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Transcript
```Lecture 1
Ch15. Simple Harmonic Motion
University Physics: Waves and Electricity
Dr.-Ing. Erwin Sitompul
http://zitompul.wordpress.com
2015
Textbook and Syllabus
Textbook:
“Fundamentals of Physics”,
Halliday, Resnick, Walker,
John Wiley & Sons, 8th Extended, 2008.
Syllabus: (tentative)
Chapter 15: Simple Harmonic Motion
Chapter 16: Transverse Waves
Chapter 17: Longitudinal Waves
Chapter 21: Coulomb’s Law
Chapter 22: Finding the Electric Field – I
Chapter 23: Finding the Electric Field – II
Chapter 24: Finding the Electric Potential
Chapter 26: Ohm’s Law
Chapter 27: Circuit Theory
Erwin Sitompul
University Physics: Waves and Electricity
1/2
Final Grade = 10% Homework + 20% Quizzes +
30% Midterm Exam + 40% Final Exam +
Extra Points
 Homeworks will be given in fairly regular basis. The average
 Homeworks are to be written on A4 papers, otherwise they
 Homeworks must be submitted on time, at least one day
before the schedule of the lecture (at least one day before
Monday IE or Tuesday EE).
 Late submission will be penalized by point deduction of –10·n,
where n is the total number of lateness made.
 But: A homework can be submitted late without penalty if a
scanned or photographed version of the homework is sent to
[email protected] at least one day before the
schedule of the class.
Erwin Sitompul
University Physics: Waves and Electricity
1/3
 There will be 3 quizzes. Only the best 2 will be counted.
 Midterm and final exam schedule will follow the schedule
Physics 2
Homework 5
Rudi Bravo
009201700008
21 March 2021
No.1. Answer: . . . . . . . .
Erwin Sitompul
University Physics: Waves and Electricity
1/4
Lecture Activities
 Extra points will be given if you solve a problem in front of the
class or if you send answer to email quizzes. You will earn 1,
2, or 3 points.
 Make up of quizzes will be held one week after the schedule
of the respective quizzes and exams.
 To maintain the integrity, the score of a make up quiz or exam,
upon discretion, can be multiplied by 0.9 (the maximum
score for a make up is 90).
How to get good grades in this class?
• Do the homeworks by yourself
• Solve problems in front of the class
• Take time to learn at home
Erwin Sitompul
University Physics: Waves and Electricity
1/5
Lecture Material
 Lectures will be held in the form of PowerPoint presentations.
 You are expected to write a note along the lectures to record
your own conclusions or materials which are not covered by
the lecture slides.
 Latest lecture slides will be available on internet. Please
check the course homepage regularly.
 The course homepage is :
http://zitompul.wordpress.com
 You are responsible to read and understand the lecture
 Quizzes, midterm exam, and final exam will be open-book. Be
sure to have your own copy of lecture slides. You are not
allowed to borrow or lend anything during quizzes or exams.
Erwin Sitompul
University Physics: Waves and Electricity
1/6
Simple Harmonic Motion
 The following figure shows a
sequence of “snapshots” of a
simple oscillating system.
 A particle is moving
repeatedly back and forth
about the origin of an x axis.
 One important property of
oscillatory motion is its
frequency, or number of
oscillations that are
completed each second.
 The symbol for frequency is f,
and its SI unit is the hertz
(abbreviated Hz).
1 hertz = 1 Hz
= 1 oscillation per second
= 1 s–1
Erwin Sitompul
University Physics: Waves and Electricity
1/7
Simple Harmonic Motion
 Related to the frequency is the period T of the motion, which
is the time for one complete oscillation (or cycle).
1
T
f
 Any motion that repeats itself at regular intervals is called
periodic motion or harmonic motion.
 We are interested here only in motion that repeats itself in a
particular way, namely in a sinusoidal way.
 For such motion, the displacement x of the particle from the
origin is given as a function of time by:
x(t )  xm cos(t   )
Erwin Sitompul
University Physics: Waves and Electricity
1/8
Simple Harmonic Motion
 This motion is called simple
harmonic motion (SHM).
 Means, the periodic motion is a
sinusoidal function of time.
 The quantity xm is called the amplitude of the motion. It is a
positive constant.
 The subscript m stands for maximum, because the amplitude
is the magnitude of the maximum displacement of the
particle in either direction.
 The cosine function varies between ±1; so the displacement
x(t) varies between ±xm.
Erwin Sitompul
University Physics: Waves and Electricity
1/9
Simple Harmonic Motion
 The constant ω is called the angular
frequency of the motion.
2

 2 f
T
  2  f

second
cycle second
 The SI unit of angular frequency is
the radian per second. To be
consistent, the phase constant Φ
2 radians  1 cycle  360
1
 radian  cycle  180
2

1
cycle  90
4

1
cycle  30
6
12
2
Erwin Sitompul
University Physics: Waves and Electricity 1/10
Simple Harmonic Motion
x(t )  xm cos(t )
x(t )  xm' cos(t )
x(t )  xm cos(2t )

x(t )  xm cos(t  )
4
Erwin Sitompul
University Physics: Waves and Electricity
1/11
Checkpoint
A particle undergoing simple harmonic oscillation of period T is
at xm at time t = 0. Is it at –xm, at +xm, at 0, between –xm and 0,
or between 0 and +xm when:
(a) t = 2T At +xm
(b) t = 3.5T
At –xm
(c) t = 5.25T
At 0
(d) t = 2.8T ?
Between 0 and +xm
0.5T
1.5T
T
Erwin Sitompul
University Physics: Waves and Electricity 1/12
Velocity and Acceleration of SHM
 By differentiating the equation of
displacement x(t), we can find an
expression for the velocity of a particle
moving with simple harmonic motion:
dx(t ) d
v(t ) 
  xm cos(t   ) 
dt
dt
v(t )   xm sin(t   )
 Knowing the velocity v(t) for simple
harmonic motion, we can find an
expression for the acceleration of the
oscillating particle by differentiating
once more:
dv(t ) d
a(t ) 
   xm sin(t   ) 
dt
dt
a(t )   2 xm cos(t   )
Erwin Sitompul
a(t )   2 x(t )
University Physics: Waves and Electricity 1/13
Plotting The Motion
Plot the following simple xm
harmonic motions:
0
(a) x1(t) = xmcosωt
(b) x2(t) = xmcos(ωt+π)
–xm
(c) x3(t) = (xm/2)cosωt
(d) x4(t) = xmcos2ωt
x
x1(t)
0.5T
T
x2(t)
m
x1(t)
0
0.5T
T
x3(t)
–xm
xm
0
–xm
Erwin Sitompul
x1(t)
0.5T
T
x4(t)
University Physics: Waves and Electricity 1/14
The Force Law for Simple Harmonic Motion
 Now we can use Newton’s second law to learn what force
must act on the particle to give it that acceleration:
F  ma  m( 2 x)
F  (m 2 ) x
 This result – a restoring force that is proportional to the
displacement but opposite in sign – is familiar.
 It is Hooke’s law for a spring:
F  kx
with the spring constant being k = mω2.
 Conclusion: Simple harmonic motion is the motion executed
by a particle, subject to a force that is proportional to the
displacement of the particle but opposite in sign.
Erwin Sitompul
University Physics: Waves and Electricity 1/15
The Force Law for Simple Harmonic Motion
 The block–spring system below forms a linear simple
harmonic oscillator (or linear oscillator, for short).
 The relation between the angular frequency ω of the simple
harmonic motion of the block is related to the spring constant
k and the mass m of the block as follows:
k

m
 For the period of the linear oscillator, after combining, we
can write:
m
T  2
k
Erwin Sitompul
1
f 
2
k
m
University Physics: Waves and Electricity 1/16
Example 1
A block whose mass m is 680 g is fastened to a spring whose
spring constant k is 65 N/m. The block is pulled a distance
x = 11 cm from its equilibrium position at x = 0 on a frictionless
surface and released from rest at t = 0.
(a) What are the angular frequency, the frequency, and the
period of the resulting motion?
k
65 N m

 9.78 rad s, f  1.557 Hz, T  0.642 s

0.68 kg
m
(b) What is the amplitude of the oscillation?
xm  11 cm
(c) What is the maximum speed vm of the oscillating block, and
where is the block when it has this speed?
vm   xm  (9.78)(0.11)  1.08 m s
Erwin Sitompul
• Where is the block?
University Physics: Waves and Electricity 1/17
Example 1
A block whose mass m is 680 g is fastened to a spring whose
spring constant k is 65 N/m. The block is pulled a distance
x = 11 cm from its equilibrium position at x = 0 on a frictionless
surface and released from rest at t = 0.
(d) What is the magnitude am of the maximum acceleration of
the block?
am   2 xm  (9.78)2 (0.11)  10.52 m s2
(e) What is the phase constant Φ for the motion?
• Why?
(f) What is the displacement function x(t) for the spring–block
system?
x(t )  xm cos(t   )  0.11 m  cos(9.78t ), t in seconds
Erwin Sitompul
University Physics: Waves and Electricity 1/18
Example 2
At t = 0, the displacement x(0) of the
block in a linear oscillator is –8.5 cm.
The block’s velocity v(0) the is
–0.920 m/s, and its acceleration a(0)
is +47 m/s2.
(a) What is the angular frequency ω of this system?
x(0)  xm cos 
v(0)   xm sin 
a(0)   2 xm cos 
Erwin Sitompul
 
a (0)
47
 
x(0)
(0.085)
University Physics: Waves and Electricity 1/19
Example 2
At t = 0, the displacement x(0) of the
block in a linear oscillator is –8.5 cm.
The block’s velocity v(0) the is
–0.920 m/s, and its acceleration a(0)
is +47 m/s2.
(b) What are the phase constant Φ and amplitude xm?
v(0)
sin 
 
  tan 
x(0)
cos 
v(0)
(0.920)
 0.460
tan   

 x(0)
(23.51)(0.085)
  24.7 ? or
  24.7  180  155.3?
Since x(0)  0  cos   0    155.3
xm 
x(0)
0.085
 0.0936 m  9.36 cm

cos  cos155.3
Erwin Sitompul
University Physics: Waves and Electricity 1/20
Example 3
A block with m = 500 g is attached to a massless spring and is
free to oscillate on a frictionless floor. The equilibrium position
is 25 cm from the origin of the x axis.
At t = 0, the block is pressed 10 cm from the equilibrium and
then released. A force of magnitude 5 N is needed to press the
spring.
F
F  kx  k 
x
5N
(0.1 m)
 50 N m

(a) How many oscillations occur in 1 minute?
  k m  (50 N m) (0.5 kg)  10rad s
10
f   2 
oscillations s  300 oscillations minute
2

 95.493 oscillations minute
Erwin Sitompul
University Physics: Waves and Electricity 1/21
Example 3
(b) Determine the block’s equation of motion.
x(t )  xequil  xm cos(t   )
xequil  0.25 m
xm  0.1 m
x(0)  0.15  0.25  0.1cos 
 cos   1,   
x(0)  0.15 m • How?
x(t )  0.25  0.1cos(10t   ), x in m, t in s
(c) Plot the motion of the block for two periods.
0.25+xm
x(t)
0.25
0
0.5T
T
1.5T
2T
0.25–xm
Erwin Sitompul
University Physics: Waves and Electricity 1/22
Homework 1: Plotting the Motions
xm
Plot the following simple
harmonic motions in three
different plots:
0
(a) xa(t) = xmcosωt
(b) xb(t) = xmcos(ωt–π/2) –xm
(c) xc(t) = xm/2cos(ωt+π/2)
xm
(d) xd(t) = 2xmcos(2ωt+π)
0
xa(t)
0.5T
T
xb(t)?
xa(t)
0.5T
T
xc(t)?
–xm
xm
0
–xm
Erwin Sitompul
xa(t)
0.5T
T
xd(t)?
University Physics: Waves and Electricity 1/23
Homework 1A: Simple Harmonic Motion
1. Plot the following simple harmonic motions in three different
plots:
(a) xa(t) = xmsinωt
(b) xb(t) = xmsin(ωt–π)
(c) xc(t) = xm/2sin(ωt+π/2)
(d) xd(t) = xm/2sin(2ωt+π/2)
2. A small car weighs 1200 kg has one spring on each wheel. It
can be assumed, that the springs are identical, and that the
mass of the car is distributed equally over the springs.
(a) Determine the spring constant of each spring if the
empty car bounces up and down 2.0 times each second.
(b) Determine the car’s oscillation frequency while carrying
four 70-kg passengers?
Erwin Sitompul
University Physics: Waves and Electricity 1/24
```