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VIBRATIONSVIBRATIONS
FreeFree
damped anddamped and forcedforced
AP3204 Topic 1 (1/2)AP3204 Topic 1 (1/2)
ExamplesExamples
4.Torsional Pendulum
2. Compound pendulum
1. Simple pendulum
3. Mixed modes
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ExamplesExamples
6. LCR circuit
5. Between two atoms
8. A mass attached to a spring on a
frictionless track moves in simple
harmonic motion.
7. piston moving in a cylinder
Q Did you notice
these examples have much in common? What are they?
Simple Harmonic MotionSimple Harmonic Motion ((s.h.m.)s.h.m.)
Definition:
Simple harmonic motion of a
mechanical system corresponds
to the oscillation of an object
between two points for an indefiniteperiod of time, withno loss in
mechanical energy.
An object exhibits simple harmonic
motion if the net external force acting
on it is a linear restoring force:
F = -kx
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Simple Harmonic MotionSimple Harmonic Motion ((s.h.m.)s.h.m.)-- Sign and feature of the forceSign and feature of the force
Q Why minus sign?Q Why minus sign?
Without the minus sign the
acceleration will continually
increase as x, so the particle
will keep on moving faster
and faster away from its
original position.
With the minus sign an
oscillation will occur.
Q What kind of force is needed?Q What kind of force is needed?
x-F k=From
Simple Harmonic MotionSimple Harmonic Motion ((s.h.m.)s.h.m.)--Mathematical expressionMathematical expression
== xdt
xd
2
2
aand Newton's 2
nd
law: ma=F
: Angular frequency, the constant in the s.h.m.
equation. It defines the time taken for an oscillation
we have
0xx 2 =+
mk
=2
T
2=
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0xx2
=+
Solution of
)sin( += tAx
Q Can you give other solutions of x?
Simple Harmonic Motion (Simple Harmonic Motion (s.h.m.)s.h.m.)--Mathematical expressionMathematical expression
)t(A +cos
tCtB sincos +
ti*ti eDDe +
QQ Check that these satisfy the equation.Check that these satisfy the equation.
QQ How are B and C related to A andHow are B and C related to A and ??
]Re[ tiAe
0xx 2 =+
Simple Harmonic Motion (Simple Harmonic Motion (s.h.m.)s.h.m.)--Alternative solutions for s.h.m.Alternative solutions for s.h.m.
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)sin( += tAx
)(
)cos(v
22xA
tAx
=
+==
)sin(2 +== tAxa &&
Simple Harmonic Motion (Simple Harmonic Motion (s.h.m.)s.h.m.)-- Velocity and accelerationVelocity and acceleration
From
Simple Harmonic MotionSimple Harmonic Motion ((s.h.m.)s.h.m.)
-- TerminologyTerminology
A few terms relative to harmonic motions:
1. The amplitude A:maximum distance that an object
moves away from its equilibrium positions.
2. The period T: the time for one complete cycle of themotion.
3. The frequency : the number of cycles or vibrationsper unit of time.
4. Phase angle : the fraction of a cycle (2); theoscillation is out of phase with some reference time.
)sin( += tAx
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Simple Harmonic MotionSimple Harmonic Motion ((s.h.m.)s.h.m.)--Graphical representationAn experimental apparatus
for demonstrating simple
harmonic motion.
- A pen attached to the
oscillating mass traces out a
sine wave on the moving
chart paper.
Q Can you find if its
trace is consistent with
its mathematical
representation?
)sin( += tAyT
2
=
y0=
The value of the
phase constant
depends on the
initial displacement
and initial velocity
of the body.
Simple Harmonic MotionSimple Harmonic Motion ((s.h.m.)s.h.m.)--Graphical representation
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)cos(v +==
tAx
The velocity graph will be a cosine curve with amplitude A.
)sin(2 +==
tAxa
The acceleration graph will be a minus sine curve with amplitude
2A.
Simple Harmonic MotionSimple Harmonic Motion ((s.h.m.)s.h.m.)--What will graphs for velocity and acceleration look like?What will graphs for velocity and acceleration look like?
Yes.
Angular velocity is the rate of rotation around a fixed point.
For instance a swinging pendulum has a variable angular
velocity but a constant angular frequency.
A mass vibrating on a spring has no angular velocity.
Only in the case of a particle moving steadily around a
circle are the two quantities correlated constants.
Q Is angular frequency different fromangular velocity?
~ v
v
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QQ Why is s.h.m. important?Why is s.h.m. important?
Any oscillation can be modelled as
being made up of a number of s.h.m. components
which can be analysed separately and
then recombined using the Principle of Superposition.
etc(t)x(t)xx(t) 21 ++=
Fourier theoremFourier theorem
Fourier theorem:Fourier theorem: Any periodic function can beAny periodic function can be
expressed as a sum of the sine and cosine functionsexpressed as a sum of the sine and cosine functions
whose frequencies increase in the ratio of naturalwhose frequencies increase in the ratio of natural
numbers. i.e.numbers. i.e.
f(t+f(t+nTnT)=f(t);)=f(t); n=0,n=0,1,1,2,2,3,3,
can be expanded in the form:can be expanded in the form:
( )
=
=
++=11
0 )sin()cos(2
1
n
n
n
n tnbtnaatf
T
2=
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( )
=
=
++=11
0 )sin()cos(2
1
n
n
n
n tnbtnaatf
where
+
=
Tt
t
n tdtntfT
a0
0
cos)(2
n=0, 1, 2,
+
=
Tt
t
n tdtntfT
b0
0
sin)(2
n=1, 2, 3,
(t0 is arbitrary, may be 0)
The important predictions of
Fourier theorem can be illustrated
by discussing the analytically simple
example of a square vibration.
ExampleExample
Modeling a Square WaveModeling a Square Wave
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Sample problem:Sample problem:
Fourier expand the function defined by the following
equation:
f(t) = -A for T/2 < t < 0
= +A for 0 < t < T/2
f(t+T) = f(t)Ans: As the function is an odd function, an = 0
[ ]nT
nn
AtdtnA
Tb )1(1
2sin2
22/
0
==
- 1.5 - 0.5- 1 1.510.50
T
t
f(t)
S3S2
S1
[ ]=
=,...3,2,1
sin)1(112
)(n
ntn
n
Atf
[ ]...5sin3sinsin4
51
31 +++= ttt
A
tA
s
sin4
1=
[ ]ttA
s
3sinsin4
31
2 +=
...
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Other cases of SHM2. Compound pendulum1. Simple pendulum
A uniform rod of
mass m and length
L is pivoted about
one end and
oscillates in a
vertical plane.
Assume its moment
of inertia is J.
3.Torsional Pendulum
A rigid body suspended by a wire attached at the top of afixed support. When the body is twisted through some small
angle , the twisted wire exerts a restoring torgue on the
body proportional to the angular displacement.
(A) Write down the equation of motion for(A) Write down the equation of motion for
each of the following cases.each of the following cases.
(B) What are their natural frequencies?(B) What are their natural frequencies?
Sample problem:Sample problem: Simple pendulum
2
2
sindt
sdmmgFs ==
L
g=)sin( += tm
Ls =
0=+
mgmL
For small : sin ~
0=+
gL
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Sample problem:Sample problem: Compound pendulum
0
2
=+
mgL
J
J
mgL
2=
)sin( += tmA uniform rod of mass m and length L
is pivoted about one end and oscillates
in a vertical plane.
Assume its moment of inertia is J.
Moment produced by gravity:
mg sinL/2
Rotation: Jd2/dt2
Sample problem:Sample problem: Torsional Pendulum
0=+
cJ
J
c=
)sin( += tm
A rigid body suspended by a
wire attached at the top of a
fixed support. When the body is
twisted through some small
angle , the twisted wire exerts
a restoring torgue on the body
proportional to the angular
displacement.
The angular displacement: -c
c is torsion constant.
According to Newtons law for
rotational motion
-c= Jd2/dt2
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Energy in anEnergy in an S.H.M.S.H.M.
The maximum KEThe maximum KE
S.H.M. represents an ideal vibration where there
is no energy loss, so the sum of KE to PE stays
constant..
Continual cycle from KE to PE and back again.
At any one time the total energy is the sum of the
kinetic energy plus the potential energy.
22
02
1Am=max(KE)
QQ What is the value ofWhat is the value of PEPEmaxmax??
Energy in anEnergy in an S.H.M.S.H.M.
QQ Where is KE a maximum?Where is KE a maximum?
QQ Where is PE a maximum?Where is PE a maximum?
Variation of potential energy, kinetic energy and total energy
with displacements for a pendulum
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Energy in anEnergy in an S.H.M.S.H.M.
How do these energies vary with time?
The variation will be assin2t andcos2t
i.e. Epot
= E0
sin2t
Ekin
= E0
cos2t
Etotal
= E0
sin2t + E0
cos2t = E0
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Sample problem:Sample problem: Oscillations on a SurfaceOscillations on a SurfaceA 0.5 kg cube connected to a light spring for which theforce constant is 20.0 N/m oscillates on a horizontal,frictionless track. (a) Calculate the maximum speed of thecube and the total energy of the system if the amplitudeof the motion is 3.0 cm. (b) What is the velocity of thecube when the displacement is 2.0 cm?
Solution:(a)
Vmax= A=0.03(20.0/0.5)1/2
=0.190m/s
E=K+U=Kmax=UmaxKmax=m(Vmax)
2=0.50.50.1902=9.010-3J
sm
k
/32.65.0
20
===
(b) x=2.cm
x=Acostsint=(1-x2/A2)1/2
V=Asint=0.036.320.745=0.141 m/s
SUMARRYSUMARRY Free vibrationsvibrations
Simple Harmonic Motion (s.h.m.)Simple Harmonic Motion (s.h.m.)
Graphical representation of system behaviourGraphical representation of system behaviour
Examples of s.h.m. systemsExamples of s.h.m. systems
Vibration energy - Energy in anEnergy in an S.H.M.S.H.M.