Oscillation is the repetitive variation, typically in time, of some measure about a central value (often a point of equilibrium) or between two or more different states. The term vibration is precisely used to describe mechanical oscillation. Familiar examples of oscillation include a swinging pendulum and alternating current.
Oscillations occur not only in mechanical systems but also in dynamic systems in virtually every area of science: for example the beating of the human heart (for circulation), business cycles in economics, predator–prey population cycles in ecology, geothermal geysers in geology, vibration of strings in guitar and other string instruments, periodic firing of nerve cells in the brain, and the periodic swelling of Cepheid variable stars in astronomy. Contents
A simple ppt yet interactive on the topic work power and energy. With smooth design and looks the ppt is very good for clearing the basics related to this topic, hope it will help you further.
A simple ppt yet interactive on the topic work power and energy. With smooth design and looks the ppt is very good for clearing the basics related to this topic, hope it will help you further.
My Learning object describes what standing waves are, how to determine where the nodes and antinodes of a standing wave are and also about the fundamental and resonant frequencies. Their is a variety of questions from multiple choice, to true and false and also a problem solving question.
Introduction to Classical Mechanics:
UNIT-I : Elementary survey of Classical Mechanics: Newtonian mechanics for single particle and system of particles, Types of the forces and the single particle system examples, Limitation of Newton’s program, conservation laws viz Linear momentum, Angular Momentum & Total Energy, work-energy theorem; open systems (with variable mass). Principle of Virtual work, D’Alembert’s principle’ applications.
UNIT-II : Constraints; Definition, Types, cause & effects, Need, Justification for realizing constraints on the system
This Unit is rely on introduction to Simple Harmonic Motion. the contents was prepared using the Curriculum of NTA level 4 at Mineral Resources Institute- Dodoma.
Presentation on Electromagnetic Induction.
Physics two presentation of CSE dept. Southeast University.
PPTX slides made by Saleh Ibne Omar.
December 2017.
Malaysia SPM syllabus Physics Chapter 7 Part 4: Electromotive force and internal resistance
Also available for hire!
Contact us for your presentation design needs: lesson / teaching, wedding, seminar, workshop, client pitch etc.
MAHARASHTRA STATE BOARD
CLASS XI and XII
CHAPTER 6
SUPERPOSITION OF WAVES
CONTENT:
Introduction
Transverse and
longitudinal waves
Displacement relation in a
progressive wave
The speed of a travelling
wave
The principle of
superposition of waves
Reflection of waves
Beats
Doppler effect
My Learning object describes what standing waves are, how to determine where the nodes and antinodes of a standing wave are and also about the fundamental and resonant frequencies. Their is a variety of questions from multiple choice, to true and false and also a problem solving question.
Introduction to Classical Mechanics:
UNIT-I : Elementary survey of Classical Mechanics: Newtonian mechanics for single particle and system of particles, Types of the forces and the single particle system examples, Limitation of Newton’s program, conservation laws viz Linear momentum, Angular Momentum & Total Energy, work-energy theorem; open systems (with variable mass). Principle of Virtual work, D’Alembert’s principle’ applications.
UNIT-II : Constraints; Definition, Types, cause & effects, Need, Justification for realizing constraints on the system
This Unit is rely on introduction to Simple Harmonic Motion. the contents was prepared using the Curriculum of NTA level 4 at Mineral Resources Institute- Dodoma.
Presentation on Electromagnetic Induction.
Physics two presentation of CSE dept. Southeast University.
PPTX slides made by Saleh Ibne Omar.
December 2017.
Malaysia SPM syllabus Physics Chapter 7 Part 4: Electromotive force and internal resistance
Also available for hire!
Contact us for your presentation design needs: lesson / teaching, wedding, seminar, workshop, client pitch etc.
MAHARASHTRA STATE BOARD
CLASS XI and XII
CHAPTER 6
SUPERPOSITION OF WAVES
CONTENT:
Introduction
Transverse and
longitudinal waves
Displacement relation in a
progressive wave
The speed of a travelling
wave
The principle of
superposition of waves
Reflection of waves
Beats
Doppler effect
Thermography test of electrical panels Thermography test of electrical panelThermography test of electrical panelThermography test of electrical panelThermography test of electrical panelThermography test of electrical panelThermography test of electrical panelThermography test of electrical panelThermography test of electrical panelThermography test of electrical panelThermography test of electrical panelThermography test of electrical panelThermography test of electrical panelThermography test of electrical panelThermography test of electrical panelThermography test of electrical panelThermography test of electrical panel
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SVC PLUS Frequency Stabilizer Frequency and voltage support for dynamic grid...Power System Operation
SVC PLUS
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Frequency and voltage support for dynamic grid stability
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About
Indigenized remote control interface card suitable for MAFI system CCR equipment. Compatible for IDM8000 CCR. Backplane mounted serial and TCP/Ethernet communication module for CCR remote access. IDM 8000 CCR remote control on serial and TCP protocol.
• Remote control: Parallel or serial interface.
• Compatible with MAFI CCR system.
• Compatible with IDM8000 CCR.
• Compatible with Backplane mount serial communication.
• Compatible with commercial and Defence aviation CCR system.
• Remote control system for accessing CCR and allied system over serial or TCP.
• Indigenized local Support/presence in India.
• Easy in configuration using DIP switches.
Technical Specifications
Indigenized remote control interface card suitable for MAFI system CCR equipment. Compatible for IDM8000 CCR. Backplane mounted serial and TCP/Ethernet communication module for CCR remote access. IDM 8000 CCR remote control on serial and TCP protocol.
Key Features
Indigenized remote control interface card suitable for MAFI system CCR equipment. Compatible for IDM8000 CCR. Backplane mounted serial and TCP/Ethernet communication module for CCR remote access. IDM 8000 CCR remote control on serial and TCP protocol.
• Remote control: Parallel or serial interface
• Compatible with MAFI CCR system
• Copatiable with IDM8000 CCR
• Compatible with Backplane mount serial communication.
• Compatible with commercial and Defence aviation CCR system.
• Remote control system for accessing CCR and allied system over serial or TCP.
• Indigenized local Support/presence in India.
Application
• Remote control: Parallel or serial interface.
• Compatible with MAFI CCR system.
• Compatible with IDM8000 CCR.
• Compatible with Backplane mount serial communication.
• Compatible with commercial and Defence aviation CCR system.
• Remote control system for accessing CCR and allied system over serial or TCP.
• Indigenized local Support/presence in India.
• Easy in configuration using DIP switches.
We have compiled the most important slides from each speaker's presentation. This year’s compilation, available for free, captures the key insights and contributions shared during the DfMAy 2024 conference.
Forklift Classes Overview by Intella PartsIntella Parts
Discover the different forklift classes and their specific applications. Learn how to choose the right forklift for your needs to ensure safety, efficiency, and compliance in your operations.
For more technical information, visit our website https://intellaparts.com
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Industrial Training at Shahjalal Fertilizer Company Limited (SFCL)MdTanvirMahtab2
This presentation is about the working procedure of Shahjalal Fertilizer Company Limited (SFCL). A Govt. owned Company of Bangladesh Chemical Industries Corporation under Ministry of Industries.
NUMERICAL SIMULATIONS OF HEAT AND MASS TRANSFER IN CONDENSING HEAT EXCHANGERS...ssuser7dcef0
Power plants release a large amount of water vapor into the
atmosphere through the stack. The flue gas can be a potential
source for obtaining much needed cooling water for a power
plant. If a power plant could recover and reuse a portion of this
moisture, it could reduce its total cooling water intake
requirement. One of the most practical way to recover water
from flue gas is to use a condensing heat exchanger. The power
plant could also recover latent heat due to condensation as well
as sensible heat due to lowering the flue gas exit temperature.
Additionally, harmful acids released from the stack can be
reduced in a condensing heat exchanger by acid condensation. reduced in a condensing heat exchanger by acid condensation.
Condensation of vapors in flue gas is a complicated
phenomenon since heat and mass transfer of water vapor and
various acids simultaneously occur in the presence of noncondensable
gases such as nitrogen and oxygen. Design of a
condenser depends on the knowledge and understanding of the
heat and mass transfer processes. A computer program for
numerical simulations of water (H2O) and sulfuric acid (H2SO4)
condensation in a flue gas condensing heat exchanger was
developed using MATLAB. Governing equations based on
mass and energy balances for the system were derived to
predict variables such as flue gas exit temperature, cooling
water outlet temperature, mole fraction and condensation rates
of water and sulfuric acid vapors. The equations were solved
using an iterative solution technique with calculations of heat
and mass transfer coefficients and physical properties.
Overview of the fundamental roles in Hydropower generation and the components involved in wider Electrical Engineering.
This paper presents the design and construction of hydroelectric dams from the hydrologist’s survey of the valley before construction, all aspects and involved disciplines, fluid dynamics, structural engineering, generation and mains frequency regulation to the very transmission of power through the network in the United Kingdom.
Author: Robbie Edward Sayers
Collaborators and co editors: Charlie Sims and Connor Healey.
(C) 2024 Robbie E. Sayers
Final project report on grocery store management system..pdfKamal Acharya
In today’s fast-changing business environment, it’s extremely important to be able to respond to client needs in the most effective and timely manner. If your customers wish to see your business online and have instant access to your products or services.
Online Grocery Store is an e-commerce website, which retails various grocery products. This project allows viewing various products available enables registered users to purchase desired products instantly using Paytm, UPI payment processor (Instant Pay) and also can place order by using Cash on Delivery (Pay Later) option. This project provides an easy access to Administrators and Managers to view orders placed using Pay Later and Instant Pay options.
In order to develop an e-commerce website, a number of Technologies must be studied and understood. These include multi-tiered architecture, server and client-side scripting techniques, implementation technologies, programming language (such as PHP, HTML, CSS, JavaScript) and MySQL relational databases. This is a project with the objective to develop a basic website where a consumer is provided with a shopping cart website and also to know about the technologies used to develop such a website.
This document will discuss each of the underlying technologies to create and implement an e- commerce website.
Welcome to WIPAC Monthly the magazine brought to you by the LinkedIn Group Water Industry Process Automation & Control.
In this month's edition, along with this month's industry news to celebrate the 13 years since the group was created we have articles including
A case study of the used of Advanced Process Control at the Wastewater Treatment works at Lleida in Spain
A look back on an article on smart wastewater networks in order to see how the industry has measured up in the interim around the adoption of Digital Transformation in the Water Industry.
2. • Oscillations of a Spring
• Simple Harmonic Motion
• Energy in the Simple Harmonic Oscillator
• Simple Harmonic Motion Related to Uniform
Circular Motion
• The Simple Pendulum
• The Physical Pendulum and the Torsion
Pendulum
• Damped Harmonic Motion
• Forced Oscillations; Resonance
3. Oscillatory Motion
Motion which is periodic in time, that
is, motion that repeats itself in time.
Examples:
• Power line oscillates when the
wind blows past it
• Earthquake oscillations move
buildings
4. If an object vibrates or
oscillates back and forth
over the same path,
each cycle taking the
same amount of time,
the motion is called
periodic. The mass and
spring system is a
useful model for a
periodic system.
Oscillations of a Spring
5. We assume that the surface is frictionless.
There is a point where the spring is neither
stretched nor compressed; this is the
equilibrium position. We measure
displacement from that point (x = 0 on the
previous figure).
The force exerted by the spring depends on
the displacement:
Oscillations of a Spring
6. • The minus sign on the force indicates that it
is a restoring force—it is directed to restore
the mass to its equilibrium position.
• k is the spring constant.
• The force is not constant, so the acceleration
is not constant either.
Oscillations of a Spring
8. • Displacement is measured from the
equilibrium point.
• Amplitude is the maximum
displacement.
• A cycle is a full to-and-fro motion.
• Period is the time required to
complete one cycle.
• Frequency is the number of cycles
completed per second.
Oscillations of a Spring
9. If the spring is hung
vertically, the only change
is in the equilibrium
position, which is at the
point where the spring
force equals the
gravitational force.
Oscillations of a Spring
10. When a family of four with a total
mass of 200 kg step into their
1200-kg car, the car’s springs
compress 3.0 cm. (a) What is the
spring constant of the car’s
springs, assuming they act as a
single spring? (b) How far will the
car lower if loaded with 300 kg
rather than 200 kg?
Oscillations of a Spring
12. Any vibrating system where the
restoring force is proportional to the
negative of the displacement is in
simple harmonic motion (SHM), and is
often called a simple harmonic
oscillator (SHO).
Simple Harmonic Motion
13. Substituting F = kx into Newton’s
second law gives the equation of
motion:
with solutions of the form:
Simple Harmonic Motion
F = -kx = m
dv
dt
= m
d
dt
dx
dt
æ
è
ç
ö
ø
÷ = m
d2
x
dt2
d2
x
dt2
+
k
m
x = 0
14. Substituting, we verify that this solution does
indeed satisfy the equation of motion, with:
The constants A and φ
will be determined by
initial conditions; A is
the amplitude, and φ
gives the phase of the
motion at t = 0.
Simple Harmonic Motion
15. The velocity can be found by differentiating the
displacement:
These figures illustrate the effect of φ:
Simple Harmonic Motion
17. Determine the period and frequency of a car
whose mass is 1400 kg and whose shock
absorbers have a spring constant of 6.5 x
104 N/m after hitting a bump. Assume the
shock absorbers are poor, so the car really
oscillates up and down.
Simple Harmonic Motion
19. The velocity and
acceleration for simple
harmonic motion can
be found by
differentiating the
displacement:
Simple Harmonic Motion
20. Simple Harmonic Motion
x = Acos(wt +f),
v = -wAsin(wt +f)
=wAcos(wt +f +
p
2
),
a = -w2
Acos(wt +f)
= -w2
Asin(wt +f +
p
2
).
21. A vibrating floor.
A large motor in a factory causes the floor
to vibrate at a frequency of 10 Hz. The
amplitude of the floor’s motion near the
motor is about 3.0 mm. Estimate the
maximum acceleration of the floor near
the motor.
Simple Harmonic Motion
22. Solution:
Assuming the motion is simple harmonic,
w = 2p f = 2p ×10Hz=6.3s-1
,
amax =w2
A = 6.3s-1
( )
2
×3.0´10-3
m=12m/s.
23. Loudspeaker.
The cone of a loudspeaker oscillates in SHM at a
frequency of 262 Hz (“middle C”). The amplitude at the
center of the cone is A = 1.5 x 10-4 m, and at t = 0, x = A.
(a) What equation describes the motion of the center
of the cone? (b) What are the velocity and
acceleration as a function of time? (c) What is the
position of the cone at t = 1.00 ms (= 1.00 x 10-3 s)?
Simple Harmonic Motion
24. Solution:
a. w = 2p f = 2p ×262Hz =1650rad/s,
x(t) = Asin(wt +f),
x(0) = Asinf = A Þ f = p 2,
x(t) = Asin(wt +p 2)
= 1.5´10-4
m( )cos 1650t( ).
25. Solution:
b. v(t) =
dx
dt
= -0.25m/s( )sin 1650t( );
a(t) =
dv
dt
= -410m/s2
( )cos 1650t( )
c. at t =1.00ms,
x = 1.5´10-4
m( )cos 1650rad/s×1.00´10-3
s( )
= -1.2´10-5
m.
26. Spring calculations.
A spring stretches 0.150 m when a 0.300-kg mass is
gently attached to it. The spring is then set up
horizontally with the 0.300-kg mass resting on a
frictionless table. The mass is pushed so that the spring
is compressed 0.100 m from the equilibrium point, and
released from rest. Determine: (a) the spring stiffness
constant k and angular frequency ω; (b) the amplitude of
the horizontal oscillation A; (c) the magnitude of the
maximum velocity vmax; (d) the magnitude of the
maximum acceleration amax of the mass; (e) the period T
and frequency f; (f) the displacement x as a function of
time; and (g) the velocity at t = 0.150 s.
Simple Harmonic Motion
28. Spring is started with a push.
Suppose the spring of the former example
(where ω = 8.08 s-1) is compressed 0.100 m
from equilibrium (x0 = -0.100 m) but is given a
shove to create a velocity in the +x direction
of v0 = 0.400 m/s. Determine (a) the phase
angle φ, (b) the amplitude A, and (c) the
displacement x as a function of time, x(t).
Simple Harmonic Motion
29. Solution:
a. x = Acos(wt +f), x0 = Acosf,
v = -wAsin(wt +f), v0 = -wAsinf,
v0
x0
= -w tanf, f = arctan
v0
-wx0
= 3.60rad.
b. A =
x0
cosf
= 0.112m.
c. x = (0.112m)cos(8.08t +3.60)
30. We already know that the potential energy of a
spring is given by:
The total mechanical energy is then:
The total mechanical energy will be
conserved, as we are assuming the system
is frictionless.
Energy in the SHO
31. If the mass is at the limits
of its motion, the energy is
all potential.
If the mass is at the
equilibrium point, the
energy is all kinetic.
We know what the potential
energy is at the turning
points:
Energy in the SHO
32. The total energy is, therefore,
And we can write:
This can be solved for the velocity as a
function of position:
where
Energy in the SHO
33. This graph shows the potential energy
function of a spring. The total energy is
constant.
Energy in the SHO
34. Energy calculations.
For the simple harmonic oscillation where k =
19.6 N/m, A = 0.100 m, x = -(0.100 m) cos 8.08t,
and v = (0.808 m/s) sin 8.08t, determine (a) the
total energy, (b) the kinetic and potential
energies as a function of time, (c) the velocity
when the mass is 0.050 m from equilibrium, (d)
the kinetic and potential energies at half
amplitude (x = ± A/2).
Energy in the SHO
36. Solution:
c. K = E -U,
1
2
mv2
=
1
2
kA2
-
1
2
kx2
,
v =
k
m
(A2
- x2
) =w A2
- x2
= 8.08Hz× (0.100m)2
-(0.050m)2
= 0.70m/s.
d. U =
1
2
kx2
=
1
2
k
A
2
æ
è
ç
ö
ø
÷
2
=
1
4
E = 2.5´10-2
J,
E = K -U = 7.3´10-2
J.
37. Doubling the amplitude.
Suppose this spring is
stretched twice as far (to x =
2A).What happens to (a) the
energy of the system, (b)
the maximum velocity of the
oscillating mass, (c) the
maximum acceleration of
the mass?
Energy in the SHO
39. If we look at the projection onto
the x axis of an object moving
in a circle of radius A at a
constant speed υM , we find that
the x component of its velocity
varies as:
This is identical to SHM.
SHO Related to Uniform Circular Motion
40. SHO Related to Uniform Circular Motion
q =q0 +wt ºwt +f,
x = Acosq = Acos(wt +f),
y = Asinq = Asin(wt +f),
v =wA,
vx = -wAsin(wt +f),
vy =wAcos(wt +f).
41. A simple pendulum
consists of a mass at
the end of a
lightweight cord. We
assume that the cord
does not stretch, and
that its mass is
negligible.
The Simple Pendulum
42. In order to be in SHM, the
restoring force must be
proportional to the negative of
the displacement. Here we
have:
which is proportional to sin θ
and not to θ itself.
However, if the angle is small,
sin θ ≈ θ.
The Simple Pendulum
43. Therefore, for small angles, we have:
where
The period and frequency are:
The Simple Pendulum
44. The Simple Pendulum
t = Ia,
t = F = -mg sinq, I = m 2
;
-mg sinq = m 2 d2
q
dt2
,
d2
q
dt2
+
g
q = 0 Þw2
=
g
.
45. So, as long as the cord can
be considered massless
and the amplitude is small,
the period does not depend
on the mass.
The Simple Pendulum
46. The Simple Pendulum
Measuring g.
A geologist uses a simple pendulum that has
a length of 37.10 cm and a frequency of 0.8190
Hz at a particular location on the Earth. What
is the acceleration of gravity at this location?
48. A physical pendulum is any real
extended object that oscillates
back and forth.
The torque about point O is:
Substituting into Newton’s
second law gives:
The Physical and Torsional Pendulum
49. The Physical and Torsional
Pendulum
For small angles, this becomes:
which is the equation for SHM, with
50. The Physical and Torsional Pendulum
Moment of inertia measurement.
An easy way to measure the moment of
inertia of an object about any axis is to
measure the period of oscillation about
that axis. (a) Suppose a nonuniform 1.0-kg
stick can be balanced at a point 42 cm
from one end. If it is pivoted about that
end, it oscillates with a period of 1.6 s.
What is its moment of inertia about this
end? (b) What is its moment of inertia
about an axis perpendicular to the stick
through its center of mass?
51. Solution:
T = 2p
I
mgh
,
I =
T
2p
æ
è
ç
ö
ø
÷
2
mgh =
1.6s
2p
æ
è
ç
ö
ø
÷
2
1.0kg×9.8m/s2
×0.42m = 0.27kg×m2
.
I = ICM + mh2
,
ICM = I - mh2
= 0.27kg×m2
-1.0kg×(0.42m)2
= 0.09kg×m2
.
52. The Physical and Torsional Pendulum
A torsional pendulum is
one that twists rather than
swings. The motion is SHM
as long as the wire obeys
Hooke’s law, with
(K is a constant that
depends on the wire.)
53. Damped harmonic motion is harmonic motion
with a frictional or drag force. If the damping
is small, we can treat it as an “envelope” that
modifies the undamped oscillation.
If
then
Damped Harmonic Motion
54. This gives
If b is small, a solution of the form
will work, with
Damped Harmonic Motion
55. If b2 > 4mk, ω’ becomes imaginary, and the
system is overdamped (C).
For b2 = 4mk, the system is critically damped
(B) —this is the case in which the system
reaches equilibrium in the shortest time.
Damped Harmonic Motion
56. There are systems in which
damping is unwanted, such as
clocks and watches.
Then there are systems in
which it is wanted, and often
needs to be as close to critical
damping as possible, such as
automobile shock absorbers
and earthquake protection for
buildings.
Damped Harmonic Motion
57. Simple pendulum with damping.
A simple pendulum has a length
of 1.0 m. It is set swinging with
small-amplitude oscillations. After
5.0 minutes, the amplitude is only
50% of what it was initially. (a)
What is the value of γ for the
motion? (b) By what factor does
the frequency, f’, differ from f, the
undamped frequency?
Damped Harmonic Motion
58. Solution:
a. 50% = e-g×5.00´60s
, g = -
ln50%
300s
= 2.3´10-3
s-1
.
b. w =
g
=
9.8m/s2
1.0m
= 3.1rad/s, ¢w = w2
-g2
,
w - ¢w
w
=
w - w2
-g2
w
=1- 1-
g2
w2
@1-(1-
g2
2w2
)
=
g2
2w2
=
(2.3´10-3
s-1
)2
2(3.1s-1
)2
= 2.7´10-7
,
f - ¢f
f
=
w - ¢w
w
= 2.7´10-7
.
59. Forced vibrations occur when there is a periodic
driving force. This force may or may not have the
same period as the natural frequency of the
system.
If the frequency is the same as the natural
frequency, the amplitude can become quite large.
This is called resonance.
Forced Oscillations; Resonance
60. The sharpness of the
resonant peak depends
on the damping. If the
damping is small (A) it
can be quite sharp; if
the damping is larger
(B) it is less sharp.
Like damping, resonance can be wanted or
unwanted. Musical instruments and TV/radio
receivers depend on it.
Forced Oscillations; Resonance
63. • For SHM, the restoring force is proportional to
the displacement.
• The period is the time required for one cycle,
and the frequency is the number of cycles per
second.
• Period for a mass on a spring:
• SHM is sinusoidal.
• During SHM, the total energy is continually
changing from kinetic to potential and back.
Summary
64. • A simple pendulum approximates SHM if its
amplitude is not large. Its period in that case is:
• When friction is present, the motion is
damped.
• If an oscillating force is applied to a SHO, its
amplitude depends on how close to the natural
frequency the driving frequency is. If it is close,
the amplitude becomes quite large. This is
called resonance.