sscp 1143 mechanics (12)periodic motion2012
TRANSCRIPT
Simple Harmonic Motion
(15-1)
In fig.a we show snapshots of a s
Simple Harmon
imple
oscillat
ic
ing
Motio
syst
n (
em
SHM)
.
The motion is periodic i.e. it repeats in time. The time needed to complete
( ) ( ) cosmx t x tω φ= +
one repetition is known as the The number of
repetitions per u
(symbol , units: s ).
nit time is called the f
perio
c
d
rquen
T
( )
1
( ) cos
y (symbol
The displacement of the particle is given by the equation:
Fig.b is a plot of ( ) versus . The quantity is called the of the
motion.
ampli
, unit hertz)
I
e
tud
m
mx
fT
t t x
f
x t x tω φ
=
= +
t gives the maximum possible displacement of the oscillating object
The quantity is called the of the oscillator. It is given by angular fre
the equatio
qu
n:
enc
y
ω2
2 fT
πω π= =
( )( ) cosmx t x tω φ= + The quantity is called the phase angle of the oscillator.
The value of is determined from the displacement (0)
and the velocity (0) at 0. In fig.a ( ) is plotted
versus for 0. ( )
x
v t x t
t x t x
φ
φ
φ
=
= =
( ) ( )
Velocity of SHM
velocity
cos
( )( ) cos sin
The quantity is called the
It expresses the maximum possible value of ( )
amplitude
m
m
m m
m
v
t
dx t dv t x t x t
dt dt
x
v t
ω
ω φ ω ω φ
ω
= = + = − +
It expresses the maximum possible value of ( )
In fig.b the velocity ( ) is plott
v t
v t ed versus for 0.
( ) sinm
t
v t x t
φ
ω ω
=
= −
( ) 2 2
2
( )( ) sin cos
The quantity is called the
Acceleration of SHM:
acceleration ampli .It expresses the maximum
possible value of a( ). In fig.c the accele
t
r
d
at
u e am
m m
m
dv t da t x t x t x
dt dt
x
t
ω ω φ ω ω ω
ω
= = − + = − = −
2
ion a( ) is plotted versus for 0.
( ) cosm
t t
a t x t
φ
ω ω
=
= −
( )
2
2 2
We saw that the acceleration of an object undergoing SHM is:
If we apply Newton's second law we get:
The Force Law for Simple Harmonic Motion
Simple harmonic motion occurs whe
a x
F ma m x m x
ω
ω ω
= −
= = − = −
The force can be wr
n the force acting on
itten as:
where is a constant. If we compare the two expressions for F we have
an object is paroportional
to the disaplacement but opposite in sign. F C
C
x= −
2
:
d anm Cω = →2
mT
Cπ=
Consider the motion of a mass attached to a spring
of srping constant than moves on a frictionless
horizontal floor as shown in the figure.
m
k
The net force on is given by Hooke's law: . If we compare this equation
with the expression we identify the constant C with the sping constant k.
We can then calculate the angular fre
F m F kx
F Cx
= −
= −
quancy and the period .
and 2 2
T
C k m mT
m m C k
ω
ω π π= = = = k
mω = 2
mT
kπ=
Example:
Example:
In the figure we show another type of oscillating system
It consists of a disc of rotational
An Angular
inertia
Simple Harmonic Oscillator; Tors
suspended from a
wire that tw
ion Pendul
ists as t
um
m ro
I
ates by an angle . The wire exerts
on the disc a
This is the angular form of Hooke's law. The constant
is called th
restoring torque
torsion constane of
the wire t
.
τ
θ
κ
κθ= −
τ κθ= −
If we compare the expression for the torque with the force equation
we realize that we identify the constant C with the torsion constant .
We can thus readily determine the angular frequen
F Cx
τ κθ
κ
= −
= −
cy and the period of the
oscillation.
We note that I is the rotational inertia of the disc about an axis that coincides wit
2 2
h
the wire. The angle is given b
C I IT
I I C
T
κω π π
κ
ω
θ
= = = =
y the equation: ( ) ( ) cosmt tθ θ ω φ= +
A simple pendulum consists of a particle of mass suspended
by a string of length from a pivot point. If the mass is
disturbed from its equilibrium position
Th
th
e Simple P
e net forc
endu
e ac
lu
n
m
ti g
m
L
on it is such that the system exectutes simple harmonic motion.
There are two forces acting on : The gravitational force and
the tension from the string. The net torque of these forces is:
g
m
r F Lτ ⊥= − = − sin Here is the angle that the thread
makes with the vertical. If 1 (say less than 5 ) then we can
mg θ θ
θ °≪makes with the vertical. If 1 (say less than 5 ) then we can
make the following approximation: where is
expressed in r
s
adians. Wit
in
h this
θ
θθ θ
°
≈
≪
( )
approximation the torque is:
If we compare the expression for Lmg
τ
τ θ τ≈ −
with the force equation we realize that we identify the constant C
with the term . We can thus readily determine the angular frequency
and the period of the oscillation. ; 2
F
C mgLT
I
Cx
Lmg
TI
ω
ω
= =
−
=
=
2I I
C mgLπ π=
In the small angle approximation we assumed that θ << 1 and used the
approximation: sinθ ≅ θ We are now going to decide what is a “small”
angle i.e. up to what angle θ is the approximation reasonably
accurate?
θθθθ (degrees) θθθθ(radians) sinθθθθ
5 0.087 0.087
10 0.174 0.174
15 0.262 0.259 (1% off)
20 0.349 0.342 (2% off)
Conclusion: If we keep θ < 10 ° we make less that 1 % error
2
2
Physical Pendul
The rotational inertia about the pivot point is equal to
Thus 2 2
A physical pendulum is an extened rigid body that is suspended
from a fixed point O and oscill
um
ates
I mL
I mLT
mgL mgLπ π= =
under the influence of gravity
The net torque sin Here is the distance between
point O and the center of mass C of the suspended body.
If we make the small angle approximation 1, we have:
mgh hτ θ
θ
= −
≪
2L
Tg
π=
If we make the small angle approximation 1, we have:θ ≪
( ) . If we compare the torques with the
force equation we realize that we identify
the constant C with the term . We can thus readily
determine period of the oscillatio
2 2
n.
mgh
F C
I IT
C
x
hmg
T
τ
π
θ
π
≈ −
=
= =
−
2
Here is the rotational
inertia about an axis through O. com
mghI
I I mh= + 2 I
Tmgh
π=
Consider an object moving on a circular path of radius
with a uniform speed . If we project th
Simple
e posit
Harmonic Motion and Unifor
ion of the
moving particle
m Circular
at point
Moti
P on the
on
mx
v
′
( )
x-axis we get point P.
The coordinate of P is: ( ) cos .
While point P executes unifrom circular motion its
projection P moves along the x-axis with simple
harmonic motion.
The speed v of point
mx t x tω φ= +
′
P is equal to . The xω′The speed v of point
( )
P is equal to . The
direction of the velocity vector is along the tangent
to the circular path. If we project the velocity
on the x-axis we get: ( ) sin
The acceleration points al
m
m
x
v
v t x t
a
ω
ω ω φ
′
= − +
�
�
( )2
ong the center O. If we project
along the x-axis we get: a( ) cos
Conclusion: Whether we look at the displacement ,the
velocity, or the acceleration , the projection of uniform
circular m
ma t x tω ω φ= − +�
otion on the x-axis diameter is SHM
The mechanical energy of a SHM
is the sum of its poten
Energy in S
tial and ki
imple Harmonic Motion
netic energies an
d
.
E
U K
( )2 2 21 1Potential energy cos
2 2
1 1 1
mU kx kx t
k
ω φ= = +
( ) ( )
( ) ( )
2 2 2 2 2 2
2 2 2 2
1 1 1Kinetic energy sin sin
2 2 2
1 1Mechanical energy cos sin
2 2
In the figure we plot the potential energy
m m
m m
kK mv m x t m x t
m
E U K kx t t kx
U
ω ω φ ω φ
ω φ ω φ
= = + = +
= + = + + + =
(green line), the kinetic energy
(red line) and the mechanical energy (black line) versus time . While and
vary with time, the energy is a constant. The energy of the oscillating object
t
K
E t U
K E
ransfers back and forth between potential and kinetic energy, while the sum of
the two remains constant
When the amplitude of an oscillating object is reduced due
to the presence of an external force the motion
Damped Simple Harmonic Motion
damp
is said to be
An example is given in the figure. A med.
ass m
attached to a spring of spring constant k oscillates vertically.
The oscillating mass is attached to a vane submerged in a
liquid. The liquid exerts a whose
magnitu
damping force
de is given by t
dF�
he equation: dF bv= −
The negative sign indicates that opposes the motion of the oscillating mass.
The parmeter is called the . The net force on m is:
damping
From Newton's second
consta
la
nt
w we
d
net
F
b
F kx bv= − −
�
2
2
2
2
have:
We substitute with and a with and get
the following differential equation: 0 d x dx
m b kxd
dx d xkx bv ma v
dt d
t t
t
d
−
+ +
=
=
−
2
2
Newton's second law for the damped harmonic oscillator:
0 The solution has the form:d x dx
m b kxdt dt
+ + =
In the picture above we plot ( ) versus . We can regard the above solutionx t t
( )/ 2 ( ) co sbt m
mx t x e tω φ− ′= +
/ 2
In the picture above we plot ( ) versus . We can regard the above solution
as a cosine function with a time-dependent amplitude . The angular
frequency of the damped harmonic oscillator
bt m
m
x t t
x e
ω
−
′
2
2
2
is given by the equation:
1 For an undamped harmonic oscillator the energy
2
If the oscillator is damped its energy is not constant but decreases with time.
If the
k=
m
damping is s
4mE kx
b
mω′ − =
/ 2
2 /
mall we can replace with By doing so we find that:
1( ) The mechanical energy decreases exponentially with time
2
bt m
m m
bt m
m
x x e
E t kx e
−
−≈
( )/ 2 ( ) co sbt m
mx t x e tω φ− ′= +
Example:
Moving
support
If an oscillating system is disturbed and then allowed
to oscillate freely the corresponding angular frequency
is calle
Forced Oscillations and Resonance
nad the The satural f me systrequ em ce ancy l. n a
ω
so
be driven as shown in the figure by a moving support that
oscillates at an arbitrary angular frequency . Such a
forced oscillator oscillates at the angular frequency
of the driving force. The
d
d
ω
ω
displacement is given by: of the driving force. The
( )
displacement is given by:
The oscillation amplitude varies
with the driving frequancy as shown in the lower figure.
The amplitude is approximatetly greatest when
( ) co
This condition
s m
d
mx t x xt
ω
φ
ω
ω
=
′= +
is called All mechanical structures
have one or more natural frequencies and if a structure is
subjected to a strong external driving force whose frequency
matches one of the natura
resona
l fre
nce.
quencies, the resulting oscillations
may damage the structure