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Chapter 13
Periodic Motionor
Oscillation
Examples:
swinging chandeliersboats bobbing at anchorguitar stringsquartz crystals in wristwatchesatoms in solidsair molecules that transmit sound…
Oscillations in real word are usually damped; that is, the motion dies out gradually, transferring mechanical energy to thermal energy
Oscillations – motions that repeat themselves
2
Periodic motionAny motion that repeat itself at regular intervals is called periodic motion or harmonic motion.
Key characteristicsPeriod (T) – time required for one cycle of periodic
motion. Units: s (seconds/cycle)Frequency (f) – number of oscillations that are
completed each second. Units: s-1 (cycle/second) that is 1 hertz = 1 Hz = 1 oscillation per second = 1 s-1
Tf 1=
Periodic motion – Simple harmonic motionSimple harmonic motion (SHM) – a special case of a periodic motion when the displacement x of the particle (object) from the origin is given as a sinusoidal function of time
)cos(2cos)( φωφπ+=⎟
⎠⎞
⎜⎝⎛ += txt
Txtx mm
SI unit for ω is the radian per second
fT
ππω 22==
3
ExamplesExamples
)cos(2cos)( φωφπ+=⎟
⎠⎞
⎜⎝⎛ += txt
Txtx mm
The velocity and acceleration of SHM
The velocity of SHM
[ ])cos()()( φω +== txdtd
dttdxtv m
)sin()( φωω +−= txtv m
The acceleration of SHM
[ ])sin()()( φωω +−== txdtd
dttdvta m
)()cos()( 22 txtxta m ωφωω −=+−=
4
ExamplesExamples
)cos()( φω += txtx m
)sin()( φωω +−= txtv m
)cos()( 2 φωω +−= txta m
CheckpointCheckpoint
)cos()( 2 φωω +−= txta m
In simple harmonic motion, the magnitude of the acceleration is: A) constant B) proportional to the displacement C) inversely proportional to the displacement D) greatest when the velocity is greatest E) never greater than g
5
The force law for simple harmonic motion
From Newton’s second law
)( 2xmmaF ω−==
For a SHM – a restoring force that is proportional to the displacement but opposite in sign
Example: Hook’s law for a spring
kxF −= then 2ωmk =
mk
=ωkmT π2=
Car springs
When a family of four people with a total mass of 200 kg step into their 1200-kg car, the car springs compress 3.0 cm
(a) What is the spring constant for the car springs, assuming that they act as a single spring?
(b) How far will the car lower if loaded with 300 kg?(c) What are the period and frequency of the car after
hitting a bump? Assume the shock absorbers are poor, so the car really oscillates up and down.
example
6
Car springs (cont.)example
HzT
f
skmT
kFx
xFkkxF
09.11
92.0105.6
)0.3000.1200(22
105.4
105.603.0
)8.9()0.200(
4
2
4
==
=⋅+
==
⋅==
⋅=⋅
===
−
frequency The
N/mkg period The
m
kg 300.0 with loaded car the if
N/mm
m/s kg then 2
ππ
Two springs
Suppose that two springs in figure have different spring constant k1 and k2. What is the frequency of oscillations of the block?
example
kxF −= mkf
π21
=
22
21
21
2121
22
11
21
)(21
21
fffm
kkf
xkkxkxkFmkf
mkf
net
+=+
=
+−=−−=
==
and
π
ππ
7
Energy in simple harmonic motion
The potential energy
)(cos21
21)( 222 φω +== tkxkxtU m
The kinetic energy
)(sin21
21)( 222 φω +== tkxmvtK m
The total mechanical energy
2
21
mkxKUE =+=
ExamplesExamples
)(cos21)( 22 φω += tkxtU m
)(sin21)( 22 φω += tkxtK m
8
Problem: energy in SHMAn object of mass m on a a horizontal frictionless surface is attached to a spring with spring constant k. The object is displaced from equilibrium of x0 horizontally and given an initial velocity of v0 back toward the equilibrium position.(a) What is the frequency of the motion?(b) What is the initial potential energy?(c) What is the initial kinetic energy?(d) What is the amplitude of the oscillation?
example
Problem: energy in SHMexample
2
2
2
12
,,,,,,
2
00
20
0
20
0
00
00
m
m
kxKUE
mvK
kxU
Tf
kmT
xKUfvxkm
=+=
=
=
== and
:Find :Given
π
0x
9
The simple pendulumThe restoring force
sL
mgLsmgmgF ==≈ θ
Comparing to kxF −=
Lmgk =
gL
kmT ππ 22 ==
)sin(θmgF =For small angles
An experiment to measure g
22
22
/5.390.1
)2(
smT
g
LTLg
=
=
=
m for
π
gLT π2=
example
10
Problem: Pendulum
A 2500 kg demolition ball sings from the end of the crane. The length of the swinging segment of cable is 17 m. (a) Find the period of the swinging, assuming that the system can be treated as a simple pendulum(b) Does the period depends on the ball’s mass?
ssm
mgLT 3.8
/8.90.1722 2 === ππ
Example
Two blocks (m=1.0 kg and M=10 kg) and a spring (k=200 N/m) are arranged on a horizontal, frictionless surface. The coefficient of static friction between the two blocks is 0.40. What amplitude of simple harmonic motion of the spring-block’s system puts the smaller block on the verge of slipping over the larger block?
mk
Mmggx
gx
xa
mgmaF
ss
s
s
59.0)(2max
max2
max2
max
maxmax
=+
==
=
=
==
μωμ
μω
ω
μ
and
then
11
Equations
)()(
)sin()()cos()(
2 txta
txtvtxtx
m
m
ω
φωωφω
−=
+−=+=
Periodic Motion
Any motion that repeat itself at regular intervals
Simple harmonic motion)cos()( φω += txtx m
fT
ππω 22==
Restoration force
cmT
mc
cxmaF
πω 2==
−==
Spring
kmTkxF π2=−=
Pendilum
gLTx
LmgF π2=−=
Energy
22
21
21 cxmvE x +=
Damped OscillationsReal-world always have some dissipative (frictional) forces.The decrease in amplitude caused by dissipative forces is called by damping.Quite often damping forces are proportional to the velocity of the oscillating object.
02
2
=++
−−=
kxdtdxb
dtxdm
bvkxF xx Second-order differential equation (we need two initial conditions)
2
2
2/
4'
)'cos()(
mb
mk
textx mbtm
−=
+= −
ω
φω
12
A pendulum + periodic external force
Model: 3 forces
• gravitational force
• frictional force is proportional to velocity
• periodic external force
)cos(,.),sin(
2
2
tFdtdmgL
dtdI
extfg
extfg
ωτθβτθτ
τττθ
=−=−=
++=
example 1
2220
202
2
,,
)cos()sin(
mLFf
mLLg
ImgL
tfdtd
dtd
====
+−−=
βαω
ωθαθωθ
)sin(202
2
θωθ−=
dtd
0 10 20 30 40 50 60-1.2
-1.0
-0.8
-0.6
-0.4
-0.2
0.0
0.2
0.4
0.6
0.8
1.0
1.2 θ(0)=1.0
θ(t)
time
dtd
dtd θαθωθ
−−= )sin(202
2
0 10 20 30 40 50-1.2
-1.0
-0.8
-0.6
-0.4
-0.2
0.0
0.2
0.4
0.6
0.8
1.0
1.2
θ(0)=1.0, α=0.1
θ(t)
time
13
example 2: beats )cos()sin(202
2
tfdtd ωθωθ
+−=
When the magnitude of the force is very large – the system is overwhelmed by the driven force (mode locking) and the are no beats
When the magnitude of the force is comparable with the magnitude of the natural restoring force the beats may occur
0 20 40 60 80 100 120-2.5
-2.0
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
θ(0)=0.2, α=0.0, f=0.2, ω=1.1
θ(
t)
time
example 3: resonance
)cos()sin(202
2
tfdtd ωθωθ
+−=
When the frequency of an external force is close to a natural frequency
0 20 40 60 80 100 120-2.5
-2.0
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
θ(0)=0.2, α=0.0, f=0.1, ω=1.0
θ(t)
time