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Name……………………
Class……………..
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The Newtonian World
Module G484.2
Circular Motion and
Oscillations
QUESTIONS AND ANSWERS
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Lesson 10 notes - Angular Measurement
Objectives
You should be able to define the radian;
You should be able to convert angles from degrees into radians and vice
versa;
Outcomes
Be able to define the radian.
Be able to convert degrees into radians and vice-versa.
Be able to understand the reasons for using radians.
Be able to solve problems involving a mixture of degrees and radians.
Be able to work out arc length.
Radians
Radians are the SI unit for angles because it is a dimensionless quantity. It is
a distance divided by a distance and so does not need other dimensions to
define it. The unit “rad” is normally put in thought to make it clear that it is an
angle.
Angles aren’t decimal like the rest of the SI system and measuring the rate of
rotation is always done in radians per second.
One radian is the size of the angle
subtended by the arc of a circle equal to
the radius.
 is the ratio of the circumference to the
diameter (c/r)
 radians = 180
1 radian = 57.3
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Extension
Arc Length
Consider a sector of a circle with angle .
Let the arc length be l .
Then, whatever fraction  is of the total angle at O, l is the same fraction of
the circumference. So,
l
 
2
(In the diagram this is about one-third.)
l    circumference
2

l    2r
2
l  rθ
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Lesson 10 questions – Angular measurement
Write the answers next to each question (show workings on a separate piece of paper
if necessary)
ALL
MOST
5
SOME
6
7
8
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Lesson 11 notes – Circular Motion
Objectives
Be able to explain that a force perpendicular to the velocity of an object will
make the object describe a circular path.
Be able to explain what is meant by centripetal acceleration.
Be able to select and apply the equations for speed and centripetal
acceleration: v = 2r/T and a = v2/r.
Outcomes
Will know that circular motion occurs because of an unbalanced force which
makes an object accelerate towards the centre of the circular path.
Be able to explain that if an object has an unbalanced force on it, there must
be acceleration and that it is called the centripetal acceleration.
Be able to apply the equations for circular speed and centripetal acceleration
to solve problems correctly.
Be able to rearrange the equations for circular speed and centripetal
acceleration.
Be able to explain what the idea of centrifugal force is and why it is imaginary.
Be able to derive the equations for circular speed and centripetal acceleration.
Centripetal Force
A circle follows a curve all the way round and
we can describe it quantitatively as well as
qualitatively. All objects that follow a curved
path must have force acting towards the centre
of that curve. We call this force the centripetal
force. (Greek: Centre seeking).
Some examples of circular motion and the associated centripetal forces are:
Planetary orbits (almost!)
Electron orbits
Centrifuge
Gramophone needle
Car cornering
Car cornering on banked track
Aircraft banking
gravitation
electrostatic force on electron
contact force (reaction) at the walls
the walls of the groove in the record
friction between road and tyres
component of gravity
horizontal component of lift on the wings
Centripetal acceleration
Newton’s 1st Law says that an object will change direction if it feels an
unbalanced force at right angles to the direction it is travelling. This is the
reason things travel in circular paths. If there is always a force at 90 degrees
to the direction of motion then an object will travel in a circle. Since velocity is
speed in a given direction if an object is travelling at a constant speed but is
constantly changing direction it must be accelerating. This is what is
happening in circular motion. The acceleration is called Centripetal
Acceleration.
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Circular velocity
The instantaneous linear velocity at a point in the circle is usually given the
letter v and measured in metres per second (m s-1).
Speed is defined as the distance / time.
For a circle, 1 complete circumference is 2r and T is the Time period for one
rotation (T)
So
v = 2r / T
Centripetal acceleration
If an object is moving in a circle at a constant speed, its direction of motion is
constantly changing. This means that its linear velocity is changing and so it
has a linear acceleration. The existence of an acceleration means that there
must also be an unbalanced force acting on the rotating object.
Consider an object of mass m moving with
constant speed (v) in a circle of radius r with
centre O.
v
v
Q
It moves from P to Q in a time t.
The change in velocity v is parallel to PO and v
= v sin

O
P
When  becomes small (that is when Q is very
close to P) sin is close to  in radians.
So v = v 
Dividing both sides by t gives:
v / t = v t
a = v2/r
a is the Centripetal Acceleration.
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Extension
Centrifugal Force
This is a difficult concept. If we have something following a circular path and
there is a force pulling it to the centre there must be an equal and opposite
force pushing out of the circle right? Wrong.
The wrong idea:
The correct idea:
Centrifugal
force
Centripetal
Force
Centripetal Force
There is no equal and opposite force pushing the object out of its circular
path.
For there to be circular motion the forces must be unbalanced. There is a
centripetal force that is proportional to a centripetal acceleration.
If there is no more centripetal force the object does not fly out of the circle
away from the centre of the circle it just carries along in a straight line out of
the circle.
Think of the following examples:
Sitting in the back seat of a car as it corners: If the car turns to the left, you
feel as if you are being thrown to the right. In fact, your bum is in contact with
the seat, and gets pulled round to the left (providing there is sufficient friction).
The upper half of your body tries to carry on in a straight line. Viewed from a
point above the car, your upper half will be seen to be trying to follow a
tangential path while the car turns to the left.
Watching a marble roll on the surface of a table in a train as the train corners:
again, if the train turns to the left, the marble will appear to drift off to the right.
It is following a straight-line path, tangential to the curve. There is no friction to
pull it to the left, so no centripetal force.
An interesting example is a helium-filled balloon inside a cornering car. The
balloon leans in towards the centre of the circle. The air in the car tries to
continue in a straight line, so it is slewing to the right inside the car. The
balloon is lighter than the air, so it gets pushed towards the lower pressure at
the centre of the circle.
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Lesson 11 questions – Circular Motion
(
/27)………%……..
1.
By describing the meanings of the terms Centripetal Force and
Centripetal Acceleration explain how an object travels around a circular
path.
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2.
In each case below, state what provides the centripetal force on the object:
a)
a car travelling at a high speed round a sharp corner;
…………………………………………………………………………………………
…………………………………………………………………………………… (1)
b)
a planet orbiting the Sun;
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c)
an electron orbiting the positive nucleus of an atom;
…………………………………………………………………………………………
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d)
clothes spinning round in the drum of a washing machine.
…………………………………………………………………………………………
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3.
The object rotates at 15 revolutions per minute. Calculate the angular speed in
radian per second.
(2)
4.
This question is about a rotating restaurant.
A high tower has a rotating restaurant that moves slowly round in a circle while
the diners are eating. The restaurant is designed to give a full 360° view of the
skyline in the two hours normally taken by diners.
a)
Calculate the angular speed in radians per second.
(2)
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b)
The diners are sitting at 20 m from the central axis of the tower.
Calculate their speed in metres per second.
(2)
c)
Do you think they will be aware of their movement relative to the
outside? Explain your answer.
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…………………………………………………………………………………… (2)
5.
An aeroplane is circling in the sky at a speed of 150ms–1. The aeroplane
describes a circle of radius 20 km. For a passenger of mass 80kg inside
this aeroplane, calculate her centripetal acceleration;
a = …………………. ms-2 (3)
6.
The diagram shows a stone tied to the end of a length of string. It is
whirled round in a horizontal circle of radius 80 cm.
The stone has a mass of 90 g and it completes 10 revolutions in a time of 8.2 s.
a) Calculate:
i)
the time taken for one revolution;
T = ……………s (1)
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ii)
the distance travelled by the stone during one revolution;
iii)
distance = …………….. m (1)
the speed of the stone as it travels in the circle;
iv)
v = ………………….. ms-1 (2)
the centripetal acceleration of the stone;
v = ………………….. ms-2 (3)
b)
What provides the centripetal force on the stone?
…………………………………………………………………………………………
…………………………………………………………………………………… (1)
c)
What is the angle between the acceleration of the stone and its
velocity?
…………………………………………………………………………………………
…………………………………………………………………………………… (1)
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Lesson 12 notes – Centripetal Force
Objectives
Explain what is meant by centripetal force.
Select and apply the equation for centripetal force: F = ma = mv2/r.
Outcomes
Be able to define Centripetal force as the force that acts towards the centre of
a circular path.
Be able to apply Newton’s 2nd Law to equations for centripetal acceleration to
get equation for centripetal force.
Be able to apply the equations for centripetal force to solve problems correctly
for different situations.
Be able to rearrange the equation for centripetal force.
Centripetal Acceleration
From last lesson we have said that:
a = v2/r
where a is the centripetal acceleration.
Centripetal Force
Applying Newton's Second Law (F = ma) gives:
F = mv2/r
This is the equation for centripetal force; you should learn to identify the
appropriate form for use in any given situation.
Worked Example:
Centripetal Force
A stone of mass 0.5 kg is swung round in a horizontal circle (on a frictionless
surface) of radius 0.75 m with a steady speed of 4 m s-1.
Calculate:
(a)
the centripetal acceleration of the stone
acceleration = v2/r = 42 / 0.75 = 21.4 m s-2
(b)
the centripetal force acting on the stone.
F = ma = mv2/r = [0.5  42] / 0.75 = 10.7 N
Notice that this is a linear acceleration and not an angular acceleration. The
angular velocity of the stone is constant and so there is no angular
acceleration.
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Lesson 12 questions – Centripetal Force
(
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1. Calculate the centripetal force in the following cases:
a)
a ball of mass 150 g is spun in a horizontal circle of radius 3m at 5 ms-1
b)
c)
Force = ………… N (2)
the Earth (mass 6x10 kg) orbits the Sun once every year (3x107 s),
orbit radius 1.5x1011 m
24
an electron (mass 9x10
radius 10-10 m
-31
Force = ………… N (2)
kg) orbits a nucleus in 1.6x10-16s, orbit
Force = ………… N (2)
2
The diagram shows a car of mass 850kg travelling on a level road in a
clockwise direction at a steady speed of 20ms–1 round a bend with radius of curvature
32m.
a)
On the diagram, draw an arrow to show the velocity of the car (label
this v) and another arrow to show the acceleration of the car (label this a).
(2)
b)
Write an equation for the centripetal acceleration a of the car moving
on a level road at a speed v round a bend of radius of curvature r.
(1)
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c)
Calculate the centripetal force acting on the car.
Force = ………………N (3)
d)
State what provides the centripetal force in (c).
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……………………………………………………………………………………… (1)
3
A rubber toy of mass 40 g is placed close to the edge of a spinning turntable.
The toy travels in a circle of radius 12 cm. The toy takes 0.85 s to complete one
revolution. For this toy, calculate:
a)
its speed;
v = …………….. ms-2 (2)
b)
the centripetal force acting on it.
F = ………………… N (3)
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Lesson 14 notes – Analysing circular motion
Objectives
Be able to use equations for centripetal acceleration and force in different
situations.
Outcomes
Be able to apply the equations for circular motion to a number of different real
life situations in 2D.
Be able to apply the equations for circular motion to a number of different real
life situations in 3D.
When we release an object swung in a circular path, it takes a net force (the
resultant of all forces) acting inward that keeps the object spinning in a circle;
if you let go, the net force is no longer inward, so the object flies outward.
For example, If a car travelling around a level curve. Where does the net
force acting toward the center of the curve come from? Static friction. So
Newton's 2nd law for this situation is determined as follows:
F = ma
and, static friction, assuming car is not skidding
Ff =
FN =
mg
Centripetal acceleration:
ac = v2/r
In this situation,
Ff = F
mg = ma = mac = mv2/r
Therefore:
g = v2/r
As a result, we see that the greater the
of the road is, the faster a car can
travel without skidding. And the car can travel in a small radius of a curve
without skidding.
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Example
If a road's static coefficient is 0.5. And the radius of the curve is 50 m. What
is the greatest speed a car can travel without skidding out of the road?
As we have derived from the curve section:
g = v2/r
As a result, we see that the greater the
of the road is, the faster a car can
travel without skidding. And the car can travel in a small radius of a curve
without skidding.
= 0.5
r = 50 m
9 = 9.8 m/s2
We have:
g = v2/r
Rearrange it:
v2 =
gr
vmax =
vmax =
vmax =15.65 m/s
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Extension
Wall of Death
A. THE DANGERS:
The frictional force needs to be high enough on the wall ( so don't try it with
wet walls )
Vehicle speed needs to stay above a calculable minimum ( so don't go too
slowly! )
Motorcycles must lean at a speed dependent angle to prevent tipping over:
the higher the speed, the smaller the needed lean angle.
Cars cannot lean but they do not need to (see why below).
Riders in cars on the wall should lean to reduce nausea and muscle strain.
THE PHYSICS CONCEPTS:

the maximum static frictional force (FF) is proportional to the normal
reaction (N), where the normal reaction is the always observed
perpendicular push back by a surface when it receives a force

the force causing motion in a circle acts towards the centre of the circle
and can be written as Fc= mv2/r, where m and v are the mass and
velocity of the rotating body and r is the radius of the rotation;

the force (down) due to gravitational attraction can be written F = mg;

the centre of mass of a body is the point that, if all the mass were
concentrated there, would behave translationally just like the real body
does;

if force is applied to a body at a distance from a pivot point, then the
body will tend to rotate about that pivot point, with the size of the
rotation determined by multiplying the force by the perpendicular
distance (the answer being a number we call "torque");

any force which does not act along a line through the centre of mass
will produce a torque (and hence a tendency to rotate);

precession (like that seen in a spinning top) needs a torque applied to
stop it.
THE PHYSICS DETAILS:
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For a body of mass m moving in a horizontal circle on a wall of death, the
normal reaction from the wall is what supplies the force necessary to obtain
motion in a circle.
So we can write: N = mv2/r
If the body does not slide down the wall, then the pull of gravity downwards
must be balanced by the frictional force (due to the ‘roughness' of the wall)
upwards.
So the force diagram looks like:
This frictional force can vary as needed up to a maximum of
.
But how much frictional force is available depends on the speed, which
means there will be a MINIMUM speed for revolution at a constant height
above the ground. This minimum speed will be the one for which the
maximum frictional force exactly equals the gravitational pull on the mass
(since any lower frictional force will not be strong enough to provide the
balance).
Higher speeds will also produce no slipping, but as the speed rises steering
and control will become more difficult.
So when
, then v is the lowest it can be (ie vMINIMUM ),
for no slipping to occur. Which means the minimum safe speed is
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.
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BUT:
The above analysis assumes a point mass (i.e. with all the mass concentrated
in a single point located at the centre of mass and all the forces acting on that
point).
However for a motorcycle on the wall of death, although the weight acts
through the centre of mass, the friction acts at the wheels (on the wall).
These three forces (friction forces on each of the tyres, a weight force through
the centre of mass) balance but are not in the same line. Which means the
motorcycle will tend to rotate and tip over.
The normal reactions from the wall (acting where the tyres touch the wall)
cannot help to balance this torque because they will not produce any turning
effect if the motorcycle is perpendicular to the wall:
But if the motorcycle leans at an angle to the vertical to the wall then the
normal reactions from the wall, will produce a tendency to rotate (a torque) in
the opposite direction.
So if the rider leans the motorcycle at the correct angle then the torques will
be equal and no rotation will occur.
For other angles of lean there will be unbalanced torques causing the
motorcycle to rotate and fall.
At these other angles of lean the rider's muscles will need to push more
strongly in order to supply the extra torque to maintain balance. The rider may
also experience nausea (because the endolymph fluid in the ear will
experience an unbalanced torque and rotate).
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With cars however the precession can't occur. Cars do not need to lean as
long as they have a relatively wide wheelbase and a low centre of mass. To
see why read the analysis below.
WHY DO CARS NOT NEED TO LEAN?
Summary : When a car is running around a vertical wall, the tyres closer to
the ground experience stronger normal reaction force than the tyres opposite
them (closer to the top of the wall). So, provided the speed is high enough,
the counterclockwise torque from the normal reaction of the lower tyres can
provide the extra balancing torque without the need to lean. Details: Consider
the diagram below showing the relevant forces and distances.
Define:
Nupper (lower) = sum of the normal reactions of the upper ( lower ) front and
rear wheels;
Fupper (lower) = sum of the friction forces of the upper ( lower ) front and rear
wheels;
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L = the wheelbase (distance between the front or back wheels, assumed the
same)
H = perpendicular distance from the car's centre of mass to the Wall;
v = speed of the centre of mass of the car;
R = radius of the circular path travelled by the centre of mass of the car;
= static friction coefficient between the Wall and the tyres.
Assume the centre of mass of the car to be on the plane that perpendicularly
bisects the wheelbase.
Force analysis reveals:
(i) total normal reaction from the Wall provides the circular motion force (Fc):
...............................Nupper + Nlower = Fc= mv2/r .........................................……
(A)
(ii) total frictional force balances the weight:
...............................Fupper + F lower = mg ......................................................……
(B)
Torques (about the car's centre of mass) yield:
...............................(N upper)(L/2) + (Fupper + Flower )(H) = (Nlower )(L/2)...........
......(C)
Hence:
...............................Nupper = (1/2) (mv2/r) – (H/L)(mg) Flower = (1/2) (mv2/r) +
(H/L)(mg)
Since Fupper can never be negative (because walls can only push, never pull),
then:
(i)
For safety, at maximum friction ,
(ii)
..............{"no flip"}
(Nupper + Nlower ) > mg, so:
................{"no slip"}
Both conditions must apply simultaneously - meaning the minimum no-lean
speed required for travel on a Wall of death of given radius is determined by
whichever is the greater of (1/
) and (2H/L).
For bodies with a wide wheelbase and a low positioned centre of mass (say,
standard cars), the ratio 2H/L will be small, so the reciprocal of the friction
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coefficient becomes the most important factor (i.e. do it on dry days only!) and
the minimum no-lean speed is easily attainable.
For bodies with a short wheelbase and a high positioned centre of mass (say,
rollerblade riders), the ratio 2H/L will be large so the minimum no-lean speed
becomes impracticably high. A motorcycle would be the limiting case of this,
since L is effectively zero and hence no no-lean speed possible.
Which is what we found when riding our real Wall of Death: cars do not lean,
motorcycles do!
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Lesson 14 questions – Examples of circular motion
(
1
/18)………..%………..
Read the short passage before answering the question below.
Figure 1.1 shows a section of a mass spectrometer. A beam of identical positivelycharged ions, all traveling at the same speed, enters an evacuated chamber through a
slit S. A uniform magnetic field directed vertically out of the plane of the diagram
causes the ions to move along a semicircular path SPT. The beam exits the chamber
through the slit at T
.
Fig 1.1
a)
i)
On Fig.1.1, draw an arrow to indicate the direction of the force
on the ion beam at P.
ii)
Name the rule you would use to verify that the ions are
positively charged.
…………………………………………………………………………………………
……………………………………………………………………………………… (1)
iii)
Explain why the ions follow a circular path in the chamber.
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……………………………………………………………………………………… (2)
b)
Describe and explain the changes to the path of the ions for a beam of
ions of greater mass but the same speed and charge.
…………………………………………………………………………………………
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…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (3)
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c)
The speed of the singly charged ions is 3.0 x 105ms-1 in the magnetic
field of flux density 0.60T. The mass of each ion is 4.0 x 10-26 kg and
the force on each ion in the beam in the magnetic field is about 3 x 1014
N.
Calculate the radius of the semicircular path.
Radius = ……………. m (3)
2
A compact disc (CD) player varies the rate of rotation of the disc in order to
keep the track from which the music is being reproduced moving at a constant linear
speed of 1.30ms-1. Calculate the rates of rotation of a 12.0cm disc when the music is
being read from
a)
The outer edge of the disc. Give your answers in both (i) rad s-1. and
-1
(ii) Rev min
i)
rate of rotation = ……………………….. rad s-1 (2)
ii) rate of rotation = ……………………….. Rev min-1 (2)
b)
A point 2.55 cm from the center of the disc. Give your answers in both
-1
(i) rad s . and (ii) Rev min-1
i)
ii)
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rate of rotation = ……………………….. rad s-1 (2)
rate of rotation = ……………………….. Rev min-1 (2)
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Lesson 15 notes – Gravitational Fields
Objectives
Be able to describe how a mass creates a gravitational field in the space
around it;
Be able to define gravitational field strength as force per unit mass;
Be able to use gravitational field lines to represent a gravitational field;
Outcomes
Be able to define gravitational field strength and explain that a gravitational
field is a region of space around a mass that a force operates.
Be able to draw field lines that describe fields about spherical objects at a
distance and close to the surface of the object. (drawings in 2D).
Gravitational Fields
All objects that have mass have an associated gravitational field. That field is
greater if the mass is greater. The field gets weaker the further you are away
from the mass. And the field is always attractive. If another mass enters that
field it will feel a force towards the other object, just as the initial object will
feel a force towards the second object because of its gravitational field.
Fields
A field in physics is defined as the force per unit …something… for
gravitational fields that something is mass.
Definition
The gravitational field strength is defined as the force per unit mass. The units
for gravitational field strength are therefore Nkg-1.
Field Lines
We cannot see or touch this field, but we can try to model it using field lines or
lines of force. In a field line diagram, the direction of the field line at a point
gives the direction of the force of attraction that would be felt by a small mass
placed there. The relative density of field lines on the diagram is an indication
of the strength of the field. Thus for a spherical mass, like the Earth, we would
have the following diagram:
The field lines are directed radially inwards,
because at any point in the Earth’s field, a body
will feel a force directed toward the centre of the
Earth. The field lines become more spread out as
the distance from the Earth increases, indicating
the diminishing strength of the field. Note that the
field is really 3-dimensional, but of course on
paper, we can only take a 2-dimensional slice of it.
This is a radial or spherical field.
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You can think of any object with mass as a gravity well distorting everything
about it so that things fall toward it.
Close to the surface of the Earth, the field lines look like:
They are directed downwards (the direction in which a body near the Earth’s
surface would feel a gravitational force), and they are parallel and equidistant
indicating that the field is constant, or uniform.
A couple of important points to note:
1. Field lines do not start or stop in empty space (even though on
diagrams they have to stop somewhere!). They end on a mass and
extend back all the way to infinity.
2. Field lines never cross. (If they did, then an object placed at the point
where they crossed would feel forces in more than one direction.
These forces could be resolved into one direction – the true direction of
the field line there.)
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Lesson 15 questions – Gravitational Fields
1
(
/8)…………%………….
Draw the gravitational field lines for the following:
a)
The Earth from a distance
(2)
b)
The Earth up close
(2)
c)
A similar sized but much denser planet than Earth from a distance
(2)
d)
A similar sized but much denser planet than Earth from up close
(2)
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Lesson 16 notes – Newton’s Law of Gravitation
Objectives
Be able to state Newton’s law of gravitation;
Be able to select and use the equation F = − GMm
r2
for the force between
two point or spherical objects;
Be able to select and apply the equation g = − GM
r2
for the gravitational
field strength of a point mass;
Be able to select and use the equation F = − GM
r2
to determine the mass
of the Earth or another similar object;
Be able to explain that close to the Earth’s surface the gravitational field
strength is uniform and approximately equal to the acceleration of free fall;
Outcomes
Be able to state Newton’s Law of Gravitation.
Be able to explain that gravitational field strength varies with distance.
Be able to explain that gravitational field strength is inversely proportional to
the square of the distance.
Be able to select and use the equation g = − GM
r2 correctly in a number of
different situations.
Be able to select and use the equation F = − GMm
r2 correctly in a number of
different situations.
Be able to explain how the gravitational field varies with distance.
Weight
What causes weight?
The gravitational attraction by the Earth.
What affects the size of the Earth’s pull on you? Why would you weigh a
different amount on the Moon?
Your mass, and its mass.
If the Earth is pulling down on you, then what else must be occurring, by
Newton’s 3rd Law?
You must be pulling up on the Earth with a force equal to your weight.
What happens to the strength of the pull of the Earth as you go further away
from it?
It gets weaker – most students guess this correctly from the incorrect
assumption that in space, astronauts are weightless!
So, in summary the force depends upon the masses of the Earth and you,
and weakens with distance. This is all embodied in Newton’s law of universal
gravitation
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We know that Weight = mass x gravity and this is true at the surface of the
Earth but as you move away from the Earth the gravitational field strength
decreases. This is not linear, it goes down inversely proportionally to the
square of the distance you are from the Earth.
ie. g is
(directly proportional to) 1/r2 where r is the radial distance.
g is also still directly proportional to the mass of the object you are standing
on, (in this case the mass of the Earth).
So now we can write that g
m(1/r2)
Since it is an attractive force it is always going to be negative and to make it
equal something we need to put in a constant of proportionality (Universal
Gravitational Constant)
We get:
g = -Gm/r2
This says that the field strength around an object goes down as you go away
from it but is bigger if the object has more mass. So the figure 9.81N/kg is a
specific figure for our Earth and our position from the centre of it.
Newton’s law of gravitation tells you how attracted you would be to an object
depending on you mass.
Newton’s law of universal gravitation
F = -Gm1m2/r2
F = gravitational force of attraction (N)
m1, m2 are the interacting masses (kg)
r is the separation of the masses (m)
G is known as the “universal gravitational constant” (NOT to be confused with
“little” g). It sets the strength of the gravitational interaction in the sense that if
it were doubled, so would all the gravitational forces.
G = 6.67  10-11 N m2 kg-2
This law applies between point masses, but spherical masses can be treated
as though they were point masses with all their mass concentrated at their
centre.
This force is ALWAYS attractive. F = -Gm1m2/r2, and the direction is given by
the fact that the force is always attractive.
Every object with a mass in the universe attracts every other according to this
law. But the actual size of the force becomes very small for objects very far
away. For example, the Sun is about one million times more massive than the
Earth, but because it’s so far away, the pull on us from the Sun is dwarfed by
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the pull on us from the Earth (which is around 1650 times greater). As the
separation of two objects increases, the separation2 increases even more,
dramatically. The gravitational force will decrease by the same factor (since
separation2 appears in the denominator of the equation). This is an example
of an “inverse square law”, so called because the force of attraction varies in
inverse proportion to the square of the separation.
We can define the field due to a body as the region of space surrounding it
where other bodies will feel a force due to it.
What extent do gravitational fields have?
The gravitational force is infinite in range, although it becomes very weak at
large distances, as it is an inverse square law. The gravitational field due to a
body is thus also infinite.
It is negative since gravity is an attractive force.
At the centre of the Earth, g = zero, because matter will be pulled equally in all
directions so that the overall force is zero.
Worked examples
Have a go at these – some hints are included for you.
Data required: G = 6.67  10-11 N m2 kg-2, mass of the Earth = 6.0  1024 kg,
radius of the Earth = 6.4  106 m, mass of the Sun = 2.0  1030 kg, average
distance from the Earth to the Sun = 1.5  1011 m.
1) Communications satellites orbit the Earth at a height of 36 000 km. How far
is this from the centre of the Earth? If such a satellite has a mass of 250 kg,
what is the force of attraction on it from the Earth?
It is (3.6 x 107 m + 6.4 x 106 m) = 4.24 x 107 m from the centre of the Earth.
(You should really give this as 4.2 x 107 m – because of the amount of
significant figures).
The force is F = Gm1m2/r2 = (6.67 x 10-11 x 6.0 x 1024 x 250)/ (4.24 x 107)2.
This gives an answer of about 56 N, which for information is about less than
the weight as a one year old toddler.
2) What is the force of attraction from the Earth on you? What do we call this
force? What is the force of attraction on the Earth from you?
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You will need to estimate their own mass in kg. If you need to convert, 1 stone
is 6.4kg (and there are 14 pounds in a stone). Then use F=Gm 1m2/r2 where r
is the radius of the Earth.
This force is usually called weight.
The force on the Earth from the you is exactly the same as your first answer,
but in the opposite direction.
3) What is the force of attraction from the Sun on you? How many times
smaller is this than the force of attraction from the Earth on you?
Again, you will need to use your own mass, and the equation F=Gm1m2/r2, but
this time r is the average distance from the Sun to the Earth. This force should
be about 1650 times less than your weight, of the order of 0.3-0.5 N. Small,
but not negligible.
4) The average force of attraction on the Moon from the Sun is 4.4  1020 N.
Taking the distance from the Sun to the Moon to be about the same as that
from the Sun to the Earth, what value of mass does this give for the Moon?
m2 = Fr2/Gm1 = (4.4 x 1020 x (1.5 x 1011)2)/(6.67 x 10-11 x 2.0 x 1030) = 7.4 x
1022 kg
5) Using the mass of the Moon you calculated in question 4, what is the pull of
the Earth on the Moon, if the Moon is 380 000 km away? How does this
compare with the pull of the Sun on the Moon?
F = Gm1m2/r2 = (6.67 x 10-11 x 6.0 x 1024 x 7.4 x 1022)/ (3.8 x 108)2 = 2.1 x 1020
N
This is actually smaller than the pull of the Sun on the Moon. So, is the Moon
is orbiting the Sun rather than the Earth? In fact, it depends on the most
useful frame of reference in a particular situation – from the Sun’s point of
view, the Moon and the Earth orbit the Sun, in a way that is affected by the
presence of the other; from the Moon’s point of view, both the Sun and the
Earth orbit the Moon, in a way that is affected by the presence of the other,
etc.
6. What is the force of attraction between two people, one of mass 80 kg and
the other 100 kg if they are 0.5m apart?
F = Gm1m2/r2
F = G x 100 x 80 / 0.52 = 2.14 x 10-6 N.
This is a very small force but it does increase as the people get closer
together!
Actually this example is not accurate because Newton's law really only applies
to spherical objects, or at least objects so far apart that they can be effectively
considered as spherical.
7. What is the force of attraction between the Earth and the Sun?
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Mass of the Sun = 2 x 1030 kg, mass of the Earth = 6 x 1024 kg, distance from
the Earth to the Sun = 1.5 x 1011 m
F = Gm1m2/r2
F = G x 2 x 1030 x 6 x 1024/ [1.5 x 1011]2 = 6.7 x 1011 N
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an enormous force!
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Lesson 16 questions – Newton’s Law of Gravitation
(
/32)………%..........
Useful data
G = 6.67  10–11 N m2 kg–2
Earth’s mass = 5.97  1024 kg
Moon’s mass = 7.34  1022 kg
Sun’s mass is 2.0  1030 kg
Radius of the Moon = 1.64  106 m
Radius of the Earth = 6.37  106 m
Earth–Moon distance = 3.8  105 km
Earth–Sun distance = 1.5  108 km
MOST
1.
You may sometimes find it difficult to get up from the sofa after
watching a TV programme. Assuming the force of gravity acts between
the centre of your body and the centre of the sofa, estimate the
attraction between you and your sofa.
(2)
2.
Calculate the size of the gravitational pull of a sphere of mass 10 kg on
a mass 2.0 kg when their centres are 200 mm apart.
What is the force of the 2.0 kg mass on the 10 kg mass?
(2)
3.
At what distance apart would two equal masses of 150 kg need to be
placed for the force between them to be 2.0  10–5 N?
(2)
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4.
Calculate the gravitational pull of the Earth on each of the following
bodies:
the Moon;
(2)
satellite A with mass 100 kg at a distance from the Earth’s centre 4.2 
7
10 m;
(2)
and satellite B mass 80 kg at a distance from the Earth’s centre 8.0 
106 m.
(2)
5.
Show that the unit for G, the universal gravitational constant, can be
expressed as m3 s–2 kg–1.
(3)
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6.
Calculate the weight of an astronaut whose mass (including spacesuit)
is 72 kg on the Moon?
(2)
What is the astronaut’s weight on Earth?
(2)
Comment on the difference.
(1)
7.
a)
Show that pull of the Sun on the Moon is about 2.2 times larger
than the pull of the Earth on the Moon.
(4)
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b)
Why then does the Moon orbit the Earth?
(2)
8.
The American space agency, NASA, plans to send a manned mission
to Mars later this century. Mars has a mass 6.42 x 1023 kg and a radius 3.38 x
106 m. G = 6.67 x 1011 N m2 kg-2 (a) The mass of a typical astronaut plus
spacesuit is 80 kg. What would be the gravitational force acting on such an
astronaut standing on the surface of Mars?
(2)
(b) State whether an astronaut on Mars would feel lighter or heavier than on
Earth.
(1)
9
Sketch a graph showing how the Earth’s gravitational field varies with
distance.
(3)
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Lesson 18 notes – Orbits
Objectives
(i) analyse circular orbits in an inverse square law field by relating the
gravitational force to the centripetal acceleration it causes;
(j) define and use the period of an object describing a circle;
(k) derive the equation T2 = 4 π2 x r3
GM
from first principles;
2
(l) select and apply the equation T = 4 π2 x r3
GM
for planets and satellites
(natural and artificial);
(m) select and apply Kepler’s third law T 2 = k r 3 to solve problems;
(n) define geostationary orbit of a satellite and state the uses of such
satellites.
Outcomes
Be able to define the period of an object describing a circle.
Be able to look at data like period, radius, gravitational field strength, to relate
gravitational force to acceleration.
Be able to define a geostationary orbit of a satellite and state the uses of such
satellites.
Be able to select and apply Kepler’s third law T 2 = k r 3 to solve problems in
different situations.
Be able to select and apply the equation T2 = 4 π x r3
GM
correctly for planets
and satellites (natural and artificial);
Be able to derive the equation T2 = 4 π x r3
GM
from first principles;
Orbits
An orbit is the (usually elliptical) path described by one celestial body in its
revolution about another. The time it takes for one complete revolution is
called the period T. The closer a body is to the object it is orbiting the faster it
will go. This is because as you move closer to an object with mass its
gravitational field increases and so the centripetal force goes up making the
orbiting object accelerate.
History
The first true laws of planetary motion were proposed by Johannes Kepler
between 1609 and 1619 as a result of his work on the twenty years of
planetary observations made by Tycho Brahe, the astronomer royal to the
King of Denmark.
Kepler used Brahe's data to come up with three laws of planetary motion
1. The planets move in ellipses with the Sun at one focus
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2. A line drawn from the planet to the Sun sweeps out equal areas in equal
times (see Figure 1)
Kepler’s 3rd Law
The ratio of the square of the period (T) of the planet about the Sun to
the cube of the mean orbit radius (r) is a constant or T2/r3 = constant.
T2 is directly proportional to r3
T2=kr3
Calculating k
The Earth is 150x109 m from the Sun and it takes one year (3.16x107 s) to
orbit the Sun.
Therefore Kepler's constant ( r3/T2) for the Solar System is
(150x109)3/(3.16x107)2 = 3.37x1018m3s-2
The geostationary satellite
A geostationary satellite has a period of exactly one day and so remains
constantly over one point on the Earth's surface. To be geostationary it must
have an orbit that lies in the
plane of the equator.
All Geosynchronous
satellites are in an orbit
42000 km from the Earth's
centre.
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Extension
Deriving T2 = 4 π2
GM
x r3
The gravitational force F exerted by the Sun on a planet is given by:
F=GMm/r2
The centripetal force for an object in orbit is given by:
F = mv2/r
The instantaneous velocity of an orbiting object is given by:
v = 2πr/T
So we have 4π2/T2 = GM/r3
and so
r3 (4π2/GM) = T2
So our constant k = (4π2/GM)
Rearranging we get: r = (goR2T2/4π2)1/3
Using this we can check that the distance of our Geostationary Satellites are
at 42000km.
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Lesson 18 questions – Orbits
(
/32)………%……….
MOST
1
A binary star is a pair of stars that move in circular orbits around their
common centre of mass. For stars of equal mass, they move is the same circular orbit,
shown by the dotted line in the diagram. In this question, consider the stars to be point
masses situated at their centres at opposite ands of a diameter of the orbit.
a)
each star.
i)
Draw on the diagram arrows to represent the force acting on
(2)
ii)
Explain why the stars must be diametrically opposite to travel
in the circular orbit.
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
……………………………………………………………………………………… (2)
b)
Newton’s law of gravitation applied to the situation in the diagram
may be expressed as
F = GM2/4R2
State what each of the symbols listed below represent
F ………………………………………………………………………………………
G ………………………………………………………………………………………
M ………………………………………………………………………………………
R …………………………………………………………………………………… (2)
c)
i)
v by v=2πR/T.
Show that the orbital period T of each star is related to its speed
(1)
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ii)
Show that the magnitude of the centripetal force required to keep each star
moving in its circular path is
F = 4π 2MR/T2
iii)
(2)
Use equations from (b) and (ii) above to show that the mass of
each star is given by
M = 16π 2R3/GT2
(2)
d)
Binary stars separated by a distance of 1x1011m have been observed
with an orbital period of 100 days. Calculate the mass of each star.
1 day = 86400s
Mass = …………………..kg (2)
2
This question is about gravitational fields. You may assume that all the mass
of the Earth, or the Moon, can be considered as a point mass at its centre.
a)
It is possible to find the mass of a planet by measuring the gravitational
field strength at the surface of the planet and knowing its radius.
i)
Define gravitational field strength, g.
…………………………………………………………………………………………..
…………………………………………………………………………………………..
……………………………………………………………………………………… (1)
ii)
Write down an expression for g at the surface of a planet in
terms of its mass M and its radius R.
(1)
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iv)
Show that the mass of the Earth is 6.0x1024kg.
Radius of Earth = 6400km.
(1)
b)
i)
Use the data below to show the value of g at the Moon’s
surface is about 1.7 Nkg-1.
Mass of Earth = 81 x mass of Moon
Radius of Earth = 3.7 x radius of Moon
(2)
ii)
Explain why a high jumper who can clear a 2m bar on Earth
should be able to clear a 7m bar on the Moon. Assume that the high jump on the
Moon takes place inside a “space bubble” where Earth’s atmospheric conditions exist.
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
……………………………………………………………………………………… (3)
c)
The distance between the centres of the Earth and the Moon is 3.8 x
108 m. Assume that the moon moves in a circular orbit about the centre of the Earth.
Estimate the period of this orbit to the nearest day.
Mass of Earth = 6.0x1024kg
1 day = 86400s
Period = ……………………… day (5)
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3
SOME
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Lesson 21 notes – SHM Intro
Objectives
(a) describe simple examples of free oscillations;
(b) define and use the terms displacement, amplitude, period, frequency,
angular frequency and phase difference;
(c) select and use the equation period = 1/frequency;
(h) explain that the period of an object with simple harmonic motion is
independent of its amplitude;
Outcomes
Be able describe simple examples of free oscillations.
Be able to define and use the terms displacement, amplitude, period,
frequency, angular frequency and phase difference.
Be able to select and use the equation period = 1/frequency.
Be able to describe that the period of an object with simple harmonic motion is
independent of its amplitude.
Oscillation
An Oscillation is the variation, typically in time, of some measure about a
central value (often a point of equilibrium and called the origin) or between
two or more different states.
Oscillation Examples
All clocks (starting with the pendulum in a grandfather clock) are essentially
simple harmonic oscillators; all transmitters and receivers of waves are
essentially simple harmonic oscillators; many aspects of engineering design
from the massive to the microscopic require a detailed knowledge of SHM
(e.g. bridges, earthquake protection of buildings, atomic force microscopes for
imaging single atoms).
In Simple Harmonic Oscillators the period is independent of the amplitude,
and we say that the motion is isochronic. This is one characteristic of a simple
harmonic oscillator.
Oscillatory motion repeats itself and thus has similarities
with circular motion which also repeatedly returns to its
starting point.
Displacement is the distance from the origin and is either
positive or negative since displacement is a vector.
Displacement is the greatest at the extremes of the
oscillation. This displacement is called the amplitude of the
oscillation.
Velocity is the greatest at the midpoint.
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The oscillating mass is in equilibrium at the midpoint of its oscillation. In a
mass-spring system the mass would hang naturally at this point and it takes a
force to push it either way of the origin.
When the mass is oscillating, its inertia carries it through the midpoint.
So we have the following pattern of motion:
The mass speeds up as it heads towards the midpoint, has its greatest speed
as it passes through the midpoint, and slows down as it continues towards the
other extreme of the oscillation. Here, it reverses and starts to accelerate
again towards the midpoint.
r
r
O
a = -kx
m
x
Of course, not all oscillations are as simple as this, but this is a particularly
simple kind, known as simple harmonic motion (SHM). It is relatively easy to
analyse mathematically, and many other types of oscillatory motion can be
broken down into a combination of SHMs.
Definitions
A summary of terms and relationships between them, shown on a diagram.
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Language to describe oscillations
Sinusoidal oscillation
Phasor picture
s = A sin t
amplitude A
+A
A
angle t
0
time t
–A
periodic time T
phase changes by 2
f turns per 2 radian
second
per turn
 = 2f radian per second
Periodic time T, frequency f, angular frequency :
f = 1/T unit of frequency Hz
 = 2f
Equation of sinusoidal oscillation:
s = A sin 2ft
s = A sin t
Phase difference /2
s = A sin 2ft
s = 0 when t = 0
sand falling from a swinging pendulum leaves
a trace of its motion on a moving track
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s = A cos 2ft
s = A when t = 0
t=0
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Lesson 21 questions – SHM Intro
(
/23)………..%………..
ALL
1
Estimate the time periods of each of the following motions and hence calculate
(to 1 s.f.) their frequencies.
a)
a child on a playground swing
period = …………….s (1)
frequency = …………… Hz (1)
b)
a baby rocked in its mother’s arms
c)
period = …………….s (1)
frequency = …………… Hz (1)
the free swing of your leg from your hip
period = …………….s (1)
frequency = …………… Hz (1)
2
The pendulum bob of a grandfather clock swings through an arc of length
196mm from end to end. The period of the swing is 2.00 s.
a)
Explain what is meant by the period of the swing.
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
……………………………………………………………………………………… (2)
b)
What is the amplitude of the swing?
Amplitude = …………… m (1)
c)
What is the frequency of the bob?
Frequency = …………… Unit ………… (2)
3
a)
What is (i) the frequency (ii) the period of:
the rise and fall of the sea
i)…………………. Unit ………. (2)
ii)…………………. Unit ……… (2)
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b)
the beat of a heart
c)
i)…………………. Unit ………. (2)
ii)…………………. Unit ……… (2)
piano strings which oscillate when middle C is played
i)…………………. Unit ………. (2)
ii)…………………. Unit ……… (2)
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Lesson 23 notes – Defining SHM
Objectives
Be able to define simple harmonic motion;
Be able to select and apply the equation a = – (2πf)2 x as the defining
equation of simple harmonic motion;
Be able to select and use x = Acos(2πft) or x = Asin(2πft) as solutions to the
equation a = – (2πf)2 x.
Outcomes
Be able to define simple harmonic motion.
Be able to select and apply the equation a = – (2πf)2 x as the defining
equation of simple harmonic motion.
Be able to select and use x = Acos(2πft) or x = Asin(2πft) as solutions to the
equation a = – (2πf)2 x.
Be able to explain why the period of an object with simple harmonic motion is
independent of its amplitude.
positive
The restoring force in SHM
Imagine a mass on a spring being displaced and then dropping through the
equilibrium position to as far down as it can go and back up again:
If we let the mass hang it will stay in the equilibrium point. The strain of the
spring equals the weight of the mass.
If we pull the mass up, the strain is less and so it will accelerate towards the
equilibrium point since gravity is still the same.
If we pull it down the strain will be greater than the weight and so it will
accelerate towards the centre again.
At the equilibrium point the forces cancel leaving no acceleration and a
maximum velocity.
If we consider up positive then any displacement, velocity or acceleration
directed downwards will be negative.
So displacement up gives a resultant force (and hence acceleration) down,
and vice versa. We call this force the restoring force (because it ties to restore
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the mass to its equilibrium position). The greater the displacement, the greater
the restoring force.
We can write this mathematically:
Restoring force F  displacement x
Since the force is always directed towards the equilibrium position, we can
say:
F -x
or
F = - kx
Where the minus sign indicates that force and displacement are in opposite
directions, and k is a constant (a characteristic of the system).
This is the necessary condition for SHM.
Now since we are dealing with vector quantities here it makes sense to use a
sign convention. We call the midpoint zero; any quantity directed to the right is
positive, to the left is negative. (For a vertical oscillation, upwards is positive.)
Think about mass-spring systems. Why might we expect a restoring force that
is proportional to displacement? (This is a consequence of Hooke’s law.)
Equations of SHM
SHM can be represented by equations. For displacement:
x = A sin 2ft
or
x = A sin t
f is the frequency of the oscillation, and is related to the period T by f = 1/T.
The amplitude of the oscillation is A.
x = A cos 2ft is also a possible solution depending on where the
oscillation starts.
Velocity: v = 2f A cos 2ft
or
v= A cos t
Acceleration: a = - (2f)2 A sin 2ft
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or
a = -2 A sin t
Comparing the equations for displacement and acceleration gives:
a = - 2x
and applying Newton’s second law gives:
F = - m 2x
These are the fundamental conditions that must be met if a mass is to
oscillate with SHM.
If, for any system, we can show that F  -x then we have shown that it
will execute SHM,
and its frequency will be given by:

 = 2f
so  is related to the restoring force per unit mass per unit displacement.
Extension
Developing equations by differentiation.
For those who like a challenge – or those studying maths you should be able to show
that differentiation of: x = A sin 2ft twice will get you to: a = - (2f)2 A sin 2ft.
Have a go! Next lesson will show you how to solve the equations graphically
and this will also help your understanding.
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Lesson 21 questions – Defining SHM
(
/17)……….%.........
ALL
1
Write an explanation which you could give to a non-scientist of what simple
harmonic motion is. Use a situation with which they will be familiar to illustrate your
explanation. 1 mark will be given for written communication.
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SOME
2
The equation defining linear s.h.m. is
a = - (constant)x
a)
What units must the constant have?
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b)
Two s.h.m.s, A and B, are similar except that the constant in A is nine
times the constant B. Describe how these s.h.m.s differ.
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3
A body oscillates with s.h.m. describe described by the equation
x = (1.6m)cos(3πs-1)t
a)
What are:
i)
the amplitude?
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ii)
the period of motion?
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b)
For t=1.5s, calculate
(i) the displacement of the body
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displacement = ………………..m (1)
(ii) the velocity
velocity = ……………….. ms-1 (2)
(iii) the acceleration of the body
acceleration = ……………….. ms-2 (2)
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Lesson 24 notes – Oscillation Graphs
Objectives
Be able to select and apply the equation vmax = (2πf)A for the maximum speed
of a simple harmonic oscillator.
Be able to describe, with graphical illustrations, the changes in displacement,
velocity and acceleration during simple harmonic motion.
Outcomes
Be able to use the equation vmax = (2πf)A for the maximum speed of a simple
harmonic oscillator correctly for different situations.
Be able to rearrange and then use the equation vmax = (2πf)A for the
maximum speed of a simple harmonic oscillator correctly for different
situations.
Be able to interpret graphical illustrations of the changes in displacement,
velocity and acceleration during simple harmonic motion;
Be able to derive the equation vmax = (2πf)A for the maximum speed of a
simple harmonic oscillator
Be able to draw graphical illustrations of the changes in displacement, velocity
and acceleration during simple harmonic motion;
Displacement – Time graphs
Recap from lesson 6 and 5 Mechanics G481.1 on displacement time graphs
and velocity time graphs.
Imagine a mass on a
spring being displaced
and then dropping
through the equilibrium
position to as far down
as it can go and back
up again:Error!
positive
SHM Graphs
If we let the mass hang it will stay in the equilibrium point. The strain of the
spring equals the weight of the mass.
If we pull the mass up, the strain is less and so it will accelerate towards the
equilibrium point.
If we pull it down the strain will be greater than the weight and so it will
accelerate towards the centre again.
At the equilibrium point the forces cancel leaving no acceleration and a
maximum velocity.
If we consider up positive then any displacement, velocity or acceleration
directed downwards will be negative.
The top graph shows displacement against time. The second graph shows the velocity
of the oscillator against time and the third graph shows the acceleration against time.
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Maximum Velocity
You can see that when the mass goes through the equilibrium point, x=0, and
velocity is maximum. Maximum velocity of the oscillating system will happen
when vmax = (2πf)A. When x is maximum, acceleration is negative maximum –
the mass is being pushed back the opposite way (acceleration is directly
proportional to negative displacement). This happens when the velocity is
zero – since it is at its maximum amplitude. This isn’t easy so take some time
going through it and drawing your own diagrams of an object oscillating about.
The dotted green lines show how the graphs line up easily. You can see that
the velocity graph is led by the displacement graph and is π/2 out of phase
with it and the acceleration graph is led by the acceleration graph and is π/2
out of phase with that one.
Explanation
If we go back to the diagram we used to define some terms and look at the
phasor diagram we can start to see clearly where the equations for SHM
come from:
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Language to describe oscillations
Sinusoidal oscillation
Phasor picture
s = A sin t
amplitude A
+A
A
angle t
0
time t
–A
periodic time T
phase changes by 2
f turns per 2 radian
second
per turn
 = 2f radian per second
Periodic time T, frequency f, angular frequency :
f = 1/T unit of frequency Hz
 = 2f
Equation of sinusoidal oscillation:
s = A sin 2ft
s = A sin t
Phase difference /2
s = A sin 2ft
s = 0 when t = 0
sand falling from a swinging pendulum leaves
a trace of its motion on a moving track
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s = A cos 2ft
s = A when t = 0
t=0
RAB Plymstock School
Motion of harmonic oscillator
velocity
force
displacement
against time against time against time
large displacement to right
right
zero velocity
mass m
large force to left
left
small displacement to right
right
small velocity
to left
mass m
small force to left
left
right
large velocity
to left
mass m
zero net force
left
small displacement to left
right
small velocity
to left
mass m
left
small force to right
large displacement to left
right
zero velocity
mass m
large force to right
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left
RAB Plymstock School
Dynamics of harmonic oscillator
How the graph continues
How the graph starts
zero initial velocity would stay
zero if no force
velocity
force changes
velocity
force of springs accelerates mass towards
centre, but less and less as the mass nears the
centre
change of velocity
decreases as
force decreases
new velocity
= initial velocity
+ change of
velocity
trace curves
inwards here
because of
inwards
change of
velocity
t
0
0
trace straight
here because no
change of
velocity
no force at centre:
no change of velocity
time
time
Extension: More SHM Graphs
You should learn to be able to understand and draw the following graphs:
+2
r
a
a
x
+r
-r
t
a = -r sin
t

-2r
v
x
v = rcos
t
+r
>
1
=
1
t
v
-r
x
t
x = rsin t
r is the amplitude in these graphs.
And if you haven’t done it go back to the extension last lesson and try it. Use
the diagrams below to help:
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Force, acceleration, velocity and displacement
Phase differences
Time traces
varies with time like:
displacement s
/2 = 90
If this is how the displacement varies
with time...
cos 2ft
... the velocity is the rate of change
of displacement...
–sin 2ft
... the acceleration is the rate of
change of velocity...
–cos 2ft
...and the acceleration tracks the force
exactly...
–cos 2ft
velocity v
/2 = 90
acceleration = F/m
same thing
zero
force F = –ks
 = 180
displacement s
... the force is exactly opposite to
the displacement...
cos 2ft
Maximum Velocity
Maximum velocity of the oscillating system will happen when vmax = (2πf)A.
Let’s look at the equations.
v= 2πfAcos(2πft) (a)
and
v=-A2πf sin(2πft) (b)
are both solutions.
Now, cos(2πft) or sin(2πft) have values of between 0 and 1,
So to get the maximum for each of these we need cos(2πft) to be equal to 1 in
(a) and sin(2πft) to be equal to 1 in (b).
cos0 = 1 and sin(π/2) = 1
(if you’re not sure of this try it on your calculator)
And if you look back at the graphs for displacement, velocity and acceleration
you will see that velocity is a maximum when this occurs.
If you can see the graphs in your head when you are tackling problems on
SHM you will find them a whole lot easier.
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Lesson 24 questions – Graphical Analysis of SHM
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/15)……….%.........
1
ALL
MOST
2
The diagram shows three sinusoidal (sine-shaped) graphs for displacement x,
velocity v and acceleration a for a simple harmonic oscillator whose Amplitude is
5nm and frequency is 300 MHz.
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a)
Explain the relationship between:
i)
the x-t and v-t graphs
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(2)
ii)
the v-t and the a-t graphs
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iii)
the x-t and the a-t graphs
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(2)
b)
Show that the maximum speed of this oscillator is 9ms-1.
(3)
c)
Illustrate your answer to (iii) by sketching a graph of x against a.
(2)
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Lesson 25 notes – Energy in SHM
Objectives
Be able to describe and explain the interchange between kinetic and potential
energy during simple harmonic motion.
Outcomes
Be able to describe and explain the interchange between kinetic and potential
energy during simple harmonic motion.
Energy in SHM
The diagram shows how the total energy of the system stays constant whilst
KE and PE change.
For the mass and spring system, the work done stretching a spring by an
amount x is the area under the force extension graph = 1/2 kx2. The PEextension graph is a parabola.
The kinetic energy will be zero at +A and a maximum when x = 0, so its graph
is an inverted version of the strain energy graph. At any position kinetic +
elastic strain energy is a constant E, where E = KEmax = PEmax.
PEmax  A2, so the total energy E of SHM is proportional to (amplitude) 2.
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Energy stored in a stretched spring
area below graph
= sum of force 
change in displacement
extra area
F1 x
F1
total area
1 Fx
2
0
x
0
unstretched
extension x
force F1
work F1 x
no force
larger force
Energy supplied
small change x
energy supplied = F x
F=0
x=0
F = kx
x
stretched to extension x by force F:
energy supplied = 12 Fx
spring obeys
Hooke’s law: F = kx
energy stored in stretched
spring = 12 kx2
Energy stored in stretched spring is 1 kx2
2
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Energy flow in an oscillator
displacement
potential energy
= 12 ks2
0
s = A sin 2ft
time
energy in stretched spring
potential energy
0
PE =
1
2
kA2 sin22ft
time
mass and
vmax
spring
oscillate
A
vmax
A
vmax
energy carried by moving mass
kinetic energy
0
KE =
1
2
2
mvmax
cos22ft
time
velocity
kinetic energy
= 12 mv2
v = vmax cos 2ft
0
vmax = 2fA
time
from spring to
moving mass
energy in
stretched spring
from spring to
moving mass
energy in
moving mass
from moving
mass to spring
from moving
mass to spring
The energy stored in an oscillator goes back and forth between stretched spring and moving
mass, between potential and kinetic energy
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Lesson 25 questions – Energy in SHM
(
1
/20)….…..%.......
This question is about a mass-spring system.
Fig1.1 shows a mass attached to two springs. The mass moves along a horizontal tube
with one spring stretched and the other compressed. An arrow marked on the mass
indicates its position on a scale. Fig 1.1 shows the situation when mass is displaced
through a distance x from its equilibrium position. The mass is experiencing an
acceleration a in the direction shown. Fig 1.2 shows a graph of the magnitude of the
acceleration against the displacement x.
fig 1.1
fig1.2
ALL
a)
i)
State one feature from each fig 1.1 and fig 1.2 which shows that
the mass performs harmonic motion when released.
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(2)
MOST
ii)
Use data from fig 1.2 to show that the frequency of the simple harmonic
oscillations of the mass is about 5Hz.
(3)
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b)
The mass-spring system of fig 1.1 can be used as a device to measure
acceleration, called an accelerometer. It is mounted on a rotating test rig, used to
simulate large g-forces for astronauts. Fig 1.3 shows the plan view of a long beam
rotating about its axis A with the astronaut seated at end B, facing towards A. The
accelerometer is parallel to the beam and is fixed under the seat 10m from A.
fig 1.3
ALL
i)
When the astronaut is rotating at a constant speed, the arrow
marked on the mass has a constant deflection. Explain why.
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ii)
Calculate the speed v of rotation of the astronaut when the
deflection is 50mm.
v = ………………. ms-1 (2)
SOME
2
A bored student (probably in a biology lesson) holds one end of a flexible
plastic ruler against the laboratory bench and flicks the other end, setting the ruler into
oscillation. The end of the ruler moves a total distance of 8.0cm as in the diagram and
makes 28 complete oscillations in 10s.
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a)
the ruler?
What are the amplitude x0 and frequency f of the motion of the end of
x0 = …………………….. m (1)
f = …………………….. Hz (1)
ALL
b)
Use x= x0cos2πft to produce a table of values of x and t for values:
t/s = 0, 0.04, 0.08, 0.12, 0.16, 0.20, 0.24, 0.28, 0.32, 0.36
(4)
Draw a graph (over the page) of x against t and use it to find the maximum speed at
the end of the ruler.
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(4)
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Lesson 26 27 notes - Damping
Objectives
Be able to describe the effects of damping on an oscillatory system.
Outcomes
Be able to define damping.
Be able to understand where damping may occur.
Be able to understand what an exponential relationship is.
Be able to describe an investigation where damping is caused (i) by the drag
of the air and (ii) by eddy currents (electromagnetic damping).
To show that for light damping, the amplitude of oscillations decays
exponentially with time.
Oscillations and damping
In S.H.M, (i) the period is independent of the amplitude and (ii) the total
energy remains constant.
In practise many objects only undergo approximate s.h.m because (i) the
restoring force is not exactly proportional to the displacement and (ii) resisting
forces oppose the motion (e.g. air resistance)
The graphs show an exponential decay.
Exponential decay of the amplitude A implies that:
Rate of decay of maximum amplitude A  present value of A.
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Large amplitude implies high maximum velocity, which implies greater drag.
Recall that PEMAX  A2, so the total energy of an oscillator  A2
When the amplitude has decayed to 1/2 its original value, the energy has
been reduced to 1/4 of the original input and so on.
In critical damping no oscillations occur:
Over damped systems bring an oscillating system to rest even quicker.
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Lesson 26 questions – Damping
(
/8)………..%...........
1
The faulty suspension system of a car is tests. The body of the stationary car is
pushed down and released. Fig 1.1 shows how the vertical displacement of the car
varies with time after it has been released.
fig 1.1
ALL
a)
i)
The graph shows light damping. Sketch on the same graph
what a critically damped system would look like.
(2)
ii)
Define simple harmonic motion.
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(2)
iii)
State two features of fig 1.1 which indicate that the car body is
oscillating in damped harmonic motion.
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(2)
b)
Use data from fig 1.1 to calculate the frequency of the car body.
frequency = …………………….. Hz (2)
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Lesson 28 29 notes – Resonance
Objectives
Be able to describe practical examples of forced oscillations and resonance;
Be able to describe graphically how the amplitude of a forced oscillation
changes with frequency near to the natural frequency of the system;
Be able to describe examples where resonance is useful and other examples
where resonance should be avoided.
Outcomes
Be able to describe practical examples of forced oscillations and resonance.
Be able to describe graphically how the amplitude of a forced oscillation
changes with frequency near to the natural frequency of the system.
Be able to describe examples where resonance is useful and other examples
where resonance should be avoided.
Resonance
Resonance is a phenomenon of forced oscillating systems. All oscillating
systems will have a natural
frequency. When the driving
force matches this frequency,
the amplitude of the oscillations
will become much greater. This
is resonance.
Let’s use an example of
swinging on a swing.
If you get yourself swinging you
will swing back and forth a
number of times in a second.
Maybe once back and forth a second. If your friend pushes you at your
maximum amplitude (when you are highest in the swing) once a second then
the amplitude will get greater and greater because this is the resonant
frequency.
If they push you before you reach the top of your swing – in other words, the
frequency of their push does not match your natural frequency – your
amplitude will decrease.
The diagram
shows
amplitude on
the y axis and
driving
frequency on
the x axis.
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The amplitude is at its greatest when the driving frequency matches the
natural frequency of the oscillator.
Damping and resonance.
Damping results in the amplitude being smaller and a wider range of driving
frequencies giving a larger amplitude as shown in the graph. Again, amplitude
on the y axis and driving frequency on the x axis.
Applications of Resonance
Stringed instruments like guitars and violins use vibrating strings to produce
resonance in a hollow wooden box. This amplifies the sound. Woodwind
instruments (e.g. clarinet) use a vibrating reed to produce resonance in a
column of air. With brass instruments (e.g. trumpet) the driving vibration is
produced by the lips of the player.
Radio receivers contain a tuning circuit consisting of a parallel inductor and
capacitor. The driving vibration comes from the incoming radio signal. The
circuit is tuned so that the resonant frequency corresponds to the frequency of
the radio station you wish to receive. The damping factor in such a circuit is
the electrical resistance.
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Lesson 28 29 questions – Examples of Resonance
(
/33)………….%..........
ALL
1
The faulty suspension system of a car is tests. The body of the stationary car is
pushed down and released. Fig 1.1 shows how the vertical displacement of the car
varies with time after it has been released.
fig 1.1
a)
i)
The graph shows light damping. Sketch on the same graph
what a critically damped system would look like.
(2)
ii)
Define simple harmonic motion.
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iii)
State two features of fig 1.1 which indicate that the car body is
oscillating in damped harmonic motion.
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b)
Use data from fig 1.1 to calculate the frequency of the car body.
frequency = …………………….. Hz (2)
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MOST
c)
To simulate the car being driven along a ridged road at different
speeds, the stationary car is oscillated up and down in simple harmonic motion by a
mechanical oscillator. The mechanical oscillator provides a movement of variable
frequency and constant amplitude. Fig 1.2 shows that graph of the vertical motion of
the car body obtained from the test.
fig 1.2
i)
Describe what resonance is.
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ii)
Use information from fig 1.2 to write down the amplitude of
the motion of the mechanical oscillator.
ii)
amplitude = ………………… mm (1)
Using your answer to (b), add the scale to the frequency axis of
fig 1.2. (1)
iii)
Two new dampers are tried on to increase the damping of the
car body a little and then a heavy damper is fitted. On fig 1.2
sketch the graph you would expect for these dampers.
(5)
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2
In this question, four marks are available for the written communication.
a)
Explain the meaning of the term resonance.
State two examples of oscillating systems in which resonance occurs; one being
useful or beneficial and the other being a nuisance or harmful. Explain their practical
significance. You may use diagrams in your answer.
State how the oscillation is driven in each case.
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b)
Describe how damping in vibrating systems affects their resonant properties.
Give an example of a practical resonant system where two of the damping effects that
you describe could be observed. A space has been left for you to draw suitable sketch
graph(s), if you wish to illustrate your answer.
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Quality of written communication (4)
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