Academic Support Service of the
Engineering Society of Queen’s
University
WORKSHOP Solutions
APSC 112 FINAL EXAM REVIEW
ENGLINKS.CA
ENGLINKS@ENGSOC.QUEENSU.CA
OSCILLATIONS AND HOOKE’S LAW
By using simple harmonic motion, it is possible to solve any form of oscillation problem – either linear or
angular.
Any equation of the form:
𝑑2 𝑥
+ 𝜔2 𝑥 = 0, 𝜔2 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
2
𝑑𝑡
is a simple harmonic motion oscillator. It has the following solution:
𝑥 (𝑡) = 𝑥𝑚 cos(𝜔𝑡 + 𝜙) 𝑜𝑟 𝜃 (𝑡) = 𝜃𝑚 cos(𝜔𝑡 + 𝜙)
𝑥𝑚 /𝜃𝑚 , 𝜔 and 𝜙 are constants, and x(t) has a maximum value when cos(𝜔𝑡 + 𝜙) = ±1.
LINEAR OSCILLATOR
1
ANGULAR OSCILLATOR
2
PENDULUMS
There are two types of pendulums: simple
pendulums and physical pendulums. A simple
pendulum is the “ball on a string”, and it acts like
exactly like a simple harmonic oscillator. A
physical pendulum is more complex – it does not
have all of its mass at a single point, so we
cannot use conservation of energy. Instead, we
use the following equation:
∑𝜏 = 𝐼𝛼 = 𝑟⃑ × 𝐹⃑
You can still get SHM with this equation, since 𝜏 depends on 𝜃 and 𝛼 =
𝑑2 𝜃
𝑑𝑡 2
3
EXAMPLE 1
4
5
6
7
8
9
WAVES
Transverse Waves: Waves that move perpendicular to the direction of propagation e.g. waves on a rope.
Longitudinal Waves: Waves that move parallel to the direction of propagation e.g. waves on a slinky.
Waves are functions of both position and time, since every point along the wave moves differently. They
have an amplitude A, a wavelength λ, a frequency f, an angular frequency 𝜔 = 2𝜋𝑓, a period T, and a
wave number 𝑘 =
2𝜋
.
𝜆
The general wave height equation is
𝑦(𝑥, 𝑡) = 𝐴𝑐𝑜𝑠(𝑘𝑥 ± 𝜔𝑡 + 𝜙)
which specifies the height of the wave at position x during time t.
Other useful equations are:
𝜆
𝑣=𝑇
This is the equation for the speed of a wave given the wavelength λ and the period T.
Additionally, a useful conversion between a wave’s period and its frequency is
𝑓=
1
𝑇
Which makes the first equation
𝑣 = 𝜆𝑓
NOTE! If a problem specifies that the wave is light, there is a high probability that the speed of
the wave is c=3*108m/s, the speed of light in a vacuum.
SPEED OF A TRANSVERSE WAVE
Note: 𝑣 = √
𝐹
𝜇
(F = force on rope, 𝜇 = density = mass/length)
10
STANDING WAVES
If both ends of a rope are fixed, the waves are standing – the endpoints of the wave are always at y=0.
The equation for these waves is:
𝑦 = 2𝐴𝑠𝑖𝑛(𝑘𝑥 )sin(𝜔𝑡)
and since the waves can’t have the endpoints moving, there are only a few possible wavelengths:
𝜆𝑛 =
2𝐿
𝑛
where n is an integer and L is the length of the rope
The points of zero height are called nodes, while the points of maximal amplitude are called antinodes.
They can occur more than once along the rope.
11
EXAMPLE 2
A horizontal string is tied at its two ends and vibrates at its fundamental mode. Calculate its maximum
transverse velocity and acceleration. If it has amplitude A and wavelength .
12
SUPERPOSITION
Waves can be added together to create a new wave.
Areas where both waves line up constructively interfere
to create bigger waves, while areas where both waves
are opposite destructively interfere to cancel each other
out.
In sound waves, these can cause beats. Beats occur
when the frequency of the two waves are very close to each other.
The “Beat Frequency” is given by 𝑓𝑏𝑒𝑎𝑡 = 𝑓1 − 𝑓2
EXAMPLE 3
One car has a motor turning at 575 rpm. If it is sitting beside a motor cycle and you hear a 2 Hz beat, what
are the possible frequencies of the motorcycle’s motor?
13
Doppler Effect
When a sound or listener is moving, sound waves tend to pile up in front of the source and spread out
behind it. The general formula for moving sources and listeners is:
𝑓𝑙𝑖𝑠𝑡𝑒𝑛𝑒𝑟 =
𝑣 ± 𝑣𝑙𝑖𝑠𝑡𝑒𝑛𝑒𝑟
𝑓
𝑣 ∓ 𝑣𝑠𝑜𝑢𝑟𝑐𝑒 𝑠𝑜𝑢𝑟𝑐𝑒
Where v is the velocity of sound in air and vsource < v.
Toward
𝑓𝑙𝑖𝑠𝑡𝑒𝑛𝑒𝑟 =
𝑣 + 𝑣𝑙𝑖𝑠𝑡𝑒𝑛𝑒𝑟
𝑓
𝑣 − 𝑣𝑠𝑜𝑢𝑟𝑐𝑒 𝑠𝑜𝑢𝑟𝑐𝑒
Away
𝑓𝑙𝑖𝑠𝑡𝑒𝑛𝑒𝑟
𝑣 − 𝑣𝑙𝑖𝑠𝑡𝑒𝑛𝑒𝑟
=
𝑓
𝑣 + 𝑣𝑠𝑜𝑢𝑟𝑐𝑒 𝑠𝑜𝑢𝑟𝑐𝑒
EXAMPLE 4
You are driving at 35m/s when you hear a police siren approaching from behind. You perceive the siren’s
frequency to be 1370Hz. The police car passes you and you now hear a frequency of 1330Hz. What is the
speed of the police car? (Assume speed of sound to be 343m/s).
14
ELECTRIC FIELDS AND POTENTIALS
COULOMB’S LAW
|𝐹⃑ | = 𝑘
𝑞1 𝑞2
𝑟2
F is the electric force between two charged particles. Note that F is repulsive
(positive) when q’s are the same sign, and attractive (negative) when q’s are
different.
F points in the direction of the line between the two particles.
⃑⃑⃑⃑⃑⃑⃑⃑⃑
𝐹𝑛𝑒𝑡 = ∑ ⃑⃑⃑
𝐹𝑖
𝑖
This force obeys the superposition principle. That is, the net electric force
acting on a particle is the vector sum of all the electric forces acting on it.
ELECTRIC FIELD
𝐸⃑⃑ =
𝐹⃑
𝑞0
The electric field is defined as the electric force per unit charge. By
convention, q0 is defined to be positive. Equivalently, a particle with charge
+q0 placed in an electric field E would feel a force F in the direction of E.
For a point charge q:
|𝐸⃑⃑ | = 𝑘
𝑞
𝑟2
E is generated by a source of charge. Electric field lines flow outward from positively charged sources,
and flow into negatively charged sources.
Calculating E in the space around the source of charge allows you to calculate the force it would exert
on an external charge q’ through F = q’E. This, positive charges want to move in the direction of E
lines, while negative charges want to move in the opposite direction.
⃑⃑⃑⃑⃑⃑⃑⃑⃑
⃑⃑⃑⃑
𝐸
𝑛𝑒𝑡 = ∑ 𝐸𝑖
𝑖
Since the electric force obeys the superposition principle, so does the electric
field. The net E at a point P is equal to the vector sum of all electric field
lines passing through P.
15
ELECTRIC POTENTIAL ENERGY
Like gravitational potential energy (given by mgh), the electric potential energy needs to be defined with
respect to a certain reference point. Thus, it is more useful to talk about a systems’ change in potential
energy—which would be the same regardless of choice of reference point—than its absolute energy:
∆𝑈 = 𝑈2 − 𝑈1 = −𝑊1→2
The difference in potential energy between point 2 and point 1 is equal
to the negative of the work that would need to be done by something
(or “someone”) external against the electric (Coulomb) force in
moving a charge from point 1 to point 2.
If the reference point is chosen to be at infinity (i.e. we let U(∞) = 0), for a collection of point charges,
2
U can be calculated from 𝑊1→2 = ∫ 𝐹⃑ ∙ 𝑑𝑠⃑ using the Coulomb force for F. This gives:
1
𝑈 = ∑ 𝑈𝑖𝑗 =
𝑖<𝑗
𝑞𝑖 𝑞𝑗
1
∑
4𝜋𝜀0
𝑟𝑖𝑗
𝑖<𝑗
if j =1:
There are no i<j.
I.e., a point
charge sitting by
itself has no
potential energy
if j = 2:
𝑈 = 𝑈12 = 𝑈21
if j = 3:
𝑈 = 𝑈12 + 𝑈13 + 𝑈23
if j = 4:
𝑈 = 𝑈12 +
𝑈13 + 𝑈23 +
𝑈14 + 𝑈24 + 𝑈34
EXAMPLE 6
Assume the magnitude of all of the charges in the initial configuration is 1nC.
16
17
ELECTRIC POTENTIAL
Electric potential is the amount of electric potential energy per unit positive charge, q0. This is similar to
how the electric field is the electric force per unit positive charge q0. Potential energy has units of Joules
(J), so potential has units of J/C, or Volts (V).
𝑉=
∆𝑉 =
𝑈
𝑞0
∆𝑈 −𝑊1→2
=
= 𝑉2 − 𝑉1
𝑞0
𝑞0
A particle with charge +q0 placed at potential level V would
have a potential energy of U. Thus, like U, V must also be
defined with respect to a reference point. If we choose V(∞)
= 0, for a collection of point charges:
𝑉 = ∑ 𝑉𝑗 =
𝑖
𝑞𝑗
1
∑
4𝜋𝜀0
𝑟𝑖
𝑖
Note that V, like U, is a scalar quantity, and thus the direction of the vector r that points from the source
charges is not required (unlike for E or F), just its magnitude, r. Also, like E, electric potential is
generated by a source of charge into the space around it. You can calculate the electric potential energy of
a charge q’ sitting at an electric potential V from U = q’V. Note that regions of highest potential energy
and potential are where the magnitude of E is largest, which is where the E lines are “closest together”,
which is close to sources with positive charge.
18
CHARGE DISTRIBUTIONS
Point charges aren’t usually found in nature. Instead, distributions of charge over lines and surface are
more realistic. For these problems, create an expression for an infinitesimal amount of charge, dq, of the
form
𝑑𝑞 = 𝜆𝑑𝑠
𝑑𝑞 = 𝜎𝑑𝐴
λ: (C/m), linear charge density.
ds: infinitesimal distance
σ: (C/m2), surface charge density.
dA: infinitesimal area
Charge densities σ and λ, are simply defined as the total amount of charge divided by the total area the
charge is spread out over (for σ) or the total linear distance the charge is spread out over (for λ). I.e., for a
disk (2-d surface) and ring (1-d curved line), both with radius R and total charge Q, we have σ = Q/πR2
and λ = Q/2πR respectively.
In many cases, components of E (ie Ex, Ey or Ez) cancel through symmetry. Then, you can calculate the
magnitude of E, and infer the direction it should point in. Note that it is generally much easier to calculate
spatial integrals with V, since it is independent of direction entirely.
19
EXAMPLE 7
20
EXAMPLE 8
21
22
23
RELATIONSHIP BETWEEN E AND V
The potential between two points 1 and 2 is equal to the line integral of E on a path from 1 to 2. In other
words, you compute the projection of E onto the spatial path that connects points 1 and 2. This
calculation is independent of the path that you choose to take, just as long as it starts on 1 and ends on 2.
Mathematically, this is represented by
2
𝑉2 − 𝑉1 = − ∫ 𝐸⃑⃑ ∙ 𝑑𝑠⃑
1
representing E as
𝐸⃑⃑ = 𝐸𝑥 𝑖⃑ + 𝐸𝑦 𝑗⃑ + 𝐸𝑧 𝑘⃑⃑
2
𝑉2 − 𝑉1 = − ∫ 𝐸𝑥 𝑑𝑥 + 𝐸𝑦 𝑑𝑦 + 𝐸𝑧 𝑑𝑧
1
The projection of E is accomplished by the dot product 𝐸⃑⃑ ∙ 𝑑𝑠⃑. This is how to “get V from E”.
Remembering that the potential is a function that can depend on 3 spatial variables, i.e. V = V(x,y,z), the
opposite form of this relationship is given by
⃑⃑𝑉 = −
𝐸⃑⃑ = −∇
𝜕𝑉
𝜕𝑉
𝜕𝑉
𝑖⃑ −
𝑗⃑ −
𝑘⃑⃑
𝜕𝑥
𝜕𝑦
𝜕𝑧
⃑⃑ is called the gradient operator. It expresses how the function
∇
is changing is each direction.
This is how to “get E from V”.
In summary, V decreases in the direction of the electric field, and conversely E points in the direction of
decreasing V. Equipotential surfaces depict shapes, or lines across which the electric potential remains the
same. If point 1 and point 2 are on an equipotential surface, i.e. V2 = V1, then no net work against the
electric (Coulomb) force is required to move from point 1 to point 2. Thus, E lines must point in the
direction perpendicular to equipotential surfaces at all points.
24
ELECTRIC DIPOLES
Two charges, with charge of equal magnitude, q, but opposite sign, that are separated by a distance d,
form an electric dipole. The electric dipole moment is defined as
𝑝⃑ = 𝑞𝑑⃑
p is useful in calculating interactions of the dipole with an external E:
𝜏⃑ = 𝑝⃑ × 𝐸⃑⃑
This torque will attempt
to align p with E.
𝜃𝑓
𝑊 = ∫ |𝜏⃑| 𝑑𝜃
𝜃𝑖
𝑊 = 𝑝⃑ ∙ 𝐸⃑⃑ |
𝜃𝑓
𝜃𝑖
This is the work done
on the dipole by E
through an angular
displacement.
𝜃𝑓
𝜃𝑖
To calculate V or E generated by a dipole, consider each charge separately and then sum the two results to
get the total V or E (principle of superposition!)
𝑈 = −𝑊 = −𝑝⃑ ∙ 𝐸⃑⃑ |
25
ELECTRIC CIRCUITS
CURRENT AND RESISTIVITY
Current is defined as the amount of charge passing through a wire per unit time. 1 Ampere is defined as 1
Coulomb per second. Explicitly,
𝑑𝑞
for steady currents, (one that doesn’t change over time): 𝑞 = 𝑖𝑡
⇒ 𝑑𝑞 = 𝑖𝑑𝑡
𝑑𝑡
The current density, J, is defined at the amount of current passing through a unit area, |𝐽⃑| = 𝑖⁄𝐴 and
points in the direction of the current.
𝑖=
Resistivity, ρ, is a built-in property of materials, and defined as the ratio of the magnitude of an external E
applied to the material and the magnitude of the current density that is observed to flow after applying
that electric field. I.e., 𝜌 = 𝐸⁄𝐽. The resistivity of a material varies with temperature, and this relationship
is described by
Where ρ is the resistivity at temperature T, and ρ0 is the resistivity at
temperature T0. The parameter α is another built-in property of the material,
called the temperature coefficient of resistance.
Note: Resistance (seen in next section), also varies with temperature in the exact same way as ρ.
𝜌 = 𝜌0 [1 + 𝛼(𝑇 − 𝑇0 )]
26
OHM’S LAW AND POWER
Ohm’s Law says the “voltage difference” (or electric potential difference) across or over a circuit element
is directly proportional to the current through that circuit element. This is mathematically summarized by
𝑉 = 𝐼𝑅
Where R is the proportionality constant, and is called the resistance of that current element. It measures
how “resistive” the element is to voltages differences (or equivalently, external electric fields). The higher
the resistance, the lower the amount of current that will be able to flow through the element for a given
potential difference, and vice-versa.
The power, in units of energy per time (i.e. 1 Watt (W) is one Joule per second), that is either generated
(P > 0) or dissipated (P < 0) by a circuit element is calculated by
𝑃 = 𝑉𝐼
or, using Ohm’s law, we can also say:
𝑃 = 𝐼2 𝑅 =
𝑉2
𝑅
CIRCUIT ANALYSIS
In this course, there are two basic circuit elements.
One supplies a rise in electric potential, or
“voltage”, and is called an electromotive force, or
EMF. These elements are usually represented by
batteries. For an EMF given by ε, the potential
rise from one end (terminal b) to the other end
(terminal a) of the device is given by 𝑉𝑎 − 𝑉𝑏 = 𝜀.
The other element is the resistor, which is a
device with a resistance R. The potential
difference between the two ends (terminals) of the
resistor is always negative when going in the
direction of the current. I.e., if current i is flowing
from terminal a to terminal b, 𝑉𝑎 − 𝑉𝑏 = −𝑖𝑅.
27
An extremely useful tactic in circuit analysis is to combine resistors and their resistances into an
equivalent resistance. There are two different ways this can be accomplished:
If a collection of resistors lie on the same branch
of a circuit, that is, if the current through each of
them is the same, then they are “in series”. The
equivalent resistance of these resistors is given by
If a collection of resistors are connected at their
terminals, that is, if the voltage difference across
each resistor in a number of different branches is
the same, then they are “in parallel”. The
equivalent resistance of these resistors is given by
For N resistors in series:
For N resistors in parallel:
𝑁
𝑅𝑒𝑞 = ∑ 𝑅𝑖 = 𝑅1 + 𝑅2 + ⋯ + 𝑅𝑁
𝑖
For 2 resistors in series:
𝑅𝑒𝑞 = 𝑅1 + 𝑅2
𝑁
1
1
1
1
1
=∑ =
+
+ ⋯+
𝑅𝑒𝑞
𝑅𝑖 𝑅1 𝑅2
𝑅𝑁
𝑖
For 2 resistors in parallel:
1
1
1
=
+
𝑅𝑒𝑞 𝑅1 𝑅2
𝑅1 𝑅2
⇒ 𝑅𝑒𝑞 =
𝑅1 + 𝑅2
28
Note: you can never have a resistor generate power. That is, for a resistor, P ≤ 0, ALWAYS. Power is
“dissipated” by resistors, and is given off as heat.
An emf, however can, both generate and absorb power. If P < 0, the battery is absorbing power, and if P
> 0, it is delivering power to the circuit.
Relationships between voltages and currents in the circuit can be found using Kirchoff’s Laws:
Kirchoffs Junction/Current Law: The sum of the currents entering a junction must equal the sum of the
currents leaving it. This is derived from the empirical law of the conservation of charge.
Kirchoffs Loop/Voltage Law: The sum of the voltage/potential rises and drops around any closed loop
(one that starts and ends at the same point) must equal zero. This is derived from the empirical
law of the conservation of energy.
A few quick notes:
-
The voltage difference between any two points a and b in the circuit is equivalent to the
voltage/potential drops and rises along any path that connects a and b.
If you calculate that a current to be negative, it just means that your initial guess for the direction
of the current is backwards. The actual current runs in the direction opposite to the one you chose,
but the magnitude of the current stays the same.
29
EXAMPLE 9
30
31
Alternative to 1b)
32
33
34
MAGNETIC FIELDS AND MAGNETIC FORCE
Unlike with electric charges, magnetic fields do not come from some “magnetic charge”. Instead, they are
caused by the motion of charges (current), and in turn they affect the motion of charges.
If we assume that a charge q moves through some magnetic field B with a velocity v, the force on the
charge is going to be
𝐹𝐵 = 𝑞 ∗ (𝑣 𝑥 𝐵)
The force on the charge is always at right angles to both the velocity and the field (this is a property of the
cross product). Calculating a cross product can be hard, but there are a few ways of doing it and a few
tricks you can keep in mind, such as the right hand rule
If there is also some electric field, the total force is given by the superposition of the electric and magnetic
forces:
𝐹 = 𝐹𝐸 + 𝐹𝐵 = 𝑞𝐸 + 𝑞(𝑣𝑥𝐵)
35
CIRCULATING CHARGES
⃑⃑ going
Suppose there is a uniform magnetic field 𝐵
into the page. If a particle q moves with velocity 𝑣⃑,
it wi
ll experience a force of magnitude |𝐹| = 𝑞𝑣𝐵. From
Newton’s second law, this force translates into
circular motion and we get that
𝐹 = 𝑚(
𝑣2
𝑞𝐵𝑟
𝑚𝑣
,𝑟 =
) = 𝑞𝑣𝐵 => 𝑣 =
𝑟
𝑚
𝑞𝐵
We can also calculate the period and angular frequency using the fact that 𝑣 = 𝜔𝑟.
If the particle also has a motion parallel to B, it will move in a helix
around the axis of B. The pitch p of the helix is the distance between
turns, and it is defined as
𝑝 = 𝑣𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑇 =
2𝜋𝑚𝑣𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙
𝑞𝐵
36
FORCE ON A CURRENT-CARRYING WIRE
A wire with constant current I in a field B will experience a force 𝐹𝐵 = 𝐼(𝐿𝑥𝐵) where L is the length and
direction of the wire. As such, wires that are free to move will do so when exposed to magnetic fields.
EXAMPLE 10
A square loop of wire of side length a carries current I. It is free to rotate along the axis as shown. What is
the torque on the loop if it is displaced on angle  inside a field B=Bzk^.
37
SOURCES OF MAGNETIC FIELDS
Magnets are of course sources of magnetic fields, but so are current-carrying wires!
BIOT-SAVART LAW
The magnetic field at some point in space depends on its
distance from currents. We will calculate these magnetic
field points in much the same way as we calculated the
electric field – by integration along the whole wire.
𝑑𝐵 =
𝜇𝑜 𝑑𝑠 𝑥 𝑟
𝐼
4𝜋
𝑟3
Each little segment ds will create a magnetic field at point P.Its direction is given by the right hand rule,
and all the bits of field dB add together by superposition to give the total field B.
𝐵=
𝜇0
𝑑𝑠 𝑥 𝑟
𝐼∫ 3
4𝜋
𝑟
EXAMPLE 11
Find the B field at P
38
39
EXAMPLE 12
𝜇 𝐼
0
Recall that an infinetly long wire creates a B field = 2𝜋𝑟
. What is the field above an infinetly long slab of
metal of thickness a that has a uniformly distributed current I.
40
MAGNETIC INDUCTION
FLUX
The flux of a surface is the amount of magnetic field that passes through it. For a flat surface you can
think of it as the number of field lines that go through the area, which is why we write
Φ𝐵 = 𝐵 ∗ 𝐴
For N turns of wire, an EMF is produced if the flux changes. This is known as Faraday’s Law.
𝜀=−
𝑑Φ𝐵
𝑑𝑡
The negative sign is important – it tells us that the EMF created tries to conserve energy. Since an EMF
generates a current, that means the current flows so as to try to keep the flux the same. If B or A increase,
the induced current goes in the direction that creates a magnetic field in the opposite direction.
Flux can change by moving a surface in and out of a field, moving a field around a wire, changing the
area of the surface, or changing the strength of the magnetic field.
EXAMPLE 13
If a wire is moving at speed v below, what is the expression for the magnetic flux? What is the magnitude
and direction of the induced current?
CCW
41
EXAMPLE 14
42
43
MOTIONAL EMF
When a conducting rod of length L moves through a magnetic
field, the positive charges are pushed to one end because they
experience a force (remember that 𝐹 = 𝑞(𝑣 × 𝐵)). This creates
and electric field in the opposite direction so that charges stay in
equilibrium. Thus, we get
𝑞𝑣𝐵 = 𝑞𝐸 => 𝐸 = 𝑣𝐵
The EMF across the capacitor is given by
𝜀 = 𝐸𝐿 = 𝑣𝐵𝐿
This is called the motional EMF, because it is caused by a rod moving in a magnetic field.
If the rod is connected from a to b by a wire, a current is produced both by the EMF and the changing
flux.
44