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Many persons get confused when given problems to solve for the terminal voltage or pd in a circuit mostly because it sounds the same as the E.M.F / total voltage of the circuit and so often loses a point or the entire question for this error. Here we will distinguish between the e.m.f and terminal voltage or p.d of a cell and show you examples of terminal voltage.

What is E.M.F?
E.m.f or electromotive force is the work or energy per unit charge that an active current producing device produces to drive current through a circuit.
In simpler terms a source that produces its own voltage has e.m.f.

What is Terminal Voltage?
Terminal voltage is the voltage / potential difference across the terminals of a cell or battery.

If the cell/battery is not connected to circuit then the terminal voltage/p.d is equal to the e.m.f of the cell or battery.
If the cell or battery is connected to a circuit then the terminal voltage/p.d is equal to the total voltage drop in the external circuit.

Cell/Battery not connected to a Circuit

Terminal P.D = E.M.F




Cell/Battery connected to a Circuit



Formula for Terminal Voltage when connected to a circuit:
EMF = v +V
EMF = Ir + IR
Therefore:
Terminal P.D = IR
Where:
v is voltage within the cell
V s voltage in external circuit
r is internal resistance
R is resistance in external circuit
I is current 
Don’t be fooled into thinking the terminal P.d would be Ir by looking at the diagram above the terminal voltage is the same as the voltage in external circuit in this case, this is because the cell is connected to a circuit and not by itself as is the case with the first diagram.
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By definition momentum is the product formed between the mass and velocity of a body. Therefore any object that has mass and velocity is capable of undergoing some form of momentum. In everyday life we have all experienced momentum at some point, a simple bouncing ball, crashing trollies or just any form of collision has some form of momentum that is taking place. Momentum falls under the Vector quantity category meaning it has both magnitude as well as direction, the opposite of vector quantity is scalar quantity; these quantities have only magnitude and they lack direction. The SI unit of momentum is kg m/s.

Formula for Momentum:


Momentum = Mass x Velocity


Here we will be using the formula above as well as the rule “Momentum before collision is equal to the momentum after collision” to solve problems involving momentum.

        Momentum Before Collision = Momentum After Collision
                      M1 x U1 + M2 x U2 = M1 x V1 + M2 x V2

    Where:

  • M1 & M2 represents mass of the two bodies colliding.
  • U1 & U2 represents the velocities of the bodies before collision.
  • V1 & V2 represents the velocities of the bodies after collision.

Problem # 1: - Finding Mass (Elastic Collision)

A car of mass 500 kg is moving at 8m/s when it hits another car. The first car stops moving after which the second car starts moving at a speed of 15m/s. What is the mass of the second car?

Answer:
Using the formula above we can easily solve for the mass of the second car:

Momentum Before Collision = Momentum After Collision
          M1 x U1 + M2 x U2 = M1 x V1 + M2 x V2


Where:

  • M1 & M2 is the mass of the first and second car respectively.
  • U1 & U2 is the velocity before collision for the first and second car respectively.
  • V1 & V2 is the velocity after collision for the first and second car respectively.


Now that we have our formula all that is left is to plug in the values and solve for the mass M2.

M1 x U1 + M2 x U2 = M1 x V1 + M2 x V2
     500 x 8 + M2 x 0 = 500 x 0 + M2 x 15
                  4000 + 0 = 0 + M2 x 15
                 4000 / 15 = M2
                   266.7kg = M2



Problem #2 –Finding Common Velocity (Inelastic Collision)

A bullet of mass 0.25kg travelling at 150km/s strikes a stationary block of mass 3kg. What is the common velocity of the bullet and block after collision?

Answer:
The first step to solving the problem is to try and find out exactly what they are asking for. The question stated that you need to find the common velocity of the bullet and the block after collision. This simply means that the objects remain together after collision and so have a common velocity. This is known as an inelastic collision, when given problems involving inelastic collisions use the formula below to solve instead of the general formula given above:

M1 x U1 + M2 x U2 = (M1 + M2) V

    Where V is the common velocity.

  • In this case M1 would be the mass of the bullet.
  • U1 is velocity of the bullet
  • M2 is mass of block
  • U2 is velocity of Block (would be zero since it is stationary)


The next step would be to plug in the values for both objects into the formula and then solve:

M1 x U1 + M2 x U2 = (M1 + M2) V
   0.25 x 150 + 3 x 0 = (0.25 + 3) V
                   37.5 + 0 = 3.25 x V


Transpose for V
37.5 + 0 = 3.25 x V
37.5/3.25=(3.25 x V) / 3.25
           V = 11.54m/s




Problem #3 -Finding Recoil Velocity

A gun of mass 4 kg fires a bullet of mass 0.15kg at 250 m/s. What is the recoil velocity of the gun after?

Answer
Using the principle of conservation of Momentum:

Momentum Before = Momentum After
                          0 = M1 x V1 – M2 x V2
              M2 x V2 = M1 x V1

  • Where M1 is the mass of the bullet
  • V1 is the velocity of the bullet
  • M2 is the mass of the gun
  • V2 is the recoil velocity of the gun

M2 x V2 = M1 x V1
   4 x V2 = 0.15 x 250
         V2 = 37.5/4
         V2 = 9.38m/s
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Assuming that we all already have some knowledge of projectile motion, we’ll not be going over much of the details involved but to focus mainly on problems involving projectile motion. If you need to read some more on projectile motion then please click here.

In previous articles we said that a body that is in projectile motion will move in two dimensions simultaneously. These bodies are resolved into their perpendicular components which includes the vertical component Vy and the horizontal component Vx. We also said that they must be dealt with simultaneously. A body in projectile motion will look somewhat like the diagram below:


To be able to solve projectile motion problems you first need to know how to resolve vectors, to read more about resolving vectors click here.

Below are some formulas associated with projectile motion, we’ll be using them as we work through the questions below. Remember that objects that possess projectile motion moves in two dimensions simultaneously (horizontal and vertical), therefore we will have formulas for both the vertical component as well as the horizontal component. These are treated separately.

Projectile Formulas for Vertical Motion:

Uy=U×sinθ
Vy = Uy + at

Where:
  • Uy represents the initial velocity of the vertical component.
  • U represents initial velocity
  • θ represents the angle formed with the horizontal.
  • Vy represents the final velocity of the vertical component.
  • a and t represents acceleration and time respectively.

Projectile Formulas for Horizontal Motion:

Ux=U×cosθ
Vx = Ux + at

Where:
  • Ux represents the initial velocity of the horizontal component.
  • U represents initial velocity
  • θ represents the angle formed with the horizontal.
  • Vx represents the final velocity of the horizontal component.
  • a and t represents acceleration and time respectively.

Note: For objects moving with projectile motion in the horizontal direction you will find that initial velocity of the horizontal component and the final velocity of the horizontal component are equal this is because there is no acceleration taking place along the horizontal.

The equations stated above for both vertical and horizontal isn’t the complete list of equations that can be used but these are the ones we will be using during our calculations here. If you look at the second equation in both cases you will realize that those equations are similar to the equations of motion. Therefore the equations that are not stated above are also equations derived from the equations of motions, all you need to do is to replace the velocities with these involved in projectile motion.

Other general Formulas:







Where:

  • V represents velocity. Use this formula in cases where they ask for the velocity or magnitude.
  • θ is the angle
  • The last formula is usually used to find the direction.
Problem # 1

A particle is projected with a speed of 25m/s at 300 above the horizontal:

    a) Find the time taken to reach the highest point of the trajectory.
    b) Find the magnitude and direction of the velocity after 2 seconds.


Answers


     a) The first part of the question asked for the time taken to reach the highest point. To make our calculations easier we’ll work out our answer through steps.

Step 1
Write down all the variables that are given. This way you will have easier access to them when you’re ready to use them in calculations also this helps you to choose the best possible formula for the question.
  • Θ = 300
  • U = 25m/s
  • Uy which is the initial velocity is = ???
  • a = -g = -10m/s2, this is because the particle is moving upward against gravity.
  • Vy = 0. In this part of the question the final velocity is at the highest point and at the highest point the particle stops moving in the vertical direction therefore the final velocity of the vertical component, Vy, is zero.
  • t =???


Step 2
The next step would be to choose the best possible formula to solve the problem. Looking at the variables we have and what we want to find (time) the best possible formula would then be:

                Vy = Uy + at

Where a is acceleration and t is time.

We already know the values for most of the variables above except the time, t, which we are trying to find and the value of one other variable that is needed to solve for time, this variable is Uy, which is the initial velocity.

To solve for the initial velocity, Uy, we simply use the formula:

Uy=U×sinθ

After filling in the values in the above equation and solve for Uy your answer should know look something like this:

Uy=U×sinθ
Uy=25×sin300
Uy=12.5m/s


Step 3
Now that we have all the values we need as well as the formula, we can now go ahead and solve for time (t):

Transpose to make time the subject


   b) This part of the question asked for the magnitude and the direction of the velocity after 2 seconds. To make our calculations allot easier we will split this question into two. First we will find the magnitude and then we’ll find the direction.

To find the magnitude we will use the formula:


If you look at the equation you’ll realize that we can’t solve for the magnitude right away because we don’t have the values for Vx or Vy. We can’t use the value we got for Vy in part (a) of the question here because that velocity was the finally velocity of the vertical when the object reached maximum height(which took 1.25seconds) but in this case we have already passed the maximum height so there’s a new final velocity for the vertical which is after 2 seconds. The initial velocity for the vertical however will remain the same.


Finding Final Velocity for Vertical, Vy:
Uy = 12.5m/s
a = -10
t = 2 sec
Vy =?

Using the formula Vy = Uy + at:

Vy = Uy + at
     = 12.5 + -10 x 2
     = -7.5m/s

Finding final velocity for Horizontal, Vx:
Remember that the Initial and final velocities for the horizontal are the same, so you can use the formula below to find the initial velocity of the horizontal, Ux, which will also be the answer for the final velocity of the horizontal, Vx:

Ux = U×cosθ
     = 25 x cos 30
     = 21.65m/s
Therefore Ux = Vx = 21.65m/s

Now that we have found both Vy and Vx we can go ahead and solve for the magnitude:


Now for the second part of the question, we must find the direction of the velocity after 2 seconds. We can use the following formula to find the direction:

Just substitute the values for Vy and Vx into the equation and then solve for the angle.
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The new cape physics unit 1 e-book is now out and you can download it here for free. This e-book contains some of the most popular topics in physics. It was created based on the cape syllabus which is also compatible with other major syllabuses such as the A-Level syllabus. It was created mainly for students going through their first year of college. It features the common cape/a-level notes most containing images and a complete guide to help you a bit more as you read. This is the first edition so not all topics were covered but you can always keep an eye out for the second edition where you’ll find all the notes you need. There are also example questions with answers in areas where most students may find physics difficult.

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Below are just some of the topics covered in this e-book:

  • Scalar and vector Quantities
  • Vectors
  • Newton’s Laws of Motion
  • Circular Motion
  • Simple Harmonic Motion
  • Etc…….


Preview



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In previous articles we discussed newton’s laws of motion and how they apply to life-like situations. In this article we’ll be using formulas/equations from the laws of motion and the laws themselves to solve various problems involving reaction force. By the end of this tutorial you should be able to:

  1. Find the reaction force in a system
  2. Apply newton’s laws of motion to various questions
  3. Find missing variables such as acceleration when given values for other variables.


Problem #1
A body of mass 7kg rest on a floor lift. Calculate the reaction force R exerted by the floor of the lift on the body if:

     a) The lift has an upward acceleration of 2m/s2.
     b) The lift has a downward acceleration of 3m/s2.
     c) The lift is moving with constant velocity.

Answers
The first thing that needs to be done is to draw a diagram to represent the information in the question, doing this makes your calculations allot easier and reduces risk of errors. In questions like these where there’s a mass there is always a weight acting downward from the body; so the first thing to be drawn on the diagram is the weight (W) acting downward. Next is the reaction force(R), the reaction force always acts perpendicularly to the body. When done your diagram should look something like this:



     a) This part of the question asked for the reaction force when the lift has an upward acceleration of 2m/s2.




When forces are moving in opposite directions you must subtract the smaller from the larger. In this question the reaction force is larger than the weight because the lift is moving in the upward direction.
We’ll use the formula derived from newton’s laws of motion:

Force = Mass x Acceleration
Your answer should know look something like this:

F = Mass × acceleration
R – W = Mass × acceleration
R – 70 = 7 x 2
R = 14 + 70
R = 84N


 Note: This is how the weight was found
Weight = mass x gravity
Weight = 7 x 10
Weight = 70N
     b) This part of the question is very similar to the one previously done. The question asked for the reaction force when given a downward acceleration of 3m/s2. In part “a” of the question the reaction force was larger because the lift was moving in an upward direction; therefore it also follows that the opposite should happen in this part of the question since the lift is now moving in the downward direction. Therefore the weight is larger than the reaction force.



F = mass x acceleration
70 – R = Mass x acceleration
70 – R= 7 x 3
70 – R = 21
70 – 21 = R
49N = R

     c) In this part of the problem you’re required to find the reaction force given that the lift is moving at a constant velocity. Since it is moving at constant velocity acceleration would be zero and no motion is taking place.

F = 0
R – W = 0
R – 70 = 0
R = 70N


Note:  
Here the reaction force is equal to the weight

Problem # 2
A body of mass 5kg is pulled up a smooth plane inclined at 300 to the horizontal by a force of 40N acting parallel to the plane. Calculate the acceleration of the body and the force exerted on it by the plane.



Answers
Before we begin calculating some other components must be added to the diagram. Remember in the previous question we said that where ever there’s a mass there is weight acting downwards. Also the weight must be resolved/split into its perpendicular components with respect to the plane. One of the components of weight is parallel to the plane while the other is perpendicular to the plane. Finally we must now add the reaction force. Remember that the reaction force must be perpendicular to the plane of the body.


You may have noticed that another angle was added to the diagram, 600, this was needed in order to find the perpendicular components of the weight. If you look at the diagram closely you’ll see that the line for the weight cuts the base of the triangle forming a 900 angle. Therefore the 60 degrees was found by subtracting the sum of the 90 degrees and 30 degrees from 180 giving 60 degrees. Because the sum of all angles in a triangle add up to 180.

Now that we have labeled our diagram we can now begin answering the question. The first part asked for the acceleration of the body.

Using newton’s second law
F = Mass x Acceleration (a)
40 – W x cos60 = 5 x a
40 – 50 x cos60 = 5 x a
40 – 50(0.5) = 5 x a
40 – 25 = 5 x a
15 = 5 x a
15/5 = a
3m/s2 = a

Note:

Remember that forces in opposite direction must be subtracted.
The next part of the question asked for the force exerted on the body by the plane, the force exerted is the reaction force. No motion is taking place in this direction therefore:

Using Newton’s first law:
F = 0
R – W x Sin60 = 0
R = W x Sin60
R = 50(0.8660)
R = 43.30N

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In previous articles we covered the topic of resolving vectors, mainly the headings: how to resolve a vector, definitions and a complete guide on how to go about splitting/resolving a vector into its perpendicular components and how to label them, in this article we will look at some examples on how to resolve vectors.

To read more on resolving vectors click here

Question # 1

Calculate the horizontal and vertical components of a 50N force which is acting 40 degrees to the horizontal.

vector
Vector


Answer: 


The question asked for the values of the horizontal and vertical components, so first you need to split the vector seen in the diagram above into its horizontal and vertical components:


resolved vector
Resolved Vector


Now that you have identified both components the next thing that needs to be done is to label the components. Look at the diagram above and find out which of the components is adjacent to the given angle and label it “F×cosθ”, where F is the force and θ is the given angle. Also find out which component is opposite to the given angle and label it “F×sinθ”. In this example the component adjacent to the given angle is the horizontal component so it is labeled 50×cos 40 degrees. The component opposite to the given angle is the vertical component so it is labeled 50×sin40 degrees.


Resolved Vector
Resolved Vector



Therefore your answer should be:

Horizontal component = 50×cos40
                                  = 38.30N

Vertical Component = 50×sin40
                               = 32.14N
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What is a Lens?
A lens can be a plastic or a glass or formed from a combination of these two such that when light passes through it the light rays get refracted.
Light rays from an object present on one side of a lens, when refracted can form an image on the other side of the lens.

Types of lens

There are two known types of lens these are:
  1. A convex/converging lens
  2. A concave/diverging lens


Convex Lens
A convex lens is a type of lens that bends/converges light in, it is also known as converging lens.
The diagram below shows how light is refracted through a convex lens.

Convex Lens



Concave Lens
A concave lens also known as a diverging lens is one that diverges/ bends light outwards.
The diagram below shows the effect on rays being refracted through a concave lens.

Concave Lens



If you look at both convex and concave lens you would notice a difference in their structures. A convex lens has more thickness in its middle than at its edges; this is what aids in converging/ bending light inward. However the concave lens is thicker at its edges than at the middle, which helps to diverge light.


“F” in both diagrams represents focal point. This is the point in a converging lens where the rays converge to and this is the point in a diverging lens such that if the rays where to be traced backward they would meet at this point.
“P” represents the principle axis. This is the imaginary line through the center of the lens.
“C” represents the optical center.
“f” represents the focal length. This is the distance measured from the mirror to the focal point.



Difference between Objects and Images


The object is the body from which the rays of light are emitted before they are refracted. The image is what is formed as a result of the rays being refracted through the lens, they can be seen where the rays of light meet or appear to meet.

Real and Virtual Images

Previously we said that images are formed when rays are refracted through a lens and also that they are formed where the rays of light meet or appear to meet. Because some images refracted actually meet at a point while some really don’t, with this fact we can place images formed into two categories:
  1. Real Images
  2. And Virtual Images
Real images are those that are formed where light rays actually meet after being refracted. These can be captured on a photographic film.
Virtual images, on the other hand, are those that are formed where light rays appears to meet but really don’t after being refracted.


What is Magnification?

The magnification is the ratio of the size of the image to the size of the corresponding object.

The formula below shows the relationship between the magnification, height of the image and height of the object:

Magnification Formula



The formula below shows the relationship between the magnification, distance of image from the lens and distance of object from the lens:

Magnification Formula



Real, Magnified and Inverted


In the example above the image formed is real, magnified and inverted.

Virtual, magnified and erect



Principle of Reversibility

The principle of reversibility states that a ray of light that takes a certain path from A to B will also take the same path from B to A.

The formula below relates focal length, image distance and object distance:

Focal length Formula

The following formula is used to find power of a lens:

Power of a Lens


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What is Diffraction?
By definition diffraction is the spreading out or scattering of waves as they pass through gaps. However certain conditions must be present for a wave to be diffracted when passing through a gap, the length of the wave and the size of the gap has to be the same or almost the same.








What is Interference?
Interference is what occurs when two or more waves interact with each other producing another wave.

Types of Interference?
There are two types of interference these are:

  1. Constructive Interference
  2. Destructive Interference

Constructive Interference
In constructive interference, when the waves interfere a larger wave is produced, this is as a result of the waves meeting at in-phase points such as a crest meeting a crest or a trough meeting a trough.



Destructive Interference
In destructive interference the two original waves will meet to produce a smaller wave. The final wave produced is as a result of the original waves meeting at out of phase points such as a crest meeting a trough. The resulting wave can also be non-existent.


















Diffraction – Interference Fringes























In the diagram above diffraction occurs at slits s1 and s2, the diffracted waves then interfere. The solid lines in the diagram represent crests; therefore in the diagram above the waves interfere at in-phase points (crest to crest) producing bright fringes or bands. In this diagram constructive interference takes place.
























In the diagram above diffraction occurs at slits s1 and s2, the diffracted waves then interfere producing dark bands on the screen. This is known as destructive interference and occurs as a result of the waves interfering at out of phase points (crest to trough), the crest is represented by the solid line while the trough is represented by the broken line.




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In some waves the vibrations can be observed moving throughout the material but in others the vibrations/ disturbances can appear to be still or not present in the material; with these facts scientists were able to categorize waves based on these differing characteristics. These are:

  1. Progressive Waves
  2. Standing/Stationary Waves

Progressive Waves
Progressive waves are those waves having visible disturbances meaning the vibrations can be seen moving throughout the material.

Standing/Stationary Waves
Standing waves are produced when two waves interfere this results in an invisible disturbance though the wave is moving throughout the material. Standing waves consist of two different points known as nodes and antinodes.

Nodes: These are points in the wave that have zero energy and displacement.
Anti-Nodes: These are points in the wave that have the highest energy or amplitude.

The distance between two consecutive nodes or anti-nodes makes half a wavelength.

Waves formed from stringed and wind instruments are stationary waves. Open ends always have an antinode while closed ends have a node.

Stationary wave in Wind Tube

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What are In-Phase Points?
Inphase points are points of a wave that are located at the same position in the wave cycle. Waves that are inphase are separated by an even multiple of half wavelength (π).Such waves will undergo constructive interference.

Inphase

The x’s in the diagram above marks points that are in-phase in the wave cycle.

What are Out of Phase Points?
Out of phase points are points that are not located in the same position in the wave cycle. Waves that are out of phase are usually separated by an odd multiple of half wavelengths (π). These waves undergo destructive interference.




Looking at the diagram above we are going to determine which are in phase against those that are out of phase.

  1. π to 3π would be inphase because the difference is an even number 2π.
  2. 2π to 4π would be inphase because the difference is also an even number 2π.
  3. 2π to 6π would be inphase because the difference is an even number 4π.
  4. 1π to 4π would be out of phase because it is separated by an odd number 3π.


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What is Longitudinal Wave?
A longitudinal wave is one in which the particles in the wave vibrate parallel to the direction in which the wave is moving. That is the particles can only move in a back or forth movement in the same direction as the wave.

Examples of Longitudinal Waves:

  1. Sound Waves
  2. Waves in Springs
  3. Vibrations in Gases
  4. Waves in Tsunami
Longitudinal wave



What is Transverse Wave?
A transverse wave is one in which the particle vibrates perpendicularly to the direction in which the wave is moving. Such waves can only move in an up or down direction with respect to the direction of the wave.

Examples of Transverse Waves?

  1. Water Waves
  2. Waves in Ropes
  3. Electromagnetic waves
  4. Guitar string that is Vibrating
Transverse Wave

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Wave
via ScienceTm
In previous lessens we looked at what is a wave. If you’re unclear on this topic click here to read more about waves.
In physics there are two known types of waves, these are:
  1. Mechanical Wave
  2. Electromagnetic wave


The differences between these two lie in their means of transferring energy. One may not see the difference between sound waves and light rays because as humans we aren’t capable of seeing such things with our naked eyes but thanks to human intelligence and through many research scientists were able to detect movement of energy in waves and thus differentiate the means by which they are transferred.

What are Mechanical Waves?
Mechanical waves are those that need a material medium with particles to enable them to move. Therefore these waves can’t “travel on their own”.

Examples of mechanical waves:

  • Sound Waves
  • Water Waves
  • Waves in springs


What is Electromagnetic wave?
Electromagnetic waves are those that do not need any material medium to enable them to move. Therefore these waves can travel on their own. These waves can travel through a vacuum.

Examples of electromagnetic waves:

  • Visible light
  • X-rays
  • Infrared rays
  • Ultraviolet rays
  • Radio waves
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Earth Image
via wikipedia

What is Gravitational Field Strength?

The gravitational field strength at any point can be defined as the force per unit mass placed at the point in the field.
Formula for Gravitational field strength:

Gravitational Field Strength

What is Geostationary Orbit?
Geostationary orbit refers to a body orbiting a fixed position above the earth. For example a satellite moving with geostationary orbit around the earth has to have the same period of rotation as the earth (24hrs) therefore allowing it to remain above the same geographic area on the earth.

What is Escape Velocity?
Escape velocity refers to the minimum velocity a body needs to escape the gravitational pull of the earth.

Formula for escape velocity:

Escape velocity

Where:
V is escape velocity
G is universal gravitational constant
me is mass of the earth
re is radius of the earth
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In previous lessons we spoke about simple harmonic motion, its definition, systems that engage in simple harmonic motion and formulas involved. Now we’ll be looking at calculating Simple harmonic motion questions.

Example # 1

S.H.M

In the diagram above a particle is moving with S.H.M with a period of 24 seconds between two points A and B. Find the time taken to travel from:
a) A to B
b) O to B
c) O to C
d) D to E

Answers:
Note that the period is the time taken to complete one revolution in this case the period is the time taken for the particle to travel from A to B and then to A again (24 seconds as stated in the question above).

a) A to B
A to B is half a period which would be = 12 second

If you want to work this question in a more detailed way, you can do this:

Distance from A to B to A again = 16m
Distance from A to B = 8m
Period (A to B to A again) = 24 seconds
Time (t) from A to B =???



b) O to B
O to B is quarter of the period so the answer would be = 6 sec

Or you can follow the above pattern:

Distance from A to B to A again = 16m
Distance from O to B = 4m
Period (A to B to A again) = 24 seconds
Time (t) from O to B =???












c) O to C
This is one eight of the period which would be = 3 seconds

Or by detailed calculation:

Distance from A to B to A again = 16m
Distance from O to C = 2m
Period (A to B to A again) = 24 seconds
Time (t) from O to C =???













d) D to E
Note with this question, if you observe the diagram carefully you will see that you weren’t given the distance between D to E directly. However this can easily be found by subtracting the distance O to D from the distance O to E and this would give you the distance from D to E.

Distance from A to B to A again = 16m
Distance from D to E = 3.5 – 3 = 0.5m
Period (A to B to A again) = 24 seconds
Time (t) from D to E =???












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What is Simple Harmonic Motion?
Simple Harmonic Motion can be defined as the motion of an object where its acceleration is directly proportional to its distance from a fixed point along a path. The acceleration is said to always be directed towards the fixed point. This fixed point is known as the equilibrium position; this is because it is where the object that is swinging freely would come to rest given that it has lost all its energy.

The diagram below shows a pendulum in simple harmonic motion:
Pendulum in S.H.M


Where:
P is the equilibrium position (where the body would come to rest)
PQ & PR gives the amplitude position which is the greatest displacement from equilibrium position.
X is the position from the equilibrium position.

When the object is swinging freely you’ll probably realize that it moves faster when passing through its equilibrium position P, this is because the body’s acceleration and velocity is greatest as it passes through the equilibrium position. Also the opposite happens as the body leaves the equilibrium position, the acceleration and velocity lessens as the body’s position, x, moves further away from the equilibrium position.

Formulas involved in simple harmonic motion:

S.H.M formulas


The diagram below shows a spring in simple harmonic motion:
Spring in S.H.M

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