) also: Period (T)
= 2π/ϖ
Frequency (f)
1
T
2
c1 c 2
2 2 1 ; A c1 c 2 ; tan
or 2 f
To have a maximum velocity: cos ωt + = 1 thus:
t 0 0 and:
for 1
k m A
V ma x
To have a max acceleration: sin t 1 thus:
t and:
180 or / 2 for 1
Amax 2 A
0
k m
A
EXAMPLE An object oscillates with SHM along the x-axis. Its position varies with time according to the eq. x = 4m cos (πt + π/4). Where t is in second and the angles in the parenthesis are in radian a.) b.) c.) d.) e.)
Determine the amplitude, frequency and period. Calc. the vel. And acc. Of the object at any time (t) Det. the position, velocity and acceleration of the object at t=1s Det. the max speed and acceleration of the object. Find the displacement of the object between t=0 and t=1s
Solution: a. From the general equation: x = A cos (ωt + ϕ) A = 4m
ϕ = π/4
ω=π
Frequency (f) = ω/2π = π/2π = ½ Period (T) b.
= 1/f
= 2 seconds
v A sin t 4m sin t
4
a 2 A cos t 2 4m cos t
4
c.
x A cos t 4m cos 1
2.83 m
4
v A sin t 4 m sin 1
m
8.89 4 s m a 2 A cos t 2 4 m cos 1 27.9 2 4 s d.
vma x A 4m 12.57
m s
2 2 ama x A 4m 39.48
e.
m s
2
2.83 m 4 x f 4 m cos 1 2.83 m 4
xi 4m cos 0
Then:
x x f xi 2.83 m 2.83m 5.66m
A
π/4
5π/4
EXAMPLE: A spring is such that it would be stretched 6 inches by a 12 lb weight. Let the weight be attached to the spring and pulled down 4 inches below the equilibrium point. If the weight is started with an upward velocity of 2 ft/sec, describe the motion and compute the period. Given:
6 in Equilibrium 12 lb
4 in
-2 ft/sec
Solution: solving for the mass:
solving for the natural frequency
slug. ft 12lbf 1 2 lbf . sec m 32.2
ft
sec 2
m 0.37 slug
solving for the spring constant: k
F x
k 24
12lbf 1 ft 6 in 12in
lbf ft
the general solution as evaluated above is: x ( t ) c1 cos t c2 sin t
o
o
k m 24
0.37 rad o 8.054 sec
using the initial values, it shows the following: c1
4in 0.33 ft in 12 ft
and
c2
2 ft / s 8.054 rad / s
0.25 ft
thus the particular solution is: x ( t ) 0.33 ft cos 8.054rad / s t 0.25 ft sin 8.054 rad / s t
0.6 0.4 0.2
C1 cos ωt
0 -0.2
C2 sin 0 5 1 5 2 5 3 5 4 5 5 5 6 5 7 5 8 5 9 . . . . . . . . . 0 1 2 3 4 5 6 7 8 . 0 . 0 . 0 . 0 . 0 . 0 . 0 . 0 . 0 0 0 0 0 0 0 0 0 0
-0.4 -0.6
the period of the motion is: T
2
n
2 8.054 rad / s
1
sec 4
x(t)
t
ω
EXAMPLE: A 50kg block moves between vertical guides as shown. The block is pulled 40mm down from its equilibrium position and released. For each spring arrangement, determine the period of vibration, the maximum velocity and maximum acceleration of the block. a. Parallel Connection: Figure:
k 1
kN
4
k 2
m
6
kN
X T Solution:
k T X T k 1 X 1 k 2 X 2 P ( k 1 k 2 ) X T P
P X T
k 1 k 2 k T kN
k T
4
k T
10
6
m kN
kN m
m
Solving for ( w ): k T
m 10
kg.m (1 )(1000 N / KN ) m N . sec 2 50kg
kN
14.14 rad / sec
Thus: a.) T
2
b.) V max
wA 14.14(0.04m) 0.57
c.) ama x
w
2 14.14
0.44 sec m
sec m w2 A (14.14) 2 (0.04m) 8 2 sec
m
X 1 X 2
b. Series Connection Figure:
k 1
4
kN
k 2
6
kN
m
m
W
50 kg Solution: P
P
X T
X T
1 1 P k 1 k 2
k 1
k 2
4 cm
X T
P
1 k 1
P
1
k 2
0.04 1 1 4
Since:
k T
or
k T
0.096KN 96N
6
P
X T
96 N 0.04
1 1
k 1
1
k 2
2,400
N m
1
1 4000
2,400
1 6000
Solving for ( ω ):
k T m
2400
50
6.93
rad sec
thus: a.) T
2
2 6.93
0.91sec
b.) V ma x A 6.93(0.04) 0.28
m sec
2 2 c.) ama x A 6.93 (0.04) 1.92
m sec
2
N m
ENERGY OF THE SIMPLE HARMONIC MOTION Assumption: the total mechanical energy of a spring-mass system is constant. Kinetic Energy (KE) 1 2
KE
2
mV
Where: V = ϖ A sin (ϖt + ϕ)
Then:
2
2
2
KE = ½ m ϖ A sin (ϖt + ϕ)
Elastic Potential Energy (U)
Wk U f s x
where :
f s
k x
2
U k x dx
dU k x dx 1
2
1
U
1
k x
2
2 Total Mechanical Energy (E)
where : x A cos t
E KE U E
1 2
2
2
m A sin
2
t 1 k A2 cos2 t 2
Since:
2
k
m 1 2 2 2 E k A sin t cos t 2 1 1 1 2 2 2 2 E k A m V m ax m A 2 2 2
Also: V
2
kx2
kA2
m
k
V
m
V
A
A
2
2
x 2
x 2
and V A sin t
EXAMPLE A block of unknown mass is attached to a spring with k=6.5 N / m. It undergoes SHM with an amplitude of 10cm. When the block is halfway between its equilibrium position and the endpoint, its speed is measured to be 30 cm/sec. Calculate a.) mass of block b.) the period c.) the maximum acceleration Solution; Given: k = 6.5 N / m v = 30 cm / sec A = 10 cm a.
2
A 30cm / sec
m
10cm 5cm 2
2
b. Period ( T )
T
2
2
3 T 2.094 sec
c. 2 a m ax A
a m ax 32 (10cm 2 )( a m ax 0.9m / sec
2
1m 100cm
m k
2
m 0.722 kg
3rad / sec
T
k
)
2
EXAMPLE: A 0.5 kg cart connected to a light spring for which the force constant is 20 N/m oscillates on a horizontal frictionless air track. A.) Calc. the total energy and max. speed of the cart if the amplitude is 3cm. B.) What is the velocity of the cart when the position is 2cm. C.) Compute the kinetic and potential energies of the system when x = 2cm.
Solution: a) E
1
E
1
2 2
E
2
kA
( 20
N m
)( 0.03m )
1 2
ma x
2
2
m ma x 2(0.009 J ) 0.5kg
0.1897
E 0.009 N m or J
b)
k m
( A
2
x 2 ) 20
N
2 m ( 0.03m kg m 0.5kg (1 ) 2 N s
0.02m 2 )
0.1414m / sec
KE
1 2 1
U
m 2 (0.5kg)(0.1414
2 0.005 J
m s
)
2
1 2 1
kx 2 ( 20
N
2 m 0.004 J
2
)(0.02 m )
m s
THE PENDULUM
ϴ
W While ϴ < 100, a pendulum exhibits a SHM
Free Body Diagram on the ‘’Pendulum Bob’’ Tension (T)
mg cos ϴ
mg sin ϴ
From the FBD is the tangential compliment of the weight (mg) therefore it is the restoring force. Applying Newton’s 2nd law. ma mg sin
Where: ma = unbalanced force m = mass of bob g = gravitational acceleration In another form:
d 2 s M 2 mg sin dt Where: S L
Thus:
d 2 m L 2 mg sin dt d 2 2
dt
g L
sin
Since: sine of very small angle is equal to the angle itself. Example:
Sin (0.0001) = 0.0001
Then: d 2
g
2
dt
L
Comparing this with: 2 d 2 x 2 dt
We see that:
2
g L
Thus the angular frequency (w)
g L
The period of rotation is T
2
2
L g
Note: the period and frequency depends on the length of string and acceleration due to gravity.
Example 1
It is suggested that an international unit of length could be defined as the length of a single pendulum having a period of exactly 1 second. How much shorter would our length unit be had this suggestion been followed?
Solution:
L
From T = 2
G 2
T2
4 L
g 2
L =
1 (9.81) 4
2
L = 0.248 m
EXAMPLE 2 The angular position of a pendulum is represented by the equation = 0.32 rad cos (wt) , where w=4.43 rad / sec. Determine the period and length of the pendulum?
Solution: ω=
g
also:
L
T =
g
2 rad 4.43rad / sec
2
L
T = 1.42 sec L =
9.81m / sec
4.43rad / sec2
L = 0.4994 m
PHYSICAL PENDULUM If a hanging object oscillates about a fixed axis that does not pass through its center of mass and the object cannot be approximated as a point mass, we cannot treat the system as a simple pendulum. In this case the system is called a physical pendulum.
Illustration:
O
d ѳ
ϲʛ
mg The torque about the axis O: mg d sin
Where: I – moment of inertia by mass angular acc d – Distance of pivot to center of mass ( Radius of Gyration) From Newton’s second laws:
Comparing this to: 2
d
I I mgd 0 Thus:
2
dt
Consequently: 2
mg d sin I
d
sin for small valueof
Then
2
dt 2
Again since:
2
dt 2
mgd I
mgd I
the period is T
2
T 2 d
2
I mgd
EXAMPLE 1 A uniform rod of mass m and length L is pivoted about one end and oscillates in vertical plane. Find the period of oscillation if the amplitude of the motion is small. SOLUTION: 2
I=
mL
where: d =
3
Thus :
L
2
I
T = 2
mgd 2
1 / 3(mL )
T = 2
mg ( L / 2) 2 L
T = 2
3g
T=1.64secs EXAMPLE 2: A physical pendulum in the form of planar body moves in SHM with a frequency of 0.45 Hz. If the pendulum has a mass of 2.2 kg and the pivot is located 0.35 m from the center of mass, determine the amount of inertia of the pendulum. SOLUTION: I =
mgd
2
Where: w = 2 f = 2 (.45) = 2.83 rad / sec Thus: I =
2.2 kg( 9.8m / sec)(0.35m ) ( 2.83rad / sec) 2
I = 0.944 kg - m 2
CENTER OF PERCUSION ( C ) The distance along the centerline such that a simple pendulum (a mass less rod pivoted at zero with mass (m) at its tip) of radius q o has the same period with a physical pendulum.
Illustration: From the moment of Inertia I md 2
r Where: d – radius of gyration From the figure: qo
mg
qo
C
I mr
Thus: Where: r=distance from pivot to C.G. d=radius of gyration q o =center of percusion So that: I
d 2
dt 2 and
n
mgr 0
mgr I
In terms of center of percussion:
g qo
I I
qo mr qo r d 2
md 2
COMMON MOMENT OF INERTIA A. About the center radial 1. Solid Cylinder or Disc
I
1 2
2
mr
r
2. Hoop about symmetrical Axis
I mR 2
R
3. Solid sphere
I
2 5
mR
4. Rod about the center
I
L
1 12
2
mL
2
B. About the specified axis 1. Solid cylinder central diameter
I
r
1 4
mR
2
L
2. Hoop about diameter
I
1 2
mR
2
3. Thin spherical shell
I
2 3
mR
2
4. Rod about End
1 2 I mL 3
L
1
R
2
mL
Example: A uniformly shaped physical pendulum has a length of 50cm. Calculate a) the natural frequency b) center of percussion and c) radius of gyration. SOLUTION: Figure:
r L
SOLUTION: From the general equation: 2
I
d 2
dt
( mgr ) 2
Thus:
d dt 2
w 2 (
2
2
I rod
where:
r
mgr ) I
mgr 2
1 / 3( mL )
Then:
2
where:
mg ( L / 2) 2
(1 / 3)( mL ) 3g 2 L 3(9.8m / sec) 2(0.5m )
5.425rad / sec
L
2
1 3
2
mL
a.) qo
I mr
2 qo (1 / 3)mL
1
m( L / 2)
2 qo L 3
qo 33.33cm
b.) d
q o r
d (33.33cm)(25cm) d 28.87cm
TORSIONAL PENDULUM
When the object is twisted through same angle , the twisted wire exert on the object restoring torque that is proportional to the angular position
k
Where: k - torsional constant ϴ - Twisting angle Thus: k I
2 d
dt 2
Rearranging the equation: d 2 dt 2
k
I
Then: 2
k I k I
And T 2
I k
EXAMPLE: A torsional pendulum is formed by taking a meter stick of mass 2 kg and attached to its center a wire with its upper end clamped, the vertical wire supports the stick as it turn in a horizontal plane. If the resulting period is 3 mins, what is the torsional constant for the wire?
Solution:
Figure
1m From : T 2
I k
where: k 4
k k
2
2
T
4 2 mL2 12T 2 2 2 4 ( 2kg )( 0.5m )
sec 123 min s 60 min
k 5.08 x 10 5 N .m
I
2
and
I m
L2 12
STIFFNESS The stiffness of a vibrating body can be related more directly to material and geometric properties. The stiffness actually gives the value of ( k )
Value of Stiffness in Different Forms: 1.)
where: E – elastic modulus of the rod (Esteel = 2.0 x 10 11 N/m) k
EA
A – cross-sectional area of the rod
ℓ - length of the rod ℓ
where: Jp – Area moment of inertia of the rod (Jrod = πd4 /32 m4) G – shear modulus of rigidity of the rod (Gcu=2.22 x 10 10 N/m2) ℓ - length of the rod
2.)
k
G JP
ѳ
3.)
d Where: G – shear modulus of the spring (Gsteel=8x1010N/m2)
4
k 2R
G d
64 n R 3
d – diameter of the spring wire n – number of turns R – radius of the coil
4.) k
3EI
3
where: E – modulus of elasticity I – area moment of inertia of the beam ℓ - length of the rod
Example A shaft is made of steel and is 2 m long with a diameter of 0.5 cm. If the disk has a polar moment of inertia J = 0.5 kg-m 2 and considering that the shear modulus of steel is G = 8 x 10 10 N/m2, calculate the frequency of the motion. Solution: 2 2
k j
G J p J
10 2 4 2 ( 8 x 10 N / m )( )( 0.5 x 10 m )
( 32 )( 2 )( 0.5kg m ) 2
2.2156 rad / s
Example A 100 lbs weight and a coil spring with modulus k = 100 lb/in are attached to the end of a wooden cantilever beam where E = 2 000 ksi and I = 0.1 in 4. Determine the length of the beam so that the natural frequency of the system will be 2 cycles per second. Neglect the mass of the spring and beam.
Solution: The weight is supported by two springs in series, namely, the beam and the coil spring. k spring 100 ft / in k beam
3EI L3
3( 2 x 106 psi )( 0.10 in4 ) L3
6 x 105 lb in2 L3
Resultant Spring Constant is: k total
( k spring )( k beam ) k spring k beam
7 6 x 10 3
L
7 6 x 10 100L3 6 x 10 5 100L3 6 x 10 5
L3
From the equation of frequency: 1 g k total f 2 W f 2
2 2
g k 2
4 W
1
2
7 g 6 x 10
4
2
1 3 5 100 100L 6 x 10
58.8 x 107 100
1 100L3 6 x 105
400L3 24 x 10 5 58.8 x 10 5
L 20.6 inches
UNDAMPED OSCILLATION Considering the differential equation: 2
m
d x 2
dt
d 2 x 2
dt
d 2 x 2
dt
kx F ( t ) k m
x F ( t )
where
2
k m
2 x F ( t )
the general solution is: x ( t ) x c x p
from SHM:
x c c1 cos t c2 sin t and y p is the particular solution of the function F(t) thus:
x ( t ) c1 cos t c2 sin t x p
Let us now apply the process by finding the solution of an undamped vibration is represented by the equation: m
d 2 x 2
dt
kx L sin t
Solution: 2
d x 2
dt
d 2 x 2
dt
k m
x
L m
sin t
2 x H sin t
2 where :
k m
and
using the auxiliary equation to solve the left hand side of the equation: m 2 + ω 2 = 0 m = ± iω
thus:
x c c1 cos t c2 sin t
H
L m
for the right hand side: F ( t ) H sin t
F(t) happens to be a particular solution of a homogeneous linear differential equation
whose auxiliary roots are: m = ± iϐ therefore the linear differential equation of F(t) is: d 2 x 2
dt
and:
2 x 0
x p C 3 cos t C 4 sin t
taking the first derivative: Dx p C 3 sin t C 4 cos t
the second derivative:
D 2 x p 2 C 3 cos t 2 C 4 sin t substituting x p to the original equation and multiplying by 1/sin ϐ t : { [ 2 C 3 cos t 2 C 4 sin t ] 2 [ C 3 cos t C 4 sin t ] H sin t }
by comparison, the cosine function = 0 and sine function =H
2 C 3 cos t 2 C 3 cos t 0 2 C 4 2 C 4 H thus: C 3 0 C 4
H
2 2
therefore:
H x p 2 sin t 2 finally:
sin t
x ( t ) c1 cos t c 2 sin t
H
2
2
1 sin t
with derivative
H cos t 2 2
x ' ( t ) c1 sin t c2 cos t
applying the initial conditions, the values of the constants are: C 1 x ( 0 )
C 2
;
H 2 2
v ( 0 )
EXAMPLE A spring is such that it is stretched 6 inches by a 12 lb weight. The 12 lb weight is pulled down 3 inches below the equilibrium point and released. If there is an impressed force of magnitude 9 sin (4t) lbs, describe the motion. Assume that the impressed force acts downward for very small time. Given:
F 6 in Equilibrium 12 lb
3 in
Solution solving for the mass of the block: W m
go gc
;
slug . ft 12lbf 1 lbf . sec 2 m 32.2 m 0.37 slug
ft
sec 2
m W
gc go
other parameters: L 9 lb f x ( 0 )
3in 12in / ft
v ( 0 ) 0
4rad / s
0.25 ft
solving for the spring constant: solving the natural frequency: k
F x
12lbf
1 ft 12in
2
lbf
k 24
24
rad 2
64.86 2 0.37 s 8.054 rad / s
6in
ft
solving for the constants: C 1 x ( 0 ) 0.25 ft
C 2
also
H
L m
9lb 0.375slug
24 ft / s 2
0 244 H 0.25 ft 2 2 8 88 2 4 4
v ( 0 )
C 3 0 C 4
H
2
2
24 8 42 2
0.5 ft
the particular solution therefore is: x ( t ) 0.25 cos 8 t 0.25 sin 8 t 0.5 sin 4 t
the graphical representation of the motion is: 0.8 0.6 0.4 0.2
c1 cos
0
C2 sin
-0.2 -0.4 -0.6 -0.8 -1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1.1
1.2
1.3
1.4
C4 sin x(t)
DAMPED OSCILLATION In many real systems, non-conservative forces, such as friction, retard the motion consequently, the mechanical energy of the system diminishes ion time, and the motion is said to be damped. Because the retarding forces can be expressed as R = -bv (where b is constant called the damping coefficient. ) from: m
d 2 x dt
d 2 x dt 2
let:
b
2
dx dt
b dx m dt
2
kx 0
k m
Equation 1 x 0
b m
n
;
2
k m
using the auxiliary equation D
2 D n 2 0
2
from quadratic equation: D
2
( 2 )
2
( 2 n ) 2
2
let: 2
d
n 2 2
then: D
2 2 2 ( 2 n2 ) 2
D 2 n2 D d 2 D1 i d D2 i d
using auxiliary eq. (imaginary roots)
x ( t ) C 1e
t
sin d t C 2 e
t
cos d t
-
General Solution
from initial value problem, where at t = 0 ; x ( 0 ) = A x ( 0 ) C 1e
( 0 )
sin d ( 0 ) C 2 e
( 0 )
cos d ( 0 )
C 2 A
thus: x ( t ) Ae cos d t t
- Particular Solution of Equation 1
EXAMPLE: A pendulum with a length of 1m is released from an initial angle of 15 . After 1sec, the amplitude has been reduced by friction to 5.5 . what is the value of b / 2m ? What is the damped frequency? Solution: (t ) Ae t cos( t )
Where at t = 0 (0) Ae ( 0 ) cos( (0)) 0 0 15 Ae cos0
A 150
At t = 1 sec (1) 150 e (1) cos( (1)) 5.5 15 e 0
e
0
cos( (1))
5. 5
15 5. 5 ln 15
1.0033
b 2 L
For the damped frequency L
2 d 2
dt
b
d dt
g 0
Using the auxiliary equation: LD bD g 0 2
From quadratic equation: D
b b 2 4 Lg 2 L
Where: d
b2 4 Lg
:
2 L
d 2
b 2 4 Lg ( 2 L ) 2
Thus: d
(
b 2 L
)
2
g L
Reversing the position of the terms inside the radical sign: d d
g L
(
9.81 1
d 2.965
b 2 L
)
2
(1.0033) 2
rad sec
DAMPING RATIO
From: m
2
dx
b
2
dt
mD
dx dt
k 0
Where : b = damping coefficient
bD k 0
2
using the quadratic formula: D
2
b m
k
D
m
0
The roots are: D
b 2m
1 2m
b
2
4mk
-
Equation 1
The determinant ( b 2 4mk ) will indicate the general solution of the differential equation Thus: Critical damping coefficient ( bcr 4mk 2 mk ) and so the damping ratio ( )
b bcr
Where:
b
2 mk
m = mass k = spring constant
Since: W o
k m
-
Multiplying both side by ( m )
m o m 2
m o
2
m
m2
m o
2
k k
m
mk
Then:
b 2 m o
Or b 2m
o
-
Equation 2
Also from the damped angular frequency:
b d m 2m k
2
b d 0 2m
2
2
d o ( 0 ) 2 2
d o (1 o ) 2 2
d o 1 2
-
Equation 3 (True Equation)
Note that equation 3 is the true while is less than 1. In case where is greater than 1, then:
d d
2
b
2
4mk
2m b ( 2m )
2
4mk ( 2m )
2
And since: b 2m k
m
o
o 2
d
o 2 o
d o 1
-
Equation 4
Substituting Equation 2 and Equation 4 to Equation 1
D o o 1
-
Equation 5
Using this equation to evaluate the damped vibration, three cases will arise. Case 1.
0 < < 1 For this case, the quantity inside the radical sign will give an imaginary root, thus using the auxiliary equation for imaginary roots. D
o d i
-
where: ( i =
1 )
The general solution is: x(t ) e
Case 2.
ot
(C 1 sin d t C 2 cos d t ) t
1
- over damped motion
For this case, the determination of equation 5 is positive, thus, it yields to: D 0 o 2 1
And the general solution is: x(t ) e ot (a1e
o 2 1
a2 e
o
2 1
)
Where: a1 a2
Case 3.
o ( 2 1) o xo 2 o 1 2
o ( 2 1) o xo 2 o 1 2
= 1
critically damped For this case, equation 3 is reduced to D
o
And the general solution is: x(t ) e
Where:
ot
(a1 a2 t )
a1 xo a2 o o xo
EXAMPLE A spring is such that it would stretch 6in by a 12lbf weight. Let the weight be attached to the spring and pulled down 4in below the equilibrium point. If the weight is started with an upward velocity of 2 ft / sec with a damping force of magnitude 0.6 V. Determine the displacement x(t) after 1.5 sec. Given:
6 in Equilibrium 4 in
V o
2
ft sec
Solution a. Damping ratio
b bcr
b
2 mk
Where: b = 0.6 W m
m
go
m W
;
gc
12lbf 1
gc go
slug . ft
lbf . sec 2
32.2
ft sec
2
m 0.37 slug
k
F x
k 24
12lbf
1 f t 12in
6in
lbf f t
Thus:
0.6 2 0.37(24)
0.1007
Since 0 1
; solution will come from Case 1
Where: k
o
o
o
8.054
m 24 0.37
rad sec
