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Transcript
```YEAR 12
PHYSICS SUMMARY
USE THIS AS A REMINDER OF CONTENT AND A WAY
OF IDENTIFYING WEAKNESS, NOT AS A REPLACEMENT
FOR REVISION
QUANTITIES AND UNITS
• You will need to
memorise the base units
and be able to derive the
other units
• Derivation can be done
simply by looking at the
unit and finding a simple
equation that uses it!
• For example, the N is
also one kgms-2 and this
is derived from F=ma
MEASUREMENTS AND UNCERTAINTIES
•
See where the vernier
scale (bottom) begins at
the main scale (this
case=10cm)
uncertainties (e.g. the
ones that have + in
them)
•
Look for the only line on
the vernier scale that is
in-line with the main
scale (this case the
second one is!)
• Multiplying or Dividing?
cm
0.5mm
0.14mm
Measurement
uncertainties (e.g. work
out the % that the + is
of the value)
• Raising to a Power?
• MULTIPLY the
percentage uncertainty
by the power
DECIMAL PLACES AND SIG FIGS
calculation
questions, always
first as is then to
an appropriate
number of sig.figs
• Usually the
appropriate
number is the
significant figures
that are provided
for values in the
question
SCALAR VS VECTOR
• You’ll be expected to be
able to identify scalar
values (no direction,
only a magnitude) and
vector (magnitude AND
direction
• Quick trick: If you draw
the value on a diagram
with an arrow it is
usually a vector
• Vectors can be
combined to make a
resultant
RESOLVING VECTORS: METHOD 1
The simplest method
for resolving a vector
is to draw a scale
diagram of it!
In this, the question could be….A boat travels 40m
east and then 30m south, what is the resultant
displacement?
DON’T FORGET: give the angle! It’s a vector after all!
This can be worked out using trigonometry
(SOHCAHTOA)
Convert the data to
an appropriate scale
(e.g. 10ms-1 being 10
s) and draw the first
vector
From the end of that
one draw the next
(e.g. 5ms-1 at 20°
from the vertical) and
then join them up!
(start to end)
RESOLVING VECTORS: METHOD 2
A
Don’t forget, draw the vector components so that
they encompass the angle! It doesn’t matter for
example if the vertical component is acting at [ A], it
can still be worked out by placing it over on the right
hand side!
Then use simple trigonometry remembering you
have the resultant and you’re calculating the
components here
This one involves
using more
mathematics but can
be completed
quicker and is often
more accurate and
precise
Either method
ALWAYS draw a
vector diagram before
you calculate!
Displacement Start Speed Final Speed Acceleration Time
SUVAT
You may need to
know how to derive
these equations so
for that, look back at
our old powerpoints!
These can ONLY be
used for UNIFORM
ACCELERATION i.e.
when acceleration is
unchanging
DO NOT FORGET the initial velocity has both
horizontal and vertical components. If you’re dealing
with the vertical motion USE THE VERTICAL
COMPONENT of velocity, not that original.
This tends (not
always) relate to
vertical motion as
gravitational
acceleration is
constant
PROJECTILES
1) Isolate the vertical/horizontal velocities
2) Use the vertical velocity to calculate the
vertical motion (usually to find ‘time’)
3) Use the value of time with the horizontal
velocity to calculate distance (s=d/t)
Things to remember:
• You consider the components separately (use SUVAT for vertical
and s=d/t for horizontal) and use the component velocities
• You can consider the halfway point (with the vertical velocity here
being zero) and then double the time if needed!
• You can consider the whole journey and use s=ut+1/2at2 with s as
zero as it hasn’t had any vertical displacement overall!
MOMENTS
A moment has the unit of Nm and is
the force multiplied by the
PERPENDICULAR distance from
the pivot
Calculate all the CLOCKWISE
moments and then calculate all the
ANTICLOCKWISE moments. If they
are equal, then the body is in
equilibrium (NOT MOVING).
Don’t forget, if the pivot is not directly
at the centre of gravity of the board,
the boards weight will contribute to
the moments!
For example, if the angle of the
ladder was to increase, the weight of
the man would act toward the left of
the floor pivot, therefore contributing
extra moment. This would make the
moments unbalanced and therefore
DISPLACEMENT-TIME GRAPHS
represents the velocity,
if its curved then you
can take an
instantaneous
velocity by using a
tangent
If the graph shows a
flat line, then the body
is stationary as there
is no displacement
occurring
If the graph returns
back to the x axis, then
the object has gone
back to its original
position!
VELOCITY-TIME GRAPHS
The gradient of a velocitytime graph represents the
acceleration
If it’s a curve, the
acceleration is not uniform
e.g.) but you can use a
tangent to calculate
instantaneous
acceleration
The area underneath the
graph represents the
displacement. Divide up
the graph into triangular
and rectangular sections to
do this! If a curve, use
sections to estimate
NEWTON’S LAWS
This plane can be described as “climbing
at a constant speed”
It may be moving but the lift=weight as it’s
a constant speed.
If all forces acting upon an object are
balanced, then the object is in
equilibrium and that means the resultant
is 0 so no acceleration occurs!
1st law: A body will remain
at rest or at a constant
motion unless acted upon
by a resultant force
2nd law: The acceleration
of an object is directly
proportional to the
resultant force acting upon
it (F=ma)
3rd law: If object A applies
a force on object B, B will
enact an equal and
opposite force on object A
NEWTON PAIRS
In this example, there’s also another at
play. That weight also acts down on the
table (contact) so the table acts upon the
book (contact)
Newton Pairs of forces are
ones that obey the
following rules:
1)
They act on different
objects
2)
They have the same
magnitude
3)
They are the same
type (e.g. gravitational)
of force
4)
They act in opposite
directions
In this example, the earth
pulls the book down
(gravitational pull/weight)
so the book pulls the
earth up (gravitational pull)
in the opposite direction
RESOLVING FORCES
When resolving forces
don’t forget what
component is actually
relevant.
vertical
In this example they want to
know the weight of the
painting. This acts vertically
downward
This MUST be being
balanced by all the upward
vertical forces. I.e. vertical
component of tension in
each cable!
Therefore you could use
30Sin(45) to work out one,
then double for both
tensions.
OBJECT ON A SLOPE
40°
40°
If an object is at equilibrium (i.e. moving at a constant speed or stationary)
on a slope of a known angle, it’s weight will be acting as a vector (at an
angle-just look, turn the image so the slope is horizontal!)
This is the ONLY downward force and is being matched by the friction and
reaction force. To work out either of those, just calculate the components of
the weight
Hint: the angle of the slope is also the angle at the top of the component
triangle (usually…slope friction=sine, contact force=cosine)
MOMENTUM
Momentum (p) = mass x velocity
BEFORE
20ms-1
500kg
AFTER
The law of conservation of
momentum states that ‘the total
momentum before a situation = the
total momentum after”
0ms-1
400kg
? ms-1
Divide up the
situation and in
the before,
calculate the
momentum of
each object
individually
together to get
the total
momentum
e.g. in this instance:
Total= (20 x 500) + ( 0 x 400)
=10,000
500 + 400 Kg
After=10,000 so then divide by
900kg (total mass) = 11.1m/s!
As the total momentum is the same before and after, to calculate the velocity
they go off at together, just divide the total momentum by the total mass!
MOMENTUM- EXPLOSIONS
0ms-1
BEFORE
Gun: 1.2kg
Bullet: 0.005kg
The momentum of both objects
together must equal 0 as both are
stationary!kgms-1
AFTER
? ms-1
400ms-1
Momentum before (P(b)) = Momentum after(P(a))
The total momentum must still be 0. This is
because the momentum of the bullet will be
e.g. in this instance:
positive and that of the gun will be negative as
Total= 0. The momentum of the bullet=400 x 0.005=
it’s moving
backward!
2
velocity
of the gun?
So: the backward momentum must= -2, so -2 divided
by the mass of the gun gives (-1/1.2)= 0.83ms-1
MOMENTUM- CHANGE
Momentum=
mass x velocity
200kg
10m/s
Momentum is a
VECTOR
quantity – it has
a size and
direction.
So this car= 2000kgm/s!
When a car crashes (for example into a wall), all its momentum will be
transferred to the it!
As such it experiences a change in momentum. This is always a set
value that is dependent on the momentum just prior to colliding
This quick change in momentum will produce a huge force (impact) on
both the car and passengers. Linked by this equation:
𝑪𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒎𝒐𝒎𝒆𝒏𝒕𝒖𝒎
𝑭𝒐𝒓𝒄𝒆 =
𝒕𝒊𝒎𝒆
WORK DONE
Work Done = Force x Distance (in direction of force)
•
•
•
When a force pushes an object it makes it move by transferring energy to
it.
We say the force “does work”.
Work done is the same as energy transferred so is measured in Joules (J).
15N
So as the man is
pushing the block 4m
with 15N, he has used
15x4=60J!
4m
GRAVITATIONAL POTENTIAL ENERGY
GPE= mass x gravitational field strength x height
Gravitational Potential Energy (GPE) is exactly the same for lifting an object as you
are moving the weight (force) a vertical distance!
This is exactly the same as the previous equation! Weight is a Force (mass x g). And
height is a distance. So GPE is the same as calculating work! Which is the same as
energy!
60kg
1.5m
So as the man is lifting
the 60kg dumbbells
1.5m with he has used
60 x 10 x 1.5=900J!
KINETIC ENERGY
GPE= mass x gravitational field strength x height
This is exactly the same as the previous equation! Weight is a
Force (mass x g). And height is a distance. So GPE is the same
as calculating work! Which is the same as energy!
60kg
1.5m
So as the man is lifting
the 60kg dumbbells
1.5m with he has used
60 x 10 x 1.5=900J!
Kinetic Energy = ½ x mass x
velocity2
Note – only
the velocity
is squared!
The amount of kinetic energy an object has
depends on its mass and how fast it is moving…
So if both of these were
travelling at 10m/s, the
car would have more
kinetic energy as its
mass is greater!
POWER
Power= energy ÷ time
Power= work done ÷ time
Power= current x voltage
Power is the rate of doing work so it’s the energy transferred from one form into
another per second. Often, questions will ask you to first work out the energy
transferred (perhaps by calculating GPE) and then you just divide by the time!
The unit is the WATT (W) and this is equivalent to Js-1
ELECTRICAL QUANTITIES 1
We defined current as the rate of flow of charge and this can be
applied to anything that is carrying a charge (charge carriers) such as
conventionally, the electron.
This can be expressed simply mathematically with the formula:
𝐶𝑢𝑟𝑟𝑒𝑛𝑡=
𝐶ℎ𝑎𝑟𝑔𝑒
𝑇𝑖𝑚𝑒
aka
𝐼=
∆𝑄
∆𝑡
So if 1C worth of charge flows by in 1s, then 1A of current is flowing! The
ampere is the base unit of current but as 1A is actually a large current,
often values are given in mA, so don’t forget to convert to amperes.
ELECTRICAL QUANTITIES 2
Potential Difference
Potential difference is the more specific term for voltage, as it highlights
the transformation of energy from electrical into another type
It is the amount of electrical energy transferred by a component into
a different form per unit charge
In order to transfer the energy, the component is said to have done
work and this is why the symbol has changed.
𝑷. 𝒅(𝑽) =
𝑬𝒏𝒆𝒓𝒈𝒚 𝑻𝒓𝒂𝒏𝒔𝒇𝒆𝒓𝒓𝒆𝒅 (𝑱)
𝑪𝒉𝒂𝒓𝒈𝒆 𝑷𝒂𝒔𝒔𝒊𝒏𝒈(𝑪)
𝑾
𝑽=
𝑸
ELECTRICAL QUANTITIES 3
Electromotive Force
This is basically the supply voltage, i.e. if you were to put a voltmeter
around the power-pack or battery etc, this would be the reading
It is the energy transferred to each coulomb of charge from a supply.
It is given symbol of epsilon.
𝑬𝒏𝒆𝒓𝒈𝒚 𝑻𝒓𝒂𝒏𝒔𝒇𝒆𝒓𝒓𝒆𝒅 (𝑱)
𝑬𝑴𝑭(𝑽) =
𝑪𝒉𝒂𝒓𝒈𝒆 𝑷𝒂𝒔𝒔𝒊𝒏𝒈(𝑪)
ε=
𝑾
𝑸
QUANTITIES AND UNITS
• Some particles are said to have an
electric charge such as the
electron which has a negative
charge.
The charge on a single electron is 1.6
x 10-19C
6.25 x 1018
electrons to one C!
• This fundamental charge on a
particle is unchangeable, remember
we can’t destroy nor create charge
(conservation of charge)
KIRCHOFF’S LAWS:
CONSERVATION OF CHARGE
This rule states that the algebraic sum of all the currents entering a
junction is equal to zero
ΣI=0
In order to conserve electrical charge (law of conservation of charge), the
sum of all charge arriving at a point (a junction), must be equal to those
leaving it!
Q(T) = Q1 + Q2 + Q3 + Q4
So for this circuit:
And as I= Q/t, and t=1
I(T) = I1 + I2 + I3 + I4
Q1
Q2
Q3
Q4
And as I=V/R, and v=1
1
𝐼𝑡𝑜𝑡𝑎𝑙
1
1
1
1
1
2
3
4
=𝐼 +𝐼 +𝐼 +𝐼
If a current is arriving we give it a positive value and if it is leaving we give it
a negative value
KIRCHOFF’S LAWS: EMF
In order for energy to be conserved in any loop of a circuit (not
necessarily the whole circuit), the sum of EMFs must equal the sum
of the potential differences around a loop.
Σε= ΣV
Note: The potential differences are a result of ohms law calculations
based upon the current and the resistance. For this reason, the P.d.
may have a negative or positive value dependent on the direction of the
current
COMBINING RESISTANCE
We can combine resistors in series, and in parallel.
Rtotal = R1 + R2 + ... + Rn
Resistors in parallel work a little differently. The general formula for computing resistance
in parallel is:
1
𝑅𝑡𝑜𝑡𝑎𝑙
=
1
𝑅1
+
1
𝑅2
+
1
𝑅3
+
1
𝑅4
Don’t forget! When you’re calculating the total resistance, you end up with the value for
1/Rtotal. So remember to finish the step by finding the inverse (1/your answer).
If you have a combination of the two, you should work out the parallel total first then
use that in series with the rest!
RESISTANCE
A wire is made up of metal ions surrounded by a sea of free
electrons which flow (a current)
If the voltage increases, the temperature of a wire will
increase. This will cause the ions to vibrate with higher
amplitude and therefore increase the frequency of
electrons colliding with the ions. This would impede flow,
thus RESISTANCE INCREASES.
RESISTANCE GRAPHS
The gradient of a voltage-current graph tells us the
resistance of that component. The flatter it is, the greater
the resistance because it means we’re getting very little
increase in current!
Current (A)
0.18
In this case, the
changing which
means the
resistance is
fixed. It’s a
fixed resistor!
0.16
0.14
0.12
0.1
0.08
0.06
0.04
0.02
0
0
2
4
6
8
10
Voltage (V)
12
14
V-I RELATIONSHIPS
FIXED
RESISTOR
DIODE
• Resistance Unchanged •
• Directly Proportional
• Flatter=Higher Resistance
•
Resistance HIGH in
one direction but
LOW in the other
Forces current to only
flow in one direction
• Used to turn AC
current into DC
current
FILAMENT BULB
increases as the PD
goes up.
• This is because the
wire is heating up
(see resistance in a
wire)
SEMICONDUCTORS
Thermistors (left) and LDRs (right) are different to
normal wire components as they are semiconductors.
Semiconductors increase then number of charge
carriers available when exposed to energy (heat or light).
This would reduce resistance and increase current!
The MORE LIGHT/MORE HEAT  RESISTANCE
DECREASES  therefore INCREASING CURRENT
DRIFT VELOCITY
A metal consists of positive metal ions with free electrons that move with
random motion as they collide with the ions.
If a source of emf is connected across a metal, the electric field set up have
a tendency to push the electrons toward the positive end of the metal.
This ‘crawling’ motion towards the positive terminal is known as the
drift velocity
CALCULATING DRIFT VELOCITY
We consider the shape of a wire to be a cylinder, the current that is flowing by
can be calculated with the formula:
𝐼 = 𝑛𝐴𝑣𝑞
With ‘v’ being the drift velocity and ‘n’ the
number density of electrons. ‘A’ is the crosssectional area and the ‘q’ is the charge on a
single electron!
RESISTIVITY
Resistivity is similar to resistance with the exception that it’s an inherent
characteristic of the material itself, i.e.
“ a quantification of a material’s ability to resist the flow of electric
current”
So this means that it is a feature of the type of metal for example, rather than
anything to do with the dimensions of this metal in a wire.
𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒(Ω) =
𝑹𝒆𝒔𝒊𝒔𝒕𝒊𝒗𝒊𝒕𝒚 Ω𝒎 × 𝑆𝑎𝑚𝑝𝑙𝑒 𝑙𝑒𝑛𝑔𝑡ℎ(𝑚)
𝐶𝑟𝑜𝑠𝑠 − 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝐴𝑟𝑒𝑎 (𝑚2
𝜌𝑙
𝑅=
𝐴
INTERNAL RESISTANCE
When a current is flowing through a circuit, the electrons still have to travel
through the cell.
A small portion of the voltage/energy will be lost here because the electrons
have to overcome the internal resistance of the cell!
Because the cell has this internal resistance, the
convention is to show the cell adjacent to a
resistor.
This resistor would have a very low value.
ε =IR + Irinternal
(note, the r is the internal resistance)
•
IR = ε - Irinternal
•
V= ε – Irinternal
So this shows that the voltage produced at the terminals of the
cell (i.e. outside of the cell) is effectively the EMF – the voltage
overcoming internal resistance
DIVIDING POTENTIAL
• You can calculate
the ratio of the
resistors
• This allows you to
work out the
individual
contribution of
each resistor to
the total resistance
• E.g. R2’s
contribution to the
total resistance
(R1+R2)
RHEOSTATS
Instead of using two different resistors we can also just
use a single rheostat/potentiometer (variable resistor)
So by moving the contact, we can alter
the p.d across the terminals (Vout)
Just think of the remaining part of the
variable resistor as R1.
So just as before, work out the individual
contribution by calculating the ratio!
Note: this can be done in terms of length
along the rheostat!
UPTHRUST
𝒎
𝝆=
𝒗
As an object is
submerged in a fluid, it
will begin to displace it.
The volume of
the object that
is submerged
is equal to the
volume of the
water
displaced
So a submerged object will be
subjected to a force equal to the
WEIGHT of the fluid displaced
(UPTHRUST)
When an object is
floating, the value of
W will equal the U
To calculate U, you first
use the density
equation to calculate
mass from the volume
of fluid displaced. This
will be equal to the
volume of the object in
the fluid! USE THE
DENSITY OF THE
FLUID not the object
You then just muliply
by gravitational field
strength to get the
weight of that water!
FLOW LINES
Fluids can be said to move in either
laminar flow or turbulent flow.
The movement velocity of a fluid (gas or liquid) can be represented by
streamlines, which are arrowed lines showing the path taken by small
regions of fluid.
a) In laminar flow, like
slow moving water,
cross over one another
as there is no abrupt
changes in speed or
direction
b) However in turbulent
flow the fluid swirls
around in vortices or
eddy currents so
streamlines cross over
FLIGHT
As the air moves faster over the top, this means that there
is a difference in pressure on the wing. This means that
there is more force applied to the bottom of the wing,
resulting in uplift (and drag)
VISCOSCITY
Viscosity is defined as ‘the magnitude of internal friction
within a fluid’ a.k.a ‘how sticky it is’ and as such how
resistant it is to flow
The higher the
temperature of a liquid,
the lower its viscosity
which lets it flow at a
faster velocity
Greek symbol: Eta
Coefficient of viscosity
(Nsm-2)
η
The higher the
temperature of a gas,
the higher its viscosity
which lets it flow at a
slower velocity
DROPPING A SPHERE
F
When a sphere moves slowly
through a liquid, the relative
movement of the liquid around the
sphere is laminar.
U
As the molecules its passing
through will stick to the surface as
it travels, a viscous drag (F) is
created
This force was shown to be related
to the radius of the sphere, the
velocity of the sphere and the
coefficient of viscosity ( η )
W
CALCULATING TERMINAL
VELOCITY OF A SPHERE IN A
LIQUID
F
U
This relationship is called
Stoke’s law
sphere (m)
Drag force (N)
Velocity of
sphere (ms-1)
It is only obeyed when the sphere is not near the wall of the
cylinder (Bernoulli’s principle at play) and when the sphere is
at terminal velocity (when F+U=W)
W
EXTENSION
Whenever a force acts on a material, the materials will be
deformed into a different size or shape.
If a material is made longer, then the force is referred to
as tension and the extra length is known as extension
If a material is shortened, then the force is referred to as
compression and the reduced size is known as negative
extension
HOOKE’S LAW
Hooke’s law states that the force
needed to extend a spring is
proportional to the extension of the
spring UNTIL the limit of
proportionality has been reached
Force applied (N) = Stiffness constant
(Nm-1) x Extension (m)
∆𝐹 = 𝑘 ∆𝑥
HOOKE’S LAW: IN SERIES
If you combine springs, you
must combine their spring
constants.
For springs in series, if a
force is applied to them, they
will have an overall greater
extension (more stretchy)
To do this you treat them like
parallel resistors….
1
𝑘𝑡𝑜𝑡𝑎𝑙
=
1
𝑘1
+
1
𝑘2
HOOKE’S LAW: IN PARALLEL
For springs in parallel, if a
force is applied to them, they
share the force and as such
are harder to extend (less
stretchy)
their stiffness constants
K(total) = K1 + K2
DEFINITIONS
Yield Point
The point at which the
material will elongate without
plastically)
Elastic Limit
The maximum extension that
a material can undergo and
removed
Strength (Ultimate tensile)
The highest stress a material can take before breaking
(the peak of the graph)
ELASTIC
ENERGY
The work done in deforming a material
before it reaches the elastic limit will be
stored internally as
elastic strain energy (Eel)
The force varies with
different extensions so to
calculate the work done we
need to use the average
force over the distance of
extension
E(el)= ½ FΔ𝑥
A.K.A : The
area under the
graph!
STRESS/STRAIN
In physics, when stress is applied to a material, the material will
be put under strain (i.e. strain is the result of stress)
This is a little like force and extension except this time, it takes into
account the area of the material; so that it’s specific to the type of the
material that is used and is not specific to a certain dimension
BRITTLE VS DUCTILE
A brittle material can often be
strong (high UTS) but will break
soon after the elastic limit
A ductile material can be
stretched into wires because
it will have a large plastic
region (After elastic limit)
DIFFERENT METALS
In this case: HC steel is strong but brittle as it has a high UTS but breaks soon
after elastic limit. Copper is malleable and ductile as it undergoes significant
plastic deformation
YOUNG’S MODULUS
Named after Thomas Young, Young’s
modulus is the measure of the stiffness
of a material but taking into account the
dimensions.
I.e. it’s like the stiffness constant but for a
stress-strain graph! Just work out the
gradient of the linear part of the graph
Young’s Modulus:
Young modulus (E) =
𝑺𝒕𝒓𝒆𝒔𝒔 (𝑷𝒂)
𝑺𝒕𝒓𝒂𝒊𝒏 (𝒏𝒐 𝒖𝒏𝒊𝒕)
STRESS/STRAIN GRAPHS
As Young’s modulus is the gradient of the stress-strain graph
line, what is the area underneath?
Well as stress =
strain =
𝐹𝑜𝑟𝑐𝑒
𝐴𝑟𝑒𝑎
and
𝑒𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛
𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ
The area underneath would
be stress x strain aka:
𝐹𝑜𝑟𝑐𝑒
𝑒𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛
x
𝐴𝑟𝑒𝑎
𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ
Which is:
𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒
𝑣𝑜𝑙𝑢𝑚𝑒
Makes sense as it’s energy
per unit volume!
NATURE OF WAVES
Frequency= 1/period
TRANSVERSE WAVES
In transverse waves, the oscillations are perpendicular
to the direction the wave is travelling.
All electromagnetic waves are transverse waves!
Maximum displacement occurs at a crest or trough with
no displacement along the rest line.
LONGITUDINAL WAVES
In longitudinal waves, the oscillations move parallel to
the direction of the wave’s travel
Sound waves are longitudinal and they still have
wavelengths which can be measured from compression to
compression
POLARISATION OF LIGHT
“restricting the vibrations of a transverse wave wholly or
partially to a specific direction”
If you polarise light,
it is restricted to a
specific direction
and if you use a
second filter and
slowly turn it, the
intensity of light
getting through will
reduce. At 90°, no
light will get through
ECHO-PULSE TECHNIQUE
Just like sonar, an emitter sets out a pulse of sound and this
reflects off the object and then is detected. The time taken is
then used along with the speed of the sound to calculate the
distance travelled. DON’T forget that’s there AND back.
NATURE OF WAVES
All points in a wave will be at some point through their cycle.
One full cycle (360°) is one full wavelength (λ) which is 2π rads
λ π
Each peak would be ¼ of a wavelength (90 °/ /
)
𝟒
𝟐
PHASE DIFFERENCE
Phase Difference is
the difference in
phase(measured in
point of a wave and
the same point on
another wave of
identical frequency
(coherent) that started
at the same time
So in this example, the top
wave is π/2 ahead of the
bottom one!
PATH DIFFERENCE
Path difference is the difference in distance that two waves
have travelled when they reach the same point (P).
Measured in wavelengths
P
Constructive=
Whole number
wavelengths path
difference
Destructive=
Half wavelengths
path difference
SUPERPOSITIONING
If two waves interfere in
phase the waves will make a
resultant with a larger
amplitude
This is known as
constructive interference
and only works when they
have phase differences of
e.g. (0, 2π, 4π)
If their phase difference is
Then they interact
destructively (e.g. π, 3π,
5π)
INTERFERENCE
REMEMBER
One full cycle (360°)
is one full wavelength (λ) which is 2π rads
Coherent = have the same wavelength and
frequency and a fixed phase difference
between them
STANDING WAVES
A standing wave is the super-positioning of two progressive
waves that are coherent in opposite directions
From node to node or antinode to antinode will be half a
wavelength. So for example the third harmonic will be three half
wavelengths (1 ½ lambda)
Maximum Kinetic
energy is absorbed
by the wave at an
antinode but none is
used at a node. This
is because at an
antinode, there is
maximum
amplitude
LENSES
Lenses can both magnify and focus. There are two types of lenses you need
to deal with in A-level.
Converging lenses are convex in shape
Diverging lenses are concave in shape
A real image is formed when the focal point is actually formed where the light
converges. A virtual image is when the focal point forms where the light
appears to focus and appears behind the lens on a diverging lens
LENS POWER
The power of a lens relates to the ability of the lens to refract the light that
passes through it.
“The stronger the power of a lens, the more it refracts light, so it
produces a focal point closer to the lens (short focal length)”
The unit of power is the dioptre(D) which is equivalent to (m-1)
1
𝑃𝑜𝑤𝑒𝑟 𝑜𝑓 𝑎 𝑙𝑒𝑛𝑠 =
𝑓𝑜𝑐𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ
The power of a converging lens is always a positive value, and that of a
diverging lens is a negative value. If lots of lenses are in combination, you
𝑇𝑜𝑡𝑎𝑙 𝑝𝑜𝑤𝑒𝑟 (𝑃𝑐) = 𝑃1 + 𝑃2 + 𝑃3
DRAWING LENS DIAGRAMSCONVERGENT
1. Draw the principle ray (horizontal to the lens then down through the focal length
of the lens)
2. Draw the optical focus ray (through the centre of the lens and through the other
side)
3. Where they cross is the point that the image is produced. If you compare the
image distance to the object distance, this will tell you the magnification
DRAWING LENS DIAGRAMSDIVERGENT
1. Draw the principle ray (horizontal to the lens then up through at an angle that
would make the line APPEAR to hit the focal length to the left of the lend
2. Draw the optical focus ray (through the centre of the lens and through the other
side)
3. Where they APPEAR to cross is the point that the image is produced. If you
compare the image distance to the object distance, this will tell you the
magnification
LENS EQUATION
There is a relationship between the object distance (u), the image distance
(v) and the focal length of the lens (f)
Object Distance (u)
Image distance (v)
Focal length(f)
Don’t forget, if you are using a diverging lens, use a negative value for focal
length or power!
1 1
1
+ =𝑓
𝑢 𝑣
REFRACTION
Refraction is the bending of light due to a change in speed which is a result of the
light travelling through a change in density
REMEMBER: the angle of
incidence and angle of
reflection are the angles
from the NORMAL to the
ray.
‘n’ is the refractive index
(how much the material
They have the following
relationship (like
equivalent ratios)
𝑆𝑖𝑛∅1
𝑆𝑖𝑛∅2
=
𝑛1
𝑛2
CRITICAL ANGLE
The critical angle is the angle of incidence that results in the ray refracting at 90°.
Any angle greater than this will result ins total internal reflection (TIR)
As ∅2 would be 90°, and
Sin90 is equal to ‘1’. At
critical angle, this can be
omitted from the equation!
They have the following
relationship (like
equivalent ratios)
𝑆𝑖𝑛∅𝑐 =
𝑛1
𝑛2
NATURE OF WAVES
Diffraction is the
as they pass through an
aperture
Diffraction is the
as they pass through an
aperture
INTERFERENCE
Peak
Trough
Wave 1
Where the peaks cross, we know the waves
must be in phase (green dots)
Wave 2
Where it is a peak for one wave but the
trough for another then the waves at that
point are out of phase (red dots)
Easy: if a point that has both line cross =
constructive and if only on one
line=destructive
INTERFERENCE PATTERN
So if a screen is placed away
from the interfering waves
you’ll get a pattern like this…
There is a central maximum
and fringes of brightness
(maxima) where light
interferes constructively
and fringes of darkness
(minima) where light
interferes destructively
If the source was sound,
you’d hear the source as
loud in the maxima and quiet
in the minima
DOUBLE SLIT EXPERIMENT
𝑆𝑥
λ=
𝐷
Note: when taking the
measurement of fringe
separation you need to
take it from the centre
of the bright fringe to
the centre of the next.
To reduce uncertainty
you should take the
measurement of a
number of fringes and
then divide by the
number (e.g. 6 fringes,
then divide by 6).
DIFFRACTION
GRATINGS
White light is a mixture of
colours and if you put it
through a diffraction grating,
the different colours
(different wavelengths) of
light are diffracted by
different amounts
Each order in the pattern
becomes a spectrum with red
on the outskirts and violet
more central (longer
wavelengths diffract more)
The central maxima will be
white as all light will pass
straight through
EVIDENCE FOR WAVE
PARTICLE DUALITY
WAVE
PARTICLE
PHOTONS
A ‘photon’ is a discrete ‘packet’ of
energy. The energy of that photon is
directly proportional to its frequency.
Frequency if not provided can be
worked out by c/λ.
Electrons in atoms exist in discrete
energy levels given by whole
numbers (e.g. n=1). If electrons move
down an energy level, they emit a
packet of energy known as a photon
There are set values for the energy of
these photons due to the set energy
levels. Each photon would therefore
have a set frequency dependent on
the transition from one energy level to
another.
EMISSION SPECTRA
If you heat a gas to a high
temperature, the electrons of the
level
They will then quickly drop back
down to their energy level and
release a photon with equivalent
energy (and as such, frequency)
If the light from a hot gas is split
through a prism, it will produce an
emission spectrum of coloured lines
on a black background
These correspond to the different
wavelengths of photons emitted. As
there are only discrete energy levels
to drop from/to, there are specific
wavelengths produced.
ABSORPTION SPECTRA
If you heat a gas to a high
temperature, the electrons of the
level
They will then quickly drop back
down to their energy level and
release a photon with equivalent
energy (and as such, frequency)
If the light from a hot gas is split
through a prism, it will produce an
emission spectrum of coloured lines
on a black background
These correspond to the different
wavelengths of photons emitted. As
there are only discrete energy levels
to drop from/to, there are specific
wavelengths produced.
CORE PRACTICAL 1
Method:
1. Drop a dowel of measured
length (l) through a guiding
pipe through light gates
2. Record the time taken for
the dowel to go through the
light gate
3. Repeat for different
heights of release
4. Use the value of time for
the dowel to pass through
to calculate the speed it
was travelling through
(s=d/t)
Graph:
1. Plot a graph of v2 against height
2. Work out the gradient (m) of the line of
best fit
Reducing uncertainty:
1. Taking measurement of height on the level
(avoid parallax error)
2. Half of range of readings for t as
uncertainty of t
3. Use a high resolution ruler (30cm ruler) for
measurement of dowel length
Dowel needs to pass straight through the
beam
CORE PRACTICAL 2
Determine the resistivity of a material
Method:
1. Get a meter rule with a
metal wire attached and
attach a crocodile clip at the
zero end
2. Use a 4mm plug at the free
end of the second lead to
make a circuit with the zero
end and a digital
multimeter set to ohms
3. Every 10cm take a reading
of resistance
4. Use a micrometer to find
the diameter of the wire
used to calculate the cross
sectional area
Graph:
1. Plot a graph of R against I and measure
2. Resistivity= m x A (A being the cross
sectional area of the wire)
Reducing uncertainty:
1. Diameter readings should be taken 5
times along the wire, in both a vertical and
horizontal direction and then an average
taken
2. Look at the connection from above to
reduce parallax error
CORE PRACTICAL 3
Determine the emf and internal resistance of a cell
Method:
1. Set up the circuit as shown
with a 100ohm resistor in
series with a cell and a
variable resistor
2. Measure the potential
difference (V) across the
variable resistor and the
current going through it.
Graph:
1. Plot a graph of V against I.
2. This gives a gradient= -r (the internal
resistance)
3. This is because V=-Ir + EMF
CORE PRACTICAL 4
Using the falling-ball method to determine the viscosity of a liquid
Method:
1. Get a variety of different
their mass, use
density=mass/volume to
measure their density
2. Place three rubber bands
around a tall tube, with the first
far down enough to ensure the
ball is moving at terminal
velocity
3. Calculate the time taken for the
ball to pass from one band to
another and measure the
distance between these bands
4. Use v=d/t to find v and then do
this for all balls of this diameter
and then again for the other
balls
Use the value of ‘v’ and ‘r’ etc in the
Stoke’s law equation to calculate the
viscosity of the liquid.
Reducing uncertainty:
1. The ball must not go near the sides of the
tube as this will introduce a new source
of drag as explained by bernoulli’s
principle.
2. When considering the ball moving past a
point, this must be viewed straight on as
to remove parallax error
CORE PRACTICAL 5
Determine the Young modulus of a material
Method:
1. Fix the bench pulley at the end of the bench. Trap
one end of the wire between the two wood blocks
and secure these to the bench approximately 3 m
from the pulley. Lay out the wire so that it passes
over the pulley and attach the slotted mass hanger
to the end. Measure the diameter d of the wire.
2. Lay the metre ruler under the wire near the pulley
and attach the sticky label to act as a length marker.
You judge the length by looking vertically down over
the edge of the paper onto the scale of the metre
ruler.
3. Measure the length of wire L from the wood blocks
to the edge of the paper.
4. Add masses to the hanger and record the position
of the marker against the metre ruler. Calculate the
extension x for each mass added.
5. You might notice significant creep occurring at
higher loads. This indicates the elastic limit has been
exceeded
Use the value of mass
against extension, the
modulus
Reducing uncertainty:
should be taken 5 times
along the wire, in both a
vertical and horizontal
direction and then an
average taken
2. Look at the connection
from above to reduce
parallax error
CORE PRACTICAL 6
Determine the speed of sound in air using an oscilloscope, a generator etc..
1. The oscilloscope will display on two traces the signal fed to the loudspeaker and the signal
received by the microphone. As the distance between the microphone and the speaker is
increased, the phase of the signals varies and the traces on the screen move past each other.
2. Place the microphone next to the oscilloscope and place the speaker about 50 cm away, facing
the microphone. Turn on the signal generator and set it to about 4 kHz. Adjust the oscilloscope
to show the microphone signal with about three cycles on the screen.
3. Connect the signal generator output to the second oscilloscope input (as well as the speaker)
and adjust the controls to display three cycles of this signal.
4. Adjust the spacing on the screen and the distance between the speaker and microphone so
that the bottom of one trace is just level with the top of the other.
5. Adjust the separation so that a trough on the top trace exactly coincides with a peak on the
lower trace. Place the metre ruler alongside the microphone and speaker and record the
distance between the microphone and speaker.
6. Move the speaker away from the microphone and observe one trace sliding over the other.
Move the speaker so that the trace has moved exactly one cycle. The troughs and peaks
should just touch again. Record the new distance between the microphone and speaker. The
difference between the two distances is one wavelength.
7. Continue to move the speaker away from the microphone and record each successive distance
where the peaks of one trace coincide with the troughs of the other.
8. Calculate a mean value for the wavelength of the sound and estimate the uncertainty in this
measurement.
9. Use one of the traces to determine the frequency of the sound. You will achieve a greater
resolution this way than using the scale on the signal generator. Estimate the uncertainty in this
measurement. You should be able to measure the frequency to three significant figures and the
wavelength to at least two.
10. Using the scale on the signal generator, halve the frequency and repeat the measurements for
frequency and wavelength. You might need to increase the separation beyond 1 m.
11. You might try this experiment at much higher and lower frequencies to observe the effect.
CORE PRACTICAL 7
Investigate the effects of length, tension and mass per unit length on the frequency of
a vibrating string
Method:
1. Attach one end of the ‘string’ to the
vibration transducer. Pass the other
end over the bench pulley and attach
the mass hanger.
2. Add masses until the total mass is
100 g.
3. Turn on the signal generator to set
the rubber oscillating. Vary the
oscillating length by moving
the vibration generator until
resonance is observed.
4. In this investigation, you are
observing standing
waves. These can occur at a variety
of resonant frequencies.
5. Can change either the length of the
string or the mass added (therefore
changing tension)
So for example: Keeping the Tension
the same (1.96N) you could plot a graph
of wavelength against (1/f) to get a value
of μ (mass per unit length)
CORE PRACTICAL 8
Investigate the effects of length, tension and mass per unit length on the frequency of
a vibrating string
Method:
1. Place the laser approximately 4 m away from
a large wall and place the diffraction grating in
front of it. Arrange for the beam to pass through
the grating at normal incidence and meet the wall
perpendicularly.
2. Measure the distance D between the grating
and the wall.
3. Turn on the laser and identify the zero order
maximum (straight through). Measure the
distance ‘s’ to the nearest two first order maxima.
Calculate the mean of these two
values. The first order is the maximum produced
according to n = 1 in the equation
nλ = d sin θ
4. Measure s for increasing orders.
5. Repeat for a diffraction grating with a different
number of slits/mm.
You can find the angle
by using trigonometry
(the distance to the wall
and then the distance
between two fringes)
We use
monochromatic light
as white light consists of
multiple frequencies and
wavelengths of light.
These will diffract by
different amounts,
causing the fringes to
be a coloured blur with
red on the outside and
violet towards the
central maxima
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