Lecture 9
Oscillatory Motion
Outline
1. Periodic motion, simple harmonic motion (oscillator)
2. Ideal spring, Hooke’s law and natural frequency of oscillation 𝜔 =
𝑘
𝑚
𝑑2
𝑥
𝑑𝑡 2
3. Equation of motion:
+ 𝜔2 𝑥 = 0 and general solution 𝑥 𝑡 =
𝐴 𝑐𝑜𝑠 𝜔𝑡 + 𝜙
1. the two arbitrary constants A, 𝜙 determined from initial
conditions 𝑥 0 = 𝑥0 and 𝑣 0 = 𝑣0
2. Simple harmonic motion = projection of uniform circular motion
with angular speed 𝜔, radius(amplitude) 𝐴, and initial phase 𝜙
4. Energy of SHM and other examples (pendulum etc.)
5. Damped harmonic oscillation
6. Forced/driven harmonic oscillation and resonance
10.1 The Ideal Spring and Simple Harmonic Motion
Applied
x
F
 kx
spring constant
Units: N/m
ออกแรงภายนอกด้วยมือที่กระทากับสปริ ง เพื่อให้สปริ งยืดหรื อหด
ในรู ปจะเห็นว่างานของแรงนี้เป็ นบวก (การกระจัดทิศทางเดียวกันกับแรง)
10.1 The Ideal Spring and Simple Harmonic Motion
HOOKE’S LAW: RESTORING FORCE OF AN IDEAL SPRING
The restoring force on an ideal spring is
Fx  k x
ส่ วนแรงคงตัวของสปริงทิศทางตรงกันข้ามและขนาดของแรงแปรผันตรงกับ ระยะยืดหรื อหดของสปริ ง 𝑥 = 𝑙 − 𝑙0
โดยเราเรี ยกค่าคงที่การแปรผันนั้นว่า 𝑘 (ค่าคงที่ของสปริ ง)
ในรู ปจะเห็นว่างานของแรงสปริ งนี้เป็ นลบเสมอ คือ การกระจัดจากตาแหน่งสมดุล (หรื อความยาวปกติของสปริ ง) ทาให้พลังงานศักย์สปริ งมี
ค่าเป็ นบวก
10.1 The Ideal Spring and Simple Harmonic Motion
Conceptual Example : Are Shorter Springs Stiffer?
ตัดสปริงให้ ส้ั นลงทำให้ สปริงเหนียวขึน้ ?
A 10-coil spring has a spring constant k. If the spring is
cut in half, so there are two 5-coil springs, what is the spring
constant of each of the smaller springs?
10.2 Simple Harmonic Motion and the Reference Circle
DISPLACEMENT
x  A cos  A cos t
10.2 Simple Harmonic Motion and the Reference Circle
x  A cos  A cos t
10.2 Simple Harmonic Motion and the Reference Circle
amplitude A: the maximum displacement
period T: the time required to complete one cycle
frequency f: the number of cycles per second (measured in Hz)
1
f 
T
2
  2 f 
T
10.2 Simple Harmonic Motion and the Reference Circle
VELOCITY
v x  vT sin    
A sin t
vmax
10.2 Simple Harmonic Motion and the Reference Circle
Example The Maximum Speed of a Loudspeaker Diaphragm
The frequency of motion is 1.0 KHz and the amplitude is 0.20 mm.
(a) What is the maximum speed of the diaphragm?
(b) Where in the motion does this maximum speed occur?
10.2 Simple Harmonic Motion and the Reference Circle
v x  vT sin    
A sin t
vmax
(a)



vmax  A  A2 f   0.20 10 3 m 2  1.0 103 Hz

 1.3 m s
(b) The maximum speed
occurs midway between
the ends of its motion.
10.2 Simple Harmonic Motion and the Reference Circle
ACCELERATION
ax  ac cos    
A 2 cos t
amax
10.2 Simple Harmonic Motion and the Reference Circle
FREQUENCY OF VIBRATION
x  A cos  t
a x   A 2 cos  t
 F  kx  ma
x
 kA  mA 2
k

m
10.2 Simple Harmonic Motion and the Reference Circle
Example A Body Mass Measurement Device
The device consists of a spring-mounted chair in which the astronaut
sits. The spring has a spring constant of 606 N/m and the mass of
the chair is 12.0 kg. The measured
period is 2.41 s. Find the mass of the
astronaut.
10.2 Simple Harmonic Motion and the Reference Circle
k

mtotal
mtotal  k  2
  2 f 
mtotal 
mastro
k
 mchair  mastro
2
2 T 
k

 mchair
2
2 T 
2

606 N m 2.41 s 

 12.0 kg  77.2 kg
4 2
2
T
10.3 Energy and Simple Harmonic Motion
A compressed spring can do work.
10.3 Energy and Simple Harmonic Motion
Welastic  F cos s  12 kxo  kxf cos 0 xo  x f 
Welastic  12 kxo2  12 kx2f
10.3 Energy and Simple Harmonic Motion
DEFINITION OF ELASTIC POTENTIAL ENERGY
The elastic potential energy is the energy that a spring
has by virtue of being stretched or compressed. For an
ideal spring, the elastic potential energy is
PEelastic  12 kx2
SI Unit of Elastic Potential Energy: joule (J)
10.3 Energy and Simple Harmonic Motion
Conceptual Example Changing the Mass of a Simple
Harmonic Oscillator
The box rests on a horizontal, frictionless
surface. The spring is stretched to x=A
and released. When the box is passing
through x=0, a second box of the same
mass is attached to it. Discuss what
happens to the (a) maximum speed
(b) amplitude (c) angular frequency.
ในรู ป ก. เรายืดสปริ งจากตาแหน่ง x(0)=A แล้วปล่อย (v(0)=0) ซึ่ง
เมื่อกล่องเคลื่อนที่ผา่ นตาแหน่งสมดุล x=0 จะเคลื่อนที่ดว้ ย 𝑣max ถ้ าเรา
ปล่อยให้อีกกล่องหนึ่งตกลงไปติดกับมวลเดิม (มองว่าเหมือนเป็ นก้อนดิน
น้ ามันหรื อทรายตกลงไปแปะก็ได้) ดังรู ป ข. ถามว่าหลังมีมวลมาติด เกิด
อะไรขึ้นกับ
1. 𝑣max
2. amplitude
3. angular frequency
Homework: Vertical Spring
A 0.20-kg ball is attached to a vertical spring. The spring constant
is 28 N/m. When released from rest, how far does the ball fall
before being brought to a momentary stop by the spring?
ให้ความยาวปกติของสปริ ง(ค่าคงที่สปริ ง k) เป็ น 𝑙0 มวล 𝑚
อยูใ่ ต้สนามโน้มถ่วง 𝑔 ถ้าปล่อยมวลจากหยุดนิ่งและความยาว
สปริ งปกติ
ถามว่า
ก. ระยะสมดุลใหม่ ℎ𝑒𝑞 อยูท่ ี่ไหน
ข. ระยะยืดมากที่สุด ℎ0 (ระยะที่ลูกบอลตกลงมาจาก
จุดเริ่ มต้นจนถึงจุดที่ลูกบอลหยุดชัว่ ขณะ)
ค. เขียน y(t), v(t)
ง. What is amplitude 𝐴 and initial
phase 𝜙?
Summary of SHM
2
  2 ƒ 
T
T  2
m
k
ƒ
Math: definition of period
(periodic function COS
and SIN)
1
2
k
m
Physics: What determines
the period/frequency of
SHM ?
x(t )  A cos (t   )
dx
v
  A sin( t   )
dt
d 2x
a  2   2 A cos( t   )
dt
General form of the solution to the differential equation:
𝑥 𝑡 = 𝐵1 cos 𝜔𝑡 + 𝐵2 sin 𝜔𝑡
arbitrary constants 𝐵1 and 𝐵2 are determined by the 2 initial conditions at
t=0
𝑥 0 = 𝑥0 , 𝑣 0 = 𝑣0
𝑣0
𝑥 𝑡 = 𝑥0 cos 𝜔𝑡 − sin 𝜔𝑡
𝜔
𝑥02 + 𝑣02 /𝜔 2
−𝑣0
𝜙 = arctan(
)
𝜔𝑥0
𝐴=
All of these results are easily understood in terms of “Energy of SHM”
𝟏 𝟐
𝟏
𝑬𝒕𝒐𝒕𝒂𝒍 = 𝒌𝒙𝒎𝒂𝒙 = 𝒎𝒗𝟐𝒎𝒂𝒙
𝟐
𝟐
𝟏
𝟏
𝟏
𝟏
𝟏
𝟏
= 𝒌𝑨𝟐 = 𝒌 𝒙 𝒕 𝟐 + 𝒎𝒗 𝒕 𝟐 = 𝒌𝒙𝟐𝟎 + 𝒎𝒗𝟐𝟎 = 𝒎𝝎𝟐 𝑨𝟐
𝟐
𝟐
𝟐
𝟐
𝟐
𝟐
SHM Example 1
From graph
1. initial conditions: x(0)=A, v(0)=0
2. What is  ?
3. Write x(t), v(t), a(t)?
•The acceleration reaches
extremes of  2A at ±A.
•The velocity reaches extremes of
 A at x = 0.
Section 15.2
SHM Example 2
From graph
1. Given initial conditions:
𝑥(0) =
0, 𝑣(0) = 𝑣0
2. What is  , A ?
3. Write x(t), v(t), a(t)?
The graph is shifted one-quarter cycle
to the right compared to the previous
graph of x(0) = A.
Section 15.2
Importance of Simple Harmonic
Oscillators
•Simple harmonic oscillators are
good models of a wide variety of
physical phenomena.
•Molecular example
– If the atoms in the molecule do
not move too far, the forces
between them can be
modeled as if there were
springs between the atoms.
– The potential energy acts
similar to that of the SHM
oscillator.
Potential energy
𝑟0
actual P.E.
Near equilibrium position 𝑟0 where U
(Potential energy) is minimum,
approximate P.E. as that of spring 𝑈 =
1
𝑈𝑚𝑖𝑛 + 𝐾 𝑟 − 𝑟0 2
2
+ higher order terms in 𝑟 − 𝑟0 3
SHM and Circular Motion
•The particle moves along the circle
with constant angular velocity 
•OP makes an angle  with the x
axis.
•At some time, the angle between OP
and the x axis will be   t + 
•The points P and Q always have the
same x coordinate.
𝑥 𝑡 = 𝐴 cos(𝜔𝑡 + 𝜙)
•This shows that point Q moves with
simple harmonic motion along the
x axis.
Section 15.4
SHM and Circular Motion, 4
•The angular speed of P is the same
as the angular frequency of simple
harmonic motion along the x axis.
•Point Q has the same velocity as the
x component of point P.
•The x-component of the velocity is
v = - A sin ( t + )
Section 15.4
SHM and Circular Motion, 5
•The acceleration of point P on the
reference circle is directed radially
inward.
•P ’s acceleration is a = 2A
•The x component is
–2 A cos (t + )
•This is also the acceleration of point
Q along the x axis.
Section 15.4
10.4 The Pendulum
A simple pendulum (ลูกตุ้มอย่ างง่ าย) consists of
a particle attached to a frictionless
pivot(จุดหมุน) by a cable of negligible mass.

g
L
mgL

I
(small angles only)
(small angles only)
d 2
g
g


sin




2
dt
L
L
10.4 The Pendulum
Example Keeping Time
เรำมักใช้ คำบของลูกตุ้มเป็ นมำตรฐำนเวลำ(นำฬิ กำคุณปู่ ) หรื อ กำรสั่ นของผลึกควอตซ์ ในนำฬิ กำ
หรื อแม้ แต่ นำฬิ กำอะตอมก็คือกำรสั่ นของอิเล็กตรอนในอะตอมนั่นเองทีเ่ รำอ่ ำนจำกสั ญญำนแสง
เลเซอร์
Determine the length of a simple pendulum that will
swing back and forth in simple harmonic motion with
a period of 1.00 s.
2
  2 f 

T
g
L
T 2g
L
4 2
T 2 g 1.00 s  9.80 m s 2 
L

 0.248 m
2
2
4
4
2
10.5 Damped Harmonic Motion
In simple harmonic motion, an object oscillated
with a constant amplitude.
In reality, friction or some other energy
dissipating mechanism is always present
and the amplitude decreases as time
passes.
This is referred to as damped harmonic
motion.
Damped Oscillation: model
•One example of damped motion
occurs when an object is attached to
a spring and submerged in a viscous
liquid.
•The damping/retarding force
due to viscosity can be modeled as
𝑏 is a constant called the
damping coefficient
𝑭𝒅𝒂𝒎𝒑𝒊𝒏𝒈 = −𝒃𝒗
Section 15.6
Damped Oscillations
•The amplitude decreases with time.
•The dashed lines represent the
envelope of the motion.
Section 15.6
Damped Harmonic Motion


F  ma
d 2x
a 2
dt
, F  kx
d 2x
 kx  m 2
dt
simple harmonic motion
พิจารณา แรงหน่วง (damping force)
dx

Fdamping  bv  
dt
dx
d 2x
 kx  b  m 2 damped harmonic motion 
dt
dt
d 2x
dx
m 2  b  kx  0
dt
dt
d 2x
dx
b
k
2
2

2



x

0
,


,


dt 2
dt
2m
m
34
10.5 Damped Harmonic Motion
•
•
•
•
SHM with no damping
Under-damped SHM
Critically-damped SHM
Over-damped SHM
A,B เป็ นค่าคงที่
10.6 Driven Harmonic Motion and Resonance
When a force is applied to an oscillating system at all times,
the result is driven harmonic motion.
Here, the driving force has the same frequency as the
spring system and always points in the direction of the
object’s velocity.
Driven(forced) harmonic motion

F  F0 cos  f t
แรงภายนอก
dx
d 2x
 kx    F0 cos  f t  m 2
dt
dt
 driven harmonic motion 


 forced

d 2x
dx
m 2    kx  F0 cos  f t
dt
dt
ผลเฉลยของสมการ
𝑥 𝑡 = 𝐴 cos(𝜔𝑓 𝑡 − 𝛼)
คาดการณ์ผลเฉลย (particular solution)ว่า วัตถุจะสัน่ ด้วยความถี่
𝜔𝑓 เดียวกันกับแรงกระตุ้น แต่แอมพลิจูด 𝐴 กับ เฟส 𝛼 จะต่างจาก
ของแรงกระตุน้ จากภายนอก
𝑣 𝑡 =
A
𝑑
𝑥 𝑡 = −𝜔𝑓 𝐴 𝑠𝑖𝑛(𝜔𝑓 𝑡 − 𝛼)
𝑑𝑡

F0
m
2
f
 02
  arctan


2
f
2
 4 2 2f
 02

2 f
37
10.6 Driven Harmonic Motion and Resonance
RESONANCE
Resonance is the condition in which a time-dependent force can transmit
large amounts of energy to an oscillating object, leading to a large amplitude
motion.
Resonance occurs when the frequency of the force matches a natural
frequency at which the object will oscillate.
Resonance
•When the frequency of the driving force is near the natural
frequency (f  0) an increase in amplitude occurs.
•This dramatic increase in the amplitude is called resonance.
•The natural frequency 0 is also called the resonance frequency
of the system.
•At resonance, the applied force is in phase with the velocity and
the power transferred to the oscillator is a maximum.
– The applied force and v are both proportional to sin (t + ).
– The power delivered is F.v
• This is a maximum when the force and velocity are in phase.
• The power transferred to the oscillator is a maximum.
Section 15.7
Resonance, cont.
•Resonance (maximum peak) occurs
when driving frequency equals the
natural frequency.
•The amplitude increases with
decreased damping.
•The curve broadens as the damping
increases.
•The shape of the resonance curve
depends on b.
Section 15.7
10.7 Elastic Deformation (ไม่ ออกสอบโดยตรง แต่ ควรรู้ว่าเป็ นทีม่ าของกฎของฮุค )
Because of these atomic-level “springs”, a material tends to
return to its initial shape once forces have been removed.
ATOMS
FORCES
10.7 Elastic Deformation
STRETCHING, COMPRESSION, AND YOUNG’S MODULUS
 L 
 A
F  Y 
 Lo 
equivalent to 𝐹
𝑌𝐴
where 𝐾 = 𝐿
= 𝐾 ΔL
0
Young’s modulus has the units of pressure: N/m2
Stress, Strain and Hooke’s Law
In general the quantity F/A is called the stress.
The change in the quantity divided by that quantity is called the
strain:
V Vo L Lo x Lo
HOOKE’S LAW FOR STRESS AND STRAIN
Stress is directly proportional to strain.
Strain is a unitless quantitiy.
SI Unit of Stress: N/m2
ส่ วนที่เกิดการยืดแบบ Plastic deformation
ส่ วนที่ประพฤติตามกฎของฮุค (แรงแปรผันตรงกับระยะยืด)
10.7 Elastic Deformation
Example
Bone Compression
In a circus act, a performer supports the combined weight (1080 N) of
a number of colleagues. Each thighbone of this performer has a length
of 0.55 m and an effective cross sectional area of 7.7×10-4 m2. Determine
the amount that each thighbone compresses under the extra weight.
10.7 Elastic Deformation
 L 
 A
F  Y 
 Lo 

FLo
540 N 0.55 m 
5
L 


4
.
1

10
m
9
2
4
2
YA 9.4 10 N m 7.7 10 m 
10.7 Elastic Deformation
SHEAR DEFORMATION AND THE SHEAR MODULUS
 x 
F  S   A
 Lo 
The shear modulus has the units of pressure: N/m2
10.7 Elastic Deformation
10.7 Elastic Deformation
Example J-E-L-L-O
You push tangentially across the top
surface with a force of 0.45 N. The
top surface moves a distance of 6.0 mm
relative to the bottom surface. What is
the shear modulus of Jell-O?
 x 
F  S   A
 Lo 
FLo
S
Ax
10.7 Elastic Deformation
FLo
S
Ax

0.45 N 0.030 m 
S
 460 N
2
3
0.070 m  6.0 10 m 
m2
10.7 Elastic Deformation
VOLUME DEFORMATION AND THE BULK MODULUS
 V 

P   B
 Vo 
The Bulk modulus has the units of pressure: N/m2
10.7 Elastic Deformation