# Energy of the Simple Harmonic Oscillator ```18-2-2015
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Faculty of Computer and Information
Basic Science department
2013/2014
Prof. Nabila.M.Hassan
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•Aims of Course:
•The graduates have to know the nature of vibration wave
motions with emphasis on their mathematical descriptions and
superposition.
•The fundamental ideas can be introduced with reference to
mechanical systems which are easy to visualize.
•The graduates have to know the nature of vibration and wave
motions with emphasis on their mathematical description and
superposition Developing the graduate's skills and creative
thought needed to meet new trends in science.
•Supplying graduates with basic attacks and strategies for
solving problems.
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1- A particle oscillates with simple harmonic motion, so that its displacement
varies according to the expression x = (5 cm)cos(2t + π/6) where x is in
centimeters and t is in seconds. At t = 0 find
(a) the displacement of the particle,
(b) its velocity, and
(c) its acceleration.
(d) Find the period and amplitude of the motion.
Solution:
The displacement as a function of time is
x(t) = A cos(ωt + φ). Here ω = 2/s, φ = π/6, and A = 5 cm.
The displacement at t = 0
is x(0) = (5 cm)cos(π/6) = 4.33 cm.
(b) The velocity at t = 0 is v(0) = -ω(5 cm)sin(π/6) = -5 cm/s.
(c) The acceleration at t = 0 is a(0) = -ω2(5 cm)cos(π/6) = -17.3 cm/s2.
(d) The period of the motion is T = π sec, and the amplitude is 5 cm.
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1- An oscillator consists of a block of mass 0.50 kg connected to a spring.
When set into oscillation with amplitude 35 cm, it is observed to repeat its
motion every 0.50 s. The maximum speed is :
(a) 4.4 m/s ,(b) 44.0 m/s ,( c) 0.44 m/s
2- A particle executes linear harmonic motion about the point x = 0. At t = 0,
it has displacement x = 0.37 cm and zero velocity. The frequency of the
motion is 0.25 Hz. The max speed of the motion equal:
(a) 0.59 cm/s ,(b) 5.9 cm/s ,( c) 0.059 cm/s
3- An oscillating block-spring system has a mechanical energy of 1.0 J,
amplitude of 0.10 m, and a maximum speed of 1.2 m/s. The force constant
of the spring is,
(a) 100 N/m ,(b) 200 N/m ,( c) 20 N/m
4- An oscillating block-spring system has a mechanical energy of 1.0 J,
amplitude of 0.10 m, and a maximum speed of 1.2 m/s. The mass of the
block is,
(a) 1.4 kg ,(b) 14.0 kg ,( c) .140 kg
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1- An oscillator consists of a block of mass 0.50 kg connected to a spring.
When set into oscillation with amplitude 35 cm, it is observed to repeat its
motion every 0.50 s. The maximum speed is :
(a) 4.4 m/s ,(b) 44.0 m/s ,( c) 0.44 m/s
2- A particle executes linear harmonic motion about the point x = 0. At t = 0,
it has displacement x = 0.37 cm and zero velocity. The frequency of the
motion is 0.25 Hz. The max speed of the motion equal:
(a) 0.59 cm/s ,(b) 5.9 cm/s ,( c) 0.059 cm/s
3- An oscillating block-spring system has a mechanical energy of 1.0 J,
amplitude of 0.10 m, and a maximum speed of 1.2 m/s. The force constant
of the spring is,
(a) 100 N/m ,(b) 200 N/m ,( c) 20 N/m
4- An oscillating block-spring system has a mechanical energy of 1.0 J,
amplitude of 0.10 m, and a maximum speed of 1.2 m/s. The mass of the
block is,
(a) 1.4 kg ,(b) 14.0 kg ,( c) .140 kg
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Content: Part II: Waves
Chapter 1
Oscillation Motion
- Motion of a spring:
- Energy of the Simple Harmonic Oscillator:
-Comparing SHM with uniform motion
- The simple pendulum:
- Damped Oscillations:
- Forced Oscillation
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•Objectives:
Student will be able to:
-Calculate the periodic time of S H M.
-- Define the damped motion
- Define the resonance.
-Compare between free, damped and
derived oscillations
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The simple pendulum:
A simple pendulum is a mechanical system that exhibits
motion consists of a small mass, which is suspended from
a light wire or string of length L. The displacement is
defined by the arc S. The net force on the bob (mass) is
tangent to the arc and equals ( F  mgsin  )
For small angles (sin θ ≈ θ ) , then the force equal
F  mgsin   mg
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The displacement is directly related to θ
S = L θ so that θ = S/ L , then
F  
m
g
L
Apply Newton's second law for motion in the
tangential direction
d 2
g
F  m 2  m 
L
dt
The equation of motion for simple pendulum
A simple oscillation of the
pendulum, the resulting force
is F  m gsin   m g
when θ is small
   max cos(t   )
 max
is the max angular position and the angular frequency

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d 2
g
 

2
L
dt
becomes

g
L
T  2
L
g
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Example (1)
a) How much energy is stored in the spring of a
toy gun that has a force constant of 50 N /m
and is compressed 0.150 m?
Sol: The energy can be found directly from
equation
1
1
2
PE el  KX  (50 N / m)( 0.150 m ) 2
2
2
 0.563N.m
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 0.563J
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b)In absence of friction and neglecting the mass of
the spring, at what speed will a 2.00 g plastic bullet
be ejected from the gun?
Sol: Since there is no friction, the potential energy is
converted into kinetic energy.
KEF  PEel
1
1
mV 2  KX 2  .563 J
2
2
 2 PEl 
v

 m 
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2
 2(0.563J ) 


3
 2  10 kg 
1
2
 23.7m / s
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Problem:
- A 20 g particle moves in simple harmonic motion with a
frequency of 3 oscillations per second and an amplitude of
5cm.
(a) Through what total distance does the particle move
during one cycle of its motion?
(b) What is its maximum speed? Where does that occur?
(c) Find the maximum acceleration of the particle. Where in
the motion does the maximum acceleration occur?
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Damped Oscillations:
Where the force is proportional to the speed of the moving object and
acts in the direction opposite the motion.
The retarding force can be expressed as:
R = - bv ( where b is a constant called damping coefficient)
and the restoring force of the system is – kx,
then we can write Newton's second law as
dx
d 2x
 kx  b
m
dt
dt 2
 F  kx  bv  ma
x
x
x
When the retarding force is small compared with the max restoring force
that is, b is small the solution is,

x(t )  Ae
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b
t
2m
cos(t   )

k
b 2
(
)
m
2m
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represent the position vs time for a
damped oscillation with decreasing
amplitude with time
The fig. shows the position as a function in time of the object oscillation
in the presence of a retarding force, the amplitude decreases in time,
this system is know as a damped oscillator. The dashed line which
defined the envelope of the oscillator curve, represent the
exponential factor
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The fig. represent position versus time:
•A - under damped oscillator
• b- critical damped oscillator
• c- Overdamped oscillator.
as the value of &quot;b&quot; increase the amplitude of the oscillations
decreases more and more rapidly.
When b reaches a critical value bc (bc / 2m  o ),
the system does
not oscillate and is said to be critically damped.
And whenbc / 2m  o
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the system is overdamped.
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Forced Oscillation:
For the forced oscillator is a damped oscillator driven by an external force
that varies periodically
Where
F (t )  Fo sin t
where ω is the angular frequency of the driving force and Fo is a constant
From the Newton's second law
dx
d 2x
 F  m a  Fo sin t  b dt  kx  m dt 2
Fo / m
A
(
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x  A cos(t   )
2
 )
2
o
2
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 b 


 m 
2
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
o  k m
is the natural frequency of the un-damped oscillator
(b=0).
The last two equations show the driving force and the
amplitude of the oscillator which is constant for a given driving
force.
 For small damping the amplitude is large when the
frequency of the driving force is near the natural frequency of
oscillation, or when ω͌ ≈ ωo the is called the resonance and
the natural frequency is called the resonance frequency.
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Amplitude versus the frequency, when the frequency of the
driving force equals the natural force of the oscillator, resonance
occurs. Note the depends of the curve as the value of the
damping coefficient b.
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Note that: The applied force F is in phase with the velocity. The rate at
which work is done on the oscillator by F equals the dot product F. v ;
this rate is the power delivered to the oscillator.
Because the product F. v is a maximum when F and v are in phase,
we conclude that at resonance the applied force is in
phase with the velocity and the power transferred to the
oscillator is a maximum.
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Summary of the chapter 1:
1- The acceleration of the oscillator object is proportional to its
position and is in the direction opposite the displacement from
equilibrium, the object moves with SHM. The position x varies with
time according to, x(t )  A cos(t   )
2- The time for full cycle oscillation is defined as the period, T  2 / 
.
For block spring moves as SHM on the frictionless surface with a period
T  2
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
 2
k
m
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and
3- The frequency is defined as the number of oscillation per second, is the
inverse of the period
1
f 1 
T 2
k
m
4- The velocity and the acceleration of SHM as a function of time are
dx
v
 A sin(t   )
dt
d 2x
a  2   2 A cos(t   )
dt
We not that the max speed is Aω , and the max acceleration is Aω2 .
The speed is zero when the oscillator is at position of x=&plusmn; A , and is a
max when the oscillator is at the equilibrium position at the equilibrium
position x=0.
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5- The kinetic energy and potential energy for simple harmonic oscillator are
given by,
K
1
1
mv 2  m 2 A 2 sin 2 (t   )
2
2
U 
1 2
1
kx  kA 2 cos 2 (t   )
2
2
The total energy of the SHM is constant of the motion and is given by
E 
1
kA 2
2
6- A simple pendulum of length L moves in SHM for small angular
displacement from the vertical, its period is
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T  2 L
g
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7- For the damping force R = - bv, its position for small damping is described by

x(t )  Ae
b
t
2m
cos(t   )

k
b 2
(
)
m
2m
8 - If an oscillator is driving with a force
F (t )  Fo sin t
it exhibits resonance, in which the amplitude is largest when driving
frequency matches the natural frequency of the oscillator.
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What is the effect on the period of a pendulum of doubling its length?
L  2L
TL  2 L / g
T2 L  2 2L / g
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T2 L  2TL  1.414TL
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Useful website
http://cnx.org/content/m15880/latest/
http://www.acs.psu.edu/drussell/Demos/SHO/massforce.html
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