Definition of Wave

8/6/2019 Definition of Wave

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CHAPTER 1

A. Definition and Types of Waves

Vibrations or oscillation is a periodic particle around the equilibrium point . There are twocommon examples of vibration that we encounter in everyday life, namely vibration and

vibration of objects on spring objects in a simple swing (example object vibration is vibration in

a simple pendulum swing).

Vibration that

occurs on an objectcaused by the

disturbance given to

the object. For the case

of pendulum vibration

and body vibration on the spring, the disorder is caused by external force. Actually there are

many examples of vibration that can be encountered in everyday life. Garputala vibrate when we

give garputala interference by hitting them. Vehicles will vibrate when the engine is turned on, in

this case the vehicle is given the disturbance. The voice that we say will not be heard if we do

not vibrate the vocal cords. Anything as beautiful as the music, if the loudspeaker that functions

as a source of sound and our eardrums as the recipient does not vibrate, it is certain we will never

hear the music.

Any disturbance which is given to an object will cause a vibration in the body and these

vibrations will propagate from one place to another place through a particular medium (medium

= broker). In this case, the event propagation of vibration from one place to another through a

medium called a particularwave. In other words, the wave is propagating vibration and vibration

is itself a source of waves.

1. Kinds of wave

Basically, a wave on the direction of vibration to its propagation direction.


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a. Mechanical Waves and Electromagnetic Waves

According to the propagation medium, waves consist of mechanical waves and

electromagnetic waves. Mechanical waves are the waves that require a propagation

medium to transmit energy, while electromagnetic waves are the waves that require

no propagation medium to transmit energy. Waves on water surface, sound waves,

waves on the string and slinky are example of mechanical waves, while the visible

light, radio waves and TV waves are examples of electromagnetic waves.

b. Transverse Waves and Longitudinal Waves

Based on vibration direction to the propagation direction, waves are distinguished

into transverse waves and longitudinal waves.

A transverse waves is a wave of which the vibration ( the motion of particles in the

medium ) is in a direction perpendicular to the direction of the wave propagation

(wave motion). For example, a string stretched horizontally and moved upward and

downward will form wave patterns as shown in the following figure.

Besides transverse wave, there is also longitudinal wave, that is wave of which the

vibration ( the motion of particles in the medium ) is in a direction parallel to the direction of

wave propagation. For example, the slinky that is vibrated horizontally and forms patterns as

follows.


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Longitudinal wave is always a mechanical wave because it is produced from the state of

maximum density and pressure (compression) and the state of minimum density and pressure

( rarefaction ) respectively.

1. Period, Frequency and Speed of Wave

In the study of wave, there are three important quantities, those are period, frequency and

speed of wave.

Frequency of wave refers to how often the particles of medium vibrate when a wave passes

through it. In brief, frequency is defined as the sum of waves occurring on a point per second,

and mathematically it can be determined as follows.

( frequency is the sum of wave divided of time )

Where:

f = frequency of wave (1/s)

n = the sum of wave

t = time (s)

f = nt


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The unit of frequency is 1/s that is equal to hertz (Hz). For example, if a slinky makes 3

vibrations in one second, then the frequency is 3 Hz, and if it makes 8 oscillations in 4 second,

then the frequency is 2 Hz.

Meanwhile, the quantity of wave period is the time for a particle to make one completecycle of vibration, therefore period can be determined as follows.

( period is time divided the sum of wave )

Where:

T = wave period (s)

Based on the above explanation, then the relationship between frequency and period can be

expressed as follows.

(Frequency is one per period ) or ( period is one per frequency )

Because wave is a disturbance that moves along a medium from one end to the order end,

then besides having frequency and period, it also has speed or velocity.

Generally, an objects speed is relates to how fast it moves and is usually expressed as the

distance travelled per unit of time. However, in wave, the speed is defined as the distance

travelled by a point on the wave ( such as a crest ) in a given time interval. The speed of wave

can be expressed mathematically by the equation as follows.

T = tn

f = 1T or T = 1f

V = st


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( velocity is speed divided of time )

If related to period ( T ), frequency (f) and wavelength (), then the speed of wave can be

determined by the following equation.

( velocity is wavelength divided of period or velocity is wavelength times frequency )

Where :

V = speed of wave (m/s)

Sample problem

1. A wave has wavelength of 2 m and speed of 16.1 m/s. determine its period and frequency.

Solution :

Period

V = T T= v = 216.1 s=0,124 s

Frequency

V = .f f = v = 16,12 = 8,05 m/s

Thus, the wave period is 0.124 s and its frequency is 8.05 Hz.

A. Waves Energy

V = T = .f


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As mentioned in the beginning of this chapter, in its propagation, a wave transfers energy

from one location to another but the medium passed through does not propagate. The amount of

energy transferred by the wave is proportional to the square of its amplitude ( E=y2 ).

Mathematically, the wave energy can be determined by the following equation.

E = k y2

E = 22 mf2 y2

E = 22 Avt f2 y2

( wave energy is half times constant times amplitude square)

(wave energy is two phi square times mass times frequency square times amplitude square )

(wave energy is two phi square times constant times frequency square times amplitude square )

Where

E = wave energy ( joule )

k = 22 Avt f2 = constant

y = amplitude (m)

m = mass (kg)

f = frequency ( Hz )

v = speed of wave (m/s)

t = epoch time of wave (s)

One example of wave with tremendous energy is tsunami. A tsunami wave can have

wavelength of 100 to 200 km, and may travel hundreds of kilometers across the deep ocean and

reaches a velocity of 725 km/hour to 800 km/hour. When a tsunami reaches the shore, this wave

can be 15 meters high or more and carry tremendous energy that can obliterate settlement and


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kill the people within. Such a disaster occurred in Aceh in late 2004 and in Pangandaran in the

middle of 2006.

The discussion about energy is always related to power and intensity, therefore, in this

case, there are power and intensity of wave. Wave power is defined as the energy of wavestransferred per unit of time, and the amount of wave power is determined as follows.

P= Et

( power is wave energy divided of time )

Where

P = power (watt)

Meanwhile, wave intensity is wave power transferred through a plane of one unit of area

perpendicular to the direction of wave propagation and mathematically it can be determined as

follows.

I = PA

( wave intensity is power divided area )

Where

I = wave intensity (W/m2)

If a certain wave emits from one source to all directions, the wave front formed is

spherical, so the surface area of the sphere at a distance of r from the source of wave is 4 r2.

When area (A) increases, then the amplitude of wave (y) decreases, therefore

y2y1= r1r2

( amplitude 2 square divided amplitude 1 square is distance 1 square divided distance 2 square )

Where

r = the distance of a certain location to the wave source (m)


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Based on the equation above, we can conclude that the farther a location from the wave

source, then the lower the amplitude of wave at that location. Because wave intensity is directly

proportional to the square of its amplitude, then the relation of wave intensity at a certain

location with the distance of the location from the wave source is determined as follows.

I2I1= r12r22

(wave intensity at location 2 divided wave intensity at location 1 is distance of location 1 square

divided distance of location 2 square ).

Where

I1 = wave intensity at location 1

I2 = wave intensity at location 2

r1 = distance of location 1 to the wave source

r2 = distance of location 2 to the wave source

Sample problem

The earthquake wave intensity in Yogyakarta which lies 106 km from the wave source is 8 x 10 6

W/m2. Calculate the intensity of wave in Cilacap which lies 212 km from the wave source.

Answer:

Yogyakarta (1)

Cilacap (2)

I2I1=r12r22

I2 = r1r22 I1


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= 1062122 ( 8 x 106 W/m2 )

= 2 x 106 W/m2..

Thus, the earthquake wave intensity in Cilacap is 2 x 106 W/m2 .

CHAPTER 2

A. Sound intensity

Intensity is carried by the wave energy per unit time, through a unit area perpendicular to the

direction of wave propagation. Energy per unit time is power hence can be said that intensity is


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the power carried by the waves, through a unit area perpendicular to the direction of wave

propagation.

1. The nature of the sound intensity

To better understand the intensity, we assume the sound source is at the center of a ball.

From the center of the ball, the sound wave will propagate in all directions. Due to propagate to

all directions, the direction of propagation of sound waves must pass through each unit

perpendicular to the surface area of the ball. When the vines, the sound waves carry energy .The

energy carried by the wave per unit time, through a unit area perpendicular to the direction of

wave propagation is known by the nickname intensity. Because the energy per unit time is power

then it could be said that intensity is the power carried by the waves, through a unit area

perpendicular to the direction of wave propagation. If the sound source emits sound waves

uniformly in all directions so that brought a wave of energy will also be divided evenly on the

surface of the ball. For example, the radius of the ball is r, the surface area of sphere = L = 4phi

r2 and power that brought the P wave is the wave intensity can be expressed through the

equation:

I = paverageL

(intensity of sound waves equals power average divided by the surface area of sphere ).

I = Pavarage4r2

(intensity of sound waves equals power average divided by four phi times by distance ).

From this equation it appears that the intensity of sound waves (I) is inversely proportional to

the square of the distance (r2). This means the farther a place from the sound source, the smaller

the intensity of the sound wave. Equation intensity sound waves are derived in greater detail the

subject of energy, power and intensity of the wave.

International system unit of power is the Joule / second. Name of Joule / sec is the Watt

(James Watt appreciate service.) Conversely broad international system unit is meters squared

(m2). Thus, an international system unit of intensity is watts per meter squared (W/m2).


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2. Loudness and intensity level

Declare loud loudness or softness of the sound for example sound louder than the cry of a

whisper. In this case the sound of screams louder than the sound of a whisper. Physical quantities

that are directly related to the loudness is intensity. The human ear can hear the average sound

that has the lowest intensity in 10-12 W/m2 (also called the threshold of hearing. The intensity

below can not be heard) and as high as about 1 W/m2 Note that the range of sound wave intensity

man could be heard from lowest to highest intensity is approximately 1012 W/m2

W/m2 = 1 trillion.

To produce a sound that 2 times greater intensity of sound required approximately 10

fold. For example 10-4 W/m2 intensity noise sounded 2 times louder than the sound intensity 10-5

W/m2. 10-5 W/m2 intensity sounds audible 2 times louder than the sound intensity 10-6 W/m2. 10-4

W/m2 intensity of sound heard 4 times louder than the sound intensity 10 -6 W/m2.

Because the range of intensity which can be detected by the ear is very wide (about 1

trillion W/m2) and hearing the sound loudness does not change directly to the intensity but

approached logarithmic then the sound intensity level expressed as a logarithmic scale.

Mathematically, sound intensity level expressed by the equation:

B =10 log II0

(intensity level equals ten times logarithm times intensity level divided by hearing

threshold ).

Where:

B = intensity level

I0 = 10-12 w/m2 = hearing threshold

I = sound intensity that be to look intensity


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Units of the international system to the level of intensity is the decibel (dB). 10 decibels =

1 bell. The word comes from the name of the bell Alexander Graham Bell (1847 - 1922),

inventor of the telephone.

Of the several examples above questions it appears that each additional intensity by a factor of

10 (example 10-12 to 10-11) the same increase the intensity level of 10 dB. Addition of intensity by

a factor of 100 = 102 (example 10-12 to 10-10) the same increase the intensity level of 20 dB.

Addition of intensity by a factor of 1000 = 103 (example 10-12 to 10-9), the same increase the

intensity level of 30 dB. Thus, sound intensity level of 40 dB, for example, is 10 times as loud or

10 times louder than the sound intensity level of 30 dB, 100 times louder than the sound intensity

level of 20 dB and 100 times louder than the noise level intensity 10 dB.

Here are some of the data intensity and sound intensity level in everyday life.

Source sound intensity level (dB) Description

Hearing Threshold 0 Threshold of hearing

Normal breathing,

rustling leaves 10 Barely audible

Whisper (average) 20 Very quiet

Libraries are quiet,

calm radio softly 40

A quiet office,

a car that did

not sound noisy 50

Normal conversation 65

Bustling traffic 70

Office noise

with ordinary mill machines 80 can damage hearing if its frequency

Trains, heavy trucks 90

Sample question


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1: Determine the level of sound intensity that has an intensity of I = 10-12 W/m2.

Answer:


Answer:


Answer:


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The electric vibrator vibrates the string by creating several variations of wavelength. At

some certain wavelength (), the detected frequency is measured by the frequency meter as value

f. the propagation speed of sound in the string is calculated by using general formula, v =.f.

After doing the experiment, Melde concluded that :

The propagation speed of sound in a string is directly proportional to the square root of the

strings length and the square root of the magnitude of string tension, and it is inversely

proportional to the square root of the strings mass.

The mathematical form of the Meldes experiment result is

( speed of sound waveis the root of force times length divided of mass )

Where :

V = speed of sound wave in the string

m = strings mass (kg)

F=mg = strings tension (N), g =earths gravitational acceleration

l = strings length (m)

The m/l the strings mass per unit length which is symbolized by . Hence, the propagation

speed of sound wave in a string based on the Meldes experiment can be expressed as

V=F

(speed of sound wave is the root of force divided permeability )

Example

V= Flm


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1. A student is performing Meldes experiment. The length of the string used is 1 meter and the

vibrator is producing a pattern of 2 wavelengths. The strings mass every 1 m is 10 gram,

whereas the measured frequency is 60 Hz (g =10 m/s2 ). Calculate the mass of the load attached

to the string.

Answer:

There are 2 waves at 1 m strings, it means that the wavelength or the length of 1 wave is =0.5

m.

With frequency f= 60 Hz, the speed of propagation in string is v=f =(0.5 m) (60 Hz) = 30 m/s

The mass of the string in every 1 meter is 10 grams; in means that the mass density of the string

is

= 10 g / 1 m = 0.01 kg/m.

F= v2 = (0.001 kg/m) (30 m/s)2 =9N

Recall F =mg, we get

m = F/g = (9 N) / (10 m/s2 ) = 0.9 kg.


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C. Doppler Effect

The phenomenon of the Doppler effect when the observer and the sound source moves

relative.

Try to observe the frequency of the bus horn when the bus ran into you and then away

from you. The frequency of the bus horn will enlarge when you close and shrink when away

from you. This event is called the Doppler effect. The frequency of a sound source that moves

relative to the observer will be different from the original source of the sound frequency when

the silence of the observer.

When the sound source approaching the observer, the frequency of observations will behigher than the original frequency sound source. When the sound source away from the observer,

the observed frequency becomes lower than the original frequency source.

At the time of the sound source approaching the observer, the observed wave fronts will

be more meetings. Due to describe the wave front wave peak, mean observed wavelength will

become shorter or increased frequency. Suppose the sound source approaching the observer with

the speed Vs, the observed wavelength is

=- vs T = v-vsfs

( lambda accent is lambda minus velocity times equals

And the frequency of the sound source approaching the observer will be observed for

fp = vv-vs fs

(2.18)


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Conversely, if the source away from the observer, the wave front distance further. The observed

wavelength is

= vs T = v+vsfs

Thus, the frequency of the sound source away from the observer will be observed for

fp = vv+vs fs

(

Where :

V = wave speed

(lambda ) = wavelength

f = frequency

T = period

We already know the frequency of observation when the sound source approaching the

observer. Now it will be shown the frequency of observation when the observer is close to the

source. Suppose an observer approaching a stationary source with velocity Vp. velocity of sound

relative to the observer is V '= V + Vp. then the observer would see two successive wave front at

the interval

t = sv'


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However, because the distance of two wave fronts is , the time required to meet the two waves,

so it is Tp

Tp = t = v'

And the observed frequency is Thus, the observed frequency if the observer approaching a

stationary source is

fp = v+vsv fs

Opposite situation occurs when the observer away from the source. The speed of sound to

the observer is V '= V-Vp. In the same way the observed frequency would be obtained if the

observer away from the stationary source is

fp = v-vsv fs

(2.19)

If equation (2.18) (2.19) combined will produce the original relationships between thefrequency of frequency source with the following observations With fs is the frequency source

fp = v-+vv-+vs fs

With fs is the frequency source, fp is the frequency observed, v is the velocity of sound, v s

moving source speed, vp and speed observer moves.


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Signs for Vp can be determined by

When the observer approached the source, a sign of positive vp.

When the observer away from the source, the negative sign of vp.

When a source close to the observer, a sign of negative vs.

If the source away from the observer, a sign of positive vs.

Sample questions:

1. A police car chase criminals car on the highway. Police cars running at 72 km / h and siren

with a frequency of 5760 Hz. Car criminals move with speed 36 km / hr. If the velocity of sound

in air 340 m / s, determine the frequency of the siren is heard by a passenger in the car criminals

when:

a. Moving police car passed by car criminals

b. Police cars moving in the direction of car criminals where the police car behind the car

criminals

Answer:

fs= 5,760 Hz , vp =36 km/jam, vs = 72 km/jam = 20 m/s, v =340 m/s

a. If the moving police car bumped into the car criminals

Listeners approached source: vp =+10 m/s

Sources close to the listener: vs =- 20 m/s

So that the observed frequency is

fp = v+vpv-vs fs = 340+10340-20x 5,670 = 6,300 Hz


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b. When the police car moving in the direction where the police car criminals are behind a

criminal then

Listeners away from the source: vp =-10 m/s

Sources close to the listener: vs =- 20 m/s

So that the observed frequency is

fp = v-vpv-vs fs = 340-10340-20x 5,670 = 5,940 Hz

ListenRead phonetically

Dictionary - View detailed dictionary
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CHAPTER 3

A. Interference of Light

The interference of light occurs when two (or more) coherent light beams are combined.

In this section we will learn about interference of two coherent lights.

Two beams of light are said to be coherent if they have a constant phase difference. A

destructive interference occurs when the phase difference is 180o. whereas a constructive

interference will occur if both beams have the same phase or their phase difference is zero.

Either destructive or constructive interference can be observed at the interference patterns

formed.

a. Youngs Double Slits Interference

Terms Interference of Light:

he light source must be coherent (Second light source has a different phase, frequencyand amplitude similar)

Thomas Young, a physicist makes two light sources from a single source of light, whichon two narrow slits.

At Youngs experiment, two coherent light sources are obtained from a monochromatic light

passing through two slits. Both coherent light sources will be combined forming some

interference patterns.
http://tienkartina.files.wordpress.com/2010/08/interferensi-7.jpeghttp://tienkartina.files.wordpress.com/2010/08/thomas-young.jpeg


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A maximum interference ( constructive ) indicated by the bright pattern which will occur

if both rays have the same phase. Recall the sinusoidal form of a travelling wave plotted by

its displacement (y) versus the distance (x). Two waves are said to be in phase if their path

difference is zero or an integer multiplication of the wavelength.

One source of light, passing through two narrow slits, so light that passes through both slits,

it is the two new light sources.

The result of interference of two beams / coherent light to produce patterns of light and dark.

mathematical formula to get the pattern of light and dark as follows:

S1 = Source of light

S2 and S3, two new light sources., d = distance between two sources of c

=angle turn, a = l = distance between two sources on the screen

Maximum Interference / light / constructive, occurs when:
http://tienkartina.files.wordpress.com/2010/08/interferensi-141.jpghttp://tienkartina.files.wordpress.com/2010/08/interferensi-141.jpghttp://tienkartina.files.wordpress.com/2010/08/interferensi-12.jpghttp://tienkartina.files.wordpress.com/2010/08/interferensi-6.jpeg


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1. In a Youngs double slit interference experiment, it is known that the distance of the slits

to the screen is 1.5 m and the wavelength used is 4 x 10 -7 m. the distance between the center

brightness and the 3rd bright line is 0.6 cm. Calculate the separation between the two slits.

Answer :

=1.5 m ,=4x10-7 m, y=0.6 cm = 6x10-3 m, n=3

For small angle, the sine value can be approximated by the tangent value. So, sin =tan =

y/ (see the scheme).

The bright lines are obtained when d sin =n . If the sine value is approximated by tangent

value, we get

d tan =n

d =ntan = 3(4x10-7 m)6x10 -3m1.5 m = 3x10-4 m.

Thus, the distance between the two slits, d =3x10-4 m.

CHAPTER 4

A. ELECTRIC FIELD AND ELECTRIC FIELD STRENGTH

1. Electric Field


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Electric field is the area around the electric charge in which other charges are in place in the

area will get a static electric force. electric field can be described by electric lines of force which

is defined as the line where its tangent direction is the direction of electrostatic force on a

positive charge that in place at that point. electric lines of force is often also called electric flux.

In the space surrounding the electrically charged object A, we meet some of the symptoms.

As an example of another charged object B can move away from or close to A (Fig. 1.). These

symptoms are caused by the workings of a force on any charged object placed in the space

around the charged object A. We call the phenomenon in the space around an object of this

electricity were charged electric field.

Figure 1: The forces acting on charges laid in the space around the charged object A

So the electric Field is space around the electrically charged objects in which electrically charged

objects other in this space will feel or an electric style Electric Field Direction.

2. Electric Field Direction
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Electric field can be described with imaginary lines called field lines (or lines of electric

force). You can see in figure 2 and figure 3 that the radial field lines out away from positive

charge and radial into the approaching negative charge

3. Strong Electric Field

Large force experienced by a charge in the electric field depends on the size of that chargeand electric field strength. strong electric field is often called the electric field intensity.

Strong Electric Field is the amount that states coloumb force per unit charge at a point.

direction of electric field strength equal to the electric force on a positive charge.

For example at point P, see the picture.

- If the point P on the given load, the load is called load tester (q), and always positively charged

- Q = Source charge

- Direction of Strong Electric Field (E), the direction of the force (F)

Electric field strength is mathematically formulated:
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(Electric Field is force divided test charge ).

or

( electric force is constant times Source charge times test charge divided distance square ).

Because of Columb force between the charge source Q and the test charge q

are as follows:

( electric force is constant times Source charge times test charge divided distance square ).

by: E = electric field strength (N / C)

Q = charge sources (C)

r = distance test load against load source (m)k = constant = = 9 109 Nm2/C2

0 = electric permittivity of vacuum = 8.85. 10-12 C2/Nm2
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B. Electric Potential Energy

1. Potential Energy

Potential energy is energy associated with forces that depend on the position or shape ofobjects and their environment. There are so many examples of potential energy in our lives. We

stretch the rubber slingshots have potential energy. Rubber slingshots to catapult stone because

of the potential energy in a stretched rubber. Similarly, the arc drawn by the archer can move the

arrows, because there is the potential energy in a stretched bow. Another example that springs is

pressed or stretched. The potential energy in the three samples is called elastic potential synergy.

Chemical energy in the foods we eat or chemical energy in fuels also include potential energy

2. Gravitational Potential Energy

The most common example of potential energy isgravitational potential energy.

Gravitational potential energy possessed by the object position relative to the earth. Each

object has a gravitational potential energy to do work if the object is moving toward Earth's

surface (example mango fruit fell from the tree).

Now let us define the gravitational potential energy of an object near the earth's surface..

Suppose we lift a stone of mass m. lift force that we give in stone at least equal to gravity acting

on the rock, namely mg (mass times the acceleration of gravity). To lift the stone from the soil

surface until it reaches a height h, then we have to do business as long as a product stone gravity

(W = mg) with height h. Remember, we lift direction parallel to the direction of movement of the

stone, ie to the top ...FA = lift force

- equation 1

(Work is lift force, work equals speed equals mass times minus gravity times speed equals minus

mass times the acceleration of gravity times high two minus high one ).

The negative sign indicates that the direction of the acceleration of gravity toward the bottom ...

W = FA .W = FA.s = (m)(-g) (s) = mg(h 2 -h 1 )


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Thus, the gravitational potential energy of an object is a product of gravity object (mg) and

height (h). H = h 2 - h 1 (height is two height minus one height ).

- equation 2

( potential energy equals mass times gravity times height ).

Based on the EP equation above, it appears that the higher (h) objects above ground level,

the greater the EP who owned the object. EP gravity depends on the vertical distance or height of

objects above a specific reference point. Usually we set the ground as a reference point whenthings began to move from ground level or movement of objects toward the ground surface.

If we combine equation 1 with equation 2:

(work is minus mass times gravity times two height minus one height ).

(work is minus two Potential energy minus one Potential energy).

(work is minus delta Potential energy).

EP = mgh


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This equation states that the work done by the force in moving objects from h 1 to h 2

(without acceleration) equal to the change in potential energy of objects between h 1 and h 2. Any

form of potential energy has a relationship with a particular style and can be expressed together

with the EP gravity. In general, changes in EP who has a relationship with a particular style,

similar to the style of the work done if the object was moved from first position to second

position. In a narrower sense, can be stated that the change in EP is the effort required by an

external force to move objects between two points, without acceleration.

So the work done W = Potential energy accretion.

3. Electrical Potential Energy

Electrical Potential Energy is the work done Coulomb force, to move the test charge + q

'from one point to another. If the point Q, located at the far infinity, so that r2 = ~ and the 1/r2 = 0

then the Electrical Potential Energy can be formulated as follows: Electrical Potential Energy of

two charges Q and q 'are:

(Potential energy equals constant times charge source times test charge divided distance).

EP including a scalar quantity

E = Electric Potential Energy Joule unit

k = constant = 9109 N m2 C-2, r = distance (m)

Q + charge source, q '= test charge (Coulomb)

4. Electrical Potential (V)

Electric potential is the potential energy per unit test charge, electric potential as the

following formula: V = Ep / q 'or as in the picture below

Ep = k Q.q/r,
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Electric potential at point P is formulated:

V = k Q / r

(Electric Potential is Constant times charge source divided distance from charge to point P ).

V = Electric Potential (Volt)

k = Constant Power = 9109 NC-2 m2

Q = charge source (Coulomb)

r = distance from charge to point P

Sample question

1.Mango fruit is ripe and inviting sense of hanging on a mango tree stalk that is 10 meters from

ground level. If the mass is 0.2 kg of mangoes, what is its potential energy?. let's say the

acceleration of gravity is 10 m / s 2.

answers:

EP = mgh

EP = (0.2 kg) (10 m / s 2) (10 m)

EP = 20 kg m 2 / s 2 = 20 Nm = 20 Joule

2.A mass of 5 kg monkey swinging from one branch to another branch 2 meters higher. What is

the potential energy changes to the monkey? g = 10 m/s 2 g = 10 m / s 2
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answers:

EP = mgh = (5 kg) (10 m / s 2) (2 m)

EP = 100 Joule

Thus, the potential energy changes monkeys = 100 Joules.

3.A stevedore is 1.50 meters tall lift a sack of rice, 50 kg mass from the surface and giving it to a

friend who was standing on the ship. If the person is located 0.5 meters just above the head

stevedore, calculate the potential energy relative to the sacks of rice:

a) surface soil

b) the head stevedore

answers:

a). EP sacks of rice relative to the surface soil

Altitude total sacks of rice from ground = 1.5 m + 0.5 m = 2 meters

Thus,

EP = mgh = (50 kg) (10 m / s 2) (2 m)

EP = 1000 Joule

b). EP sacks of rice relative to the head stevedore

The position of sacks of rice were measured from the head stevedore is 0.5 meters.

EP = mgh = (50 kg) (10 m / s 2) (0.5 m)

EP = 250 Joule


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C. CAPASITORS

Capacitor or condenser are electrical components that are used to store electrical

charge .And consists of two conductors separated by insulating material (dielectric material),

each conductor called a puck. Symbols used to display a capacitor in an electrical circuit is

In normal use, one piece given a positive charge and the chips were given an equal

negative charge. Between these two pieces are created by an electric field directed toward the

chip to

chip negative positive.

Capacitors are useful for:

1. choose the radio frequency receiver

2. filters in the power supply (power supply).

3. extinguish the fire flower on the car ignition system

4. save energy in electronic ignition circuit.


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In practice there are various types of capacitors, including paper capacitors, electrolytic

capacitors, and variable capacitor, such as the image above.

1. Capacity parallel chip capacitors

Capacitance or capacity is a measure of the amount of electric charge stored (or separated) to

a predetermined electric potential. The most common form of charge storage device is a

capacitor two plates / plate / chip. If the charges on the plate / plate / chip is + Q and-Q, and V is

the voltage between the plates, then the capacitance is:

(capacity of the capacitor equals electric charge divided of volt ).

C = capacity of the capacitor unit in the SI (farad abbreviated F), 1 farad = 1 coulomb / volt.

Other units of F (microfarad) 1 F = 10-6 F

Q = electric charge Coulomb units, and V = potential difference units Volt.
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o = vacuum permittivity

A= Chip cross-sectional area (m2)

d = distance between the two pieces (m)

2. Energy stored in capacitor

Capacitors store energy in the form of an electric field, energy stored in the capacitor (potential

energy) W expressed by:

(Energy stored in capacitor is half electric charge stored in capacitor square divided Capacity is

half Capacity of unit capacitor times of volt is half Capacity times volt square ).

W = Energy stored in capacitor unit joule (J)

Q = electric charge stored in capacitor unit Coulomb (C)

C = Capacity of unit capacitor farad (F)

V = potential difference between two pieces of unit volt (V)

Example Problem

1. A 300 F capacitor is connected to a battery 50 Volt, keeping Determine the magnitude

of charge on-chip capacitors.

Completion:

Given: C = 300 x 10-6 M, V = 50 V

Asked: q = ....?
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Answer: C = q / C q = CV = (3, 00 X 10-4) (50) = 1.5 x 10-2 C or 15 MC

2. A chip capacitor has a capacity of 1.3 F when loaded and 6.5 x 10-7 C

between the two pieces there are strong magnetic fields 200 NC-1. What is the distance

between keeping the second.

Given : C = 1.3 x 10-6 M, q = 6.5 x 10-7 C, and E = 200 NC-1

Asked: d = ....?

Answer: V = E. d d = V / E, from the equation C = q / V V = q / C

d = q / CE = q = 6.5 x 10-7 / (1.3 x 10-6) (200) = 2.5 x 10-3 m or 0.25 cm

3. Series Capacitors

1. Series circuit

Two or more capacitors can be arranged in series with the ends which are connected

sequentially-connected as shown below

In this series the charge stored on the capacitor will be equal, so Q total equal to the

charge on the capacitor 1, capacitor 2 and capacitor 3, consequently the potential difference of

each capacitor will be inversely proportional to the capacity of capacitors, in accordance with the

equation

Q = CV
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(electric charge is Capacity times of volt )

In the series circuit potential difference = voltage source = total voltage E = V tot, will be

divided into three parts. This explanation can be inferred from the properties owned by the

following series:

a. Q total = Q1 = Q2 = Q3 ( total electric charge is one electric charge is two electric charge is

three electric charge ).

b. E = Vtot = V1 + V2 + V3 (potential difference equals total voltage is one voltage is two

voltage is three voltage ).

c. 1/Cs = 1/C1 + 1/C2 + 1/C3 (one per capacitor series equals one per capacitor one plus one per

capacitor two plus one per capacitor three).

2. Parallel circuit

Parallel circuit is a combination of two capacitors or more with the same poles together as shown

below

In this circuit a potential difference of the ends of the capacitor will be the same because

the same position. As a result, the charge stored is proportional to the capacitors. Total charge

stored equal to the total. From his description it can be concluded owned properties parallel as

follows:
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a. Q total = Q1 + Q2 + Q3

( total electric charge is one electric charge plus two electric charge plus three electric charge ).

b. E = V total = V1 + V2 + V3

(potential difference equals total voltage is one voltage plus two voltage plus three voltage ).

c. Cp = C1 + C2 + C3

(parallel capacitor is one capacitor plus two capacitor plus three capacitor ).

CHAPTER 5

Magnetic Field

A magnet is always so having two poles, namely the one called the north pole (U) / (N = North),

the other thesouth pole (S) /(S = South), as in the types of magnets following picture:

In the vicinity of the magnet have magnetic fields that affect other magnets and objects made of

certain materials. Magnetic fieldis the space around the magnet so that other magnets are still

having style. magnet Space around the magnet can be described as an imaginary line as the

magnetic field lines that direction out of the north magnetic pole and into the south magneticpole
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Similar pole, north pole by north pole or south pole south poles repeleach other. Twopoles are

not similar, with the north pole south pole will attract each other, as shown below:

Our Earth, it also has a magnetic field like in the picture below
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Magnetic field not only caused by permanent magnets. The fact indicates that the permanent

magnets are not the only source of magnetic field. This was first discovered by Hans Christian

Oersted in 1820.

1. Magnetic induction (B) in the vicinity of electricity have current Conductor

Consider the following experiment pictures ..!

Figure (a-1) wire is not electrified, so the magnetic needle does not deviate.

Figure (a-2). Electrified wire with a downward direction, the magnetic needle to the right.

Figure (a-3). Electrified wire with upward direction, the magnetic needle to deviate to the left.

Conclusion Oersted experiment, that in the vicinity of the electrified wire have a magnetic field.

To determine the direction of the magnetic field (B), using right hand rule as shown below!

1. thumb indicates the direction of flow (i)

2. Fourth finger indicates the direction of the magnetic field (B)

To determine the value or formula of the magnetic induction

around the fast-flowing electric wire as in the picture below
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( magnetic induction at point p equals vacuum permeability times strong electric current divided

two times phi times wire radius circle ).

Bp = magnetic induction at point p (Wb/m2 =Tesla=T )

i =strong electric current (Ampere=A)

a =wire radius circle (m)

0 =vacuum permeability (4.10-7 W b/Am )

2. Magnetic induction (B) around the fast-flowing circular electrical wire.

The picture. A circular wire later electrified, then there is a

circular wire axis magnetic field direction as the picture. The direction of the magnetic field

using the right hand rule:

1. thumb indicates the direction of the magnetic field (B)

2. finger indicates the direction of the circular flow (i)
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(magnetic induction at point p equals vacuum permeability times strong electric current times

angle between distance x and r sin divided two times distance of point p to the wire square ).

Bp = magnetic induction at point p (W b/m

2

)i =strong electric current (Ampere=A)

a =wire radius circle (m)

r =distance of point p to the wire (m)

=angle between distance x and r sin

If point P is on the center circle, then the formula becomes

( magnetic induction equals vacuum permeability times strong

electric current divided two times wire radius circle ).

Example Problem :

1. compute the magnetic field strength at the point, a distance of 6 cm length of wire that

carries current A

Answer :

B =0 I2r = 4x10-7 x 92 x x 0.06= 30 x 10-6 T
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so, the magnetic field at a point 6 cm away from the wire is 30 T.

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