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 Contents

3.6 Further mechanics and thermal physics 4

3.6.1 Periodic motion 4
3.6.2 Thermal physics 30

3.7 Fields and their consequences 48

3.7.1 Fields 48
3.7.2 Gravitational fields 48
3.7.3 Electric fields 61
3.7.4 Capacitance 68
3.7.5 Magnetic fields 79

3.8 Nuclear physics 96

3.8.1 Radioactivity 96

Examination preparation
Practical and mathematical skills 125
Using logarithms and logarithmic graphs 127
Data and formulae 130
Practice exam-style questions 132
Answers 145

Glossary 150

Index 158

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 AQA Physics A-level Year 2

3.6 Further mechanics and thermal physics

3.6.1 Periodic motion

3.6.1.1 Circular motion

Angular measure
You have previously studied the motion of objects which move in a
straight line. You have used the equations of linear motion to describe
such movement in terms of the time taken, t, the displacement, s, the
acceleration, a, and the initial and final velocities, u and v.
When an object moves in a circular path, the displacement may not be
important, since after each full circle the displacement is zero. It is often
more useful to consider the total angle, u, that has been turned through.
The SI unit used to measure angles is the radian. The radian is defined
using a circle. The angle in radians, u, at the centre of a circle is the ratio of
the arc length, s, to the radius of the arc, r:

arcc length
ar s
 ( in radians)  
radi
dius of arc
dius ar r

Definition
One radian is the angle subtended at the centre of a circle by an arc that
is equal in length to the radius.

Fig 1
When the arc length is equal
to the radius, the angle u
is equal to one radian. r

θ s

s r
θ= r

If we consider a full circle, the arc length is the circumference, s = 2pr, so:
2 r
 ( radians)   2
r
In other words, there are 2p radians in a full circle. As there are 360° in a
full circle, we can use this to convert radians to degrees (see Table 1).

Table 1
Radians Degrees
Conversions between radians
and degrees 2p 360°
1 360/2p = 57.3°

4 2p/360 = 0.017 1°

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 Section 6

Example
Calculate the angle in radians that the Earth spins through in one hour.
Answer
The Earth turns through a full circle, 2p radians, in approximately
24 hours. So in one hour:
2
Angle  
24
 0.262 rad

Rotational frequency and angular speed

The rate at which an object turns is often given in terms of the number
of full circles that it completes in a given time. This is the rotational
frequency, f, and it is typically quoted in revolutions per minute (rpm). For
example, the spin speed of a washing machine may be 1000 rpm, whilst a
DVD spins at between 500 and 1500 rpm. However, the SI unit of rotational
frequency is the hertz, Hz, which is the number of revolutions per second.
The angular speed, , is the angle turned through in one second,
measured in radians per second, rad s21. Since there are 2p radians in one
revolution, the angular speed is 2p  the angular frequency, i.e.

 5 2pf

Linear speed
Even though an object is moving in a circle, we can still define its ‘linear’ Essential Notes
speed as the distance covered per unit time. The speed, v, of an object All points on a rotating object
moving in circular motion depends on the radius of the circle, r, as well have the same angular speed,
as on the angular speed, . For example, think of two children on a but points at different distances
roundabout (Fig 2). They both have the same angular speed, but the child from the axis of rotation have
further from the centre of the roundabout (the axis of rotation) travels different linear speeds.
further in the same time, and so has a greater linear speed.

Fig 2
Children A and B have the same
angular speed, but their linear
B speeds are different.

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 AQA Physics A-level Year 2

The angular speed, , is linked to the linear speed, v, by the equation:

v5r

where r = radius (distance from rotational axis).

This can be derived from the definition of the radian;
s

r
Suppose that this is the angle turned through in a certain time, t, the
angular speed,  is:
 s s
=  but  v,
t rt t
v
so ω
r
or v5r

Example
Find the rotational frequency and the angular velocity of the Earth due
to its daily rotation about its axis. Find the linear speed of a point on
the equator, given that the radius of the Earth is 6.4  106 m.
Answer
Rotational frequency is the number of revolutions per second:
1
f =  1.16
16  1
1005 H
Hzz
24  60  60

The angular speed is given by

ω  2 f  2  1.16  105  7.27  105 rads1

The speed of a point on the equator is

v  r ω  6.4  106 m  7.2

277 1
1005 rrad
ad s1  465 m s1

Centripetal acceleration
Acceleration is defined as the rate of change of velocity; this could be
a change in the magnitude (the speed) or a change in the direction of
the velocity. Since an object moving in a circular path is continuously
Essential Notes changing the direction of its motion, it must be continuously accelerating,
Remember the difference even if its speed is constant.
between vector and scalar The acceleration of an object in circular motion is always directed
quantities. Velocity is a vector towards the centre of the circle. It is known as centripetal acceleration.
quantity; it has a direction as The magnitude of this centripetal acceleration depends on how quickly
well as a magnitude. Speed is a the direction is changing. In a short time, t, the object moves through an
scalar quantity; it is defined by angle u and the velocity changes from v1 to v2 (see Fig 4). The difference
a magnitude only. in velocity, v, can be found by subtracting v1 from v2. From the triangle in
Fig 4, v 5 v  u.

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 Section 6

Fig 3
When a car goes round a roundabout
velocity or a corner, it is accelerating, even if
it maintains a constant speed.

acceleration
Notes
Objects that are travelling
on a circular path may
also change their speed:
this results in an angular
acceleration. However,
work at A-level is limited to
uniform circular motion, that
is motion at constant speed.

The acceleration of an object moving on a circular path depends upon its Notes
speed, v, and the radius of curvature of the path. Acceleration is the rate
The derivation of the
of change of velocity. The change of velocity, Δv, in time Δt, for an object
expression for centripetal
moving in a circle is vΔu, see Fig 4. So acceleration equals
acceleration is given here for
information. It will not be
v2 required in an exam.
a  rω 2 
r

Fig 4
Deriving an expression for
centripetal acceleration
v2

∆θ
–v1 v2

r v1

∆θ
r For small angles
θ radians ≈ sin θ ,
so ∆θ ≈ ∆v/v
NB. The magnitude of
v1 = v2 = v

Definition
The centripetal acceleration of an object moving in a circle is given by
a = rw 2 = v 2/r.

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 AQA Physics A-level Year 2

Centripetal force
Essential Notes Since an object moving in a circle is accelerating there must be a resultant
Note that the force is force acting on it. The resultant force could come from gravity, as in
centripetal (towards the centre), the case of a satellite orbiting the Earth, or from the tension in a string,
NOT ‘centrifugal’(away from as in the case of a conker being whirled round. The resultant force is
the centre). People talk about centripetal; it acts towards the centre of the circle.
centrifugal force ‘throwing’ The size of the centripetal force that is needed to keep a mass, m, moving
them off their feet as a bus in a circle with a radius, r, at a velocity v, is given by Newton’s Second Law
takes a sharp corner, or being in the form F 5 ma. The centripetal acceleration is v2/r, so the centripetal
responsible for spin-drying force has to be
clothes. Using centrifugal force
to explain these phenomena mv 2
F   mr ω 2
is a good way to make your r
teacher apoplectic. The person
Centripetal force:
on the bus, or the water drops
in a drier, are simply obeying • increases with mass, so that a larger force is needed to make a larger
Newton’s First Law; i.e. mass move at the same speed in a circle;
continuing to move in a straight • increases with the square of the speed; this means that it takes four times
line, until acted on by a force. as much friction to keep a car on the road if you take a corner twice as fast;
• decreases as the radius increases, so the force increases if the circle gets
smaller.

(a) satellite

Notes
F
It is a common mistake in
circular motion problems
For a satellite, it is
to invent an extra force Earth gravity that provides
called the centripetal force. the centripetal force
Remember that this is not
an extra force but simply
the resultant of the real
forces acting on the object in
circular motion. Stick to real (b) (c)
lift lift × sinθ
forces, with real causes.
R

θ
lift × cos θ θ

R cos θ
weight

For an aircraft that is turning For a cyclist on a banked track, it is

(banking), it is a component a component of the reaction force
of the lift that provides the that leads to a centripetal force
Fig 5 centripetal force
Examples of circular motion

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 Section 6

When an object moves on a circular path, the force (and therefore the
acceleration) are always at right angles to the velocity. In fact the force is
always at right angles to the velocity. Because there is no motion in the
direction of the force, there is no work done by the force. A satellite that
is well above the Earth’s atmosphere, like the Moon, can keep orbiting
without any energy transfer taking place.

Fig 6
Force and velocity for
circular motion

velocity

force

The force and velocity are always at right angles to each other.

Example
A conical pendulum is a mass on a string that is whirled round in a
horizontal circle.

r
mass, m
ω
Fig 7 ω

(a) Copy Fig 7 and mark in the forces that are acting on the mass.
(b) Identify the force or forces that are causing circular motion.
(c) If a 2 kg mass suspended on a 1 m long string is whirled around in a
circle of radius 0.25 m, how fast will it be travelling?
(d) If the string broke, which direction would the mass move in at first?

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 AQA Physics A-level Year 2

Answer
(a) The forces that act on the mass are its weight, W 5 mg, and the
tension, T, in the string.

T cos θ
T

T sin θ

Fig 8 mg

(b) Horizontally there is a resultant force of T sin u that acts towards the
centre of the circle. This is the force that causes circular motion.
(c) The resultant force is T sin u 5 mv 2/r.
We can find the tension by considering the vertical forces. Since the
mass is moving in a horizontal circle, the vertical forces must balance:
T cos u 5 mg.

We can find the angle from

r 0.25
si  
sin   0.25, so   sin1 0.25  14.5
sin
l 1
Taking g 5 9.81 N kg21,

2 kg  9.81N kg1
T  = 20.3 N
coss 14.5°
This gives
rT si
sin
n 0.2
25 20.3 N  sin 14.5°
5 m  20
v2    0.635 m 2 s2
m 2 kg

The velocity is
v  0.797 m s1

(d) The mass would fly off at a tangent to the circle.

Example
Jon bought nylon sewing thread, with an advertised breaking stress of
70 MPa. He wanted to test this but only had one mass: 1 kg. He tied
the mass to the thread and whirled it around in a horizontal circle,
gradually increasing the speed of rotation until the thread broke.
(a) Explain how Jon’s method could lead to an estimate of breaking
stress.
(b) Estimate how fast the mass was moving when the thread broke.
(c) Apart from timing the rotations, can you suggest a way for Jon to get
an estimate of the velocity of the mass, when the thread breaks?

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 Section 6

Answer
(a) The tension in the thread provides the centripetal force = mr2. The
tension will increase (proportional to 2) as the mass orbits faster. The
force required to break the thread is: = 70  106  cross-sectional area.
We need to estimate the radius of the thread, say 0.5 mm.
The breaking force = πr 2  70  106 = 55 N. This would occur at
 = √ (F / mR ), where R is the radius of the circle, say 1 m.
(b) This gives  = √ (55 / 1  1)= 7.4 rads−1 = 1.2 rev per second or 70
rpm. The mass would be travelling at a linear speed of v = r  = 1  7.4
m s−1 = 7.4 m s−1.
(c) Jon could estimate the speed of the mass at the instant the thread
broke, by treating it as a projectile and measuring the range. s = ut
(horizontally) and find t from the vertical distance fallen, s = ½ at 2.

3.6.1.2 Simple harmonic motion (SHM)

A car bouncing on its suspension, a child on a playground swing and
the vibrations of a water molecule are all examples of oscillations. An
oscillation is a repetitive, to-and-fro motion about a fixed position. This
sort of motion is caused by a resultant force that is always directed to the
same point. This resultant force, which changes direction as the object
oscillates, is referred to as a restoring force.

Fig 9
The restoring force on a mass
(a) (b) (c)
suspended from a spring

T m F

T mg
m

mg
m F

mg
T > mg T = mg T < mg
resultant force, F, no resultant force resultant force, F,
is upwards downwards

Simple harmonic motion, or SHM, is a special case of oscillatory motion.

If the magnitude of the restoring force on a body is proportional to its
distance from the equilibrium position, we say that it moves with simple

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 AQA Physics A-level Year 2

harmonic motion. Assuming that the mass of the object doesn’t change,
the acceleration will follow the same pattern.

Essential Notes Definition

The negative sign arises The acceleration, a, of an object moving with simple harmonic motion,
because the acceleration and is always proportional to its displacement, x, from a fixed point. The
the displacement are always in acceleration is always directed towards that fixed point.
opposite directions, so that a a~2x
positive displacement causes
a negative acceleration, and
vice versa.
acceleration, a/m s–2
Fig 10
Acceleration against displacement
for an object moving with SHM

a∝–x

–x/m displacement, x/m

–a/m s−2

Fig 11
Acceleration (a) and displacement (x)
for a pendulum

Essential Notes
a –a
It can be difficult to realise that
the pendulum has its greatest x=0
a=0
acceleration when it isn’t
moving at all. Conversely, the
pendulum has no acceleration
–x x
when it is moving fastest.

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 Section 6

A swinging pendulum is an example of a body moving with SHM. The

restoring force, and the acceleration, is greatest when the pendulum
is furthest from equilibrium, when the displacement is equal to its
maximum value or amplitude, A. The acceleration drops to zero when the
displacement of the pendulum is zero.

Definitions
Displacement is the distance from a fixed point in a certain direction. It
is a vector quantity.
Amplitude is the magnitude of the maximum displacement. It is a
scalar quantity.

Period, frequency and acceleration

Oscillations are often referred to as periodic motion. The vibrations of
a guitar string or the motion of a piston in a car engine are examples of
periodic motion. In each case a pattern of motion is repeated over and
over again. The time taken to complete one full cycle of motion is called
the period, T.

Fig 12
Periodic motion
displacement/m

time/s

period, T

The number of oscillations that an object completes in one second is

called the frequency of the oscillation. Frequency is measured in hertz, Hz.
Frequency is related to period by the expression:

1 1
freque
freq
equency
nc  f 
period
ri
riod T

High-frequency oscillations involve high accelerations, since the body

has to speed up and slow down many times per second. For simple
harmonic oscillations the acceleration is linked to the frequency and the
displacement by the equation

a  ( 2 f )2 x
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 AQA Physics A-level Year 2

Notes Example
Remember from circular The piston in a car engine moves with a motion that is approximately
motion that the angular simple harmonic. One cycle of motion takes 0.017 s and the piston
speed is given by  = 2πf. moves through a total distance of 100 mm. Calculate the maximum
acceleration of the piston.
Answer
1
The piston’s frequency is: f   58.8 Hz
0.017 s

The period of the motion is 0.017 s. The piston moves through a

maximum displacement of 100 mm. This will be equal to 50 mm (half
the total movement of the piston), so

a   2 x  ( 2f
 f )2 x  ( 2  58.8)2  0.05  6800 ms2 (to
o 2 s.f.)

in a direction opposite to that of the displacement.

Displacement and time

Bodies that move with simple harmonic motion have a displacement that
varies sinusoidally with time.

Fig 13
Displacement against time for a
x
simple harmonic oscillator
A
displacement/m

0
time/s

–A

Essential Notes –x

To convert degrees to radians

you need to remember that The equation that describes the graph in Fig 13 is
there are 2p radians in a circle.
So 2p radians 5 360°. From this x  Acos(
A  t ) = A cos(
os ( 2 ft )
ft
you can work out that 1 radian
5 360/2p 5 57.3°. See Table 1,
page 1. where x is the displacement, A is the amplitude, f is the frequency and t is
the time. The term 2pft has units of radians (see page 1).

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 Section 6

Example
(a) Plot a displacement–time graph to show three cycles of an oscillation
that has an amplitude of 6 mm and a frequency of 5000 Hz.
(b) Calculate the displacement after 0.24 ms.
Answer
(a) You need to use the equation
x  Acos(
A  t ) = A cos(
os ( 2 ft )
ft

x  0.006 cos (10 000  t )

Now you need to choose suitable values of t to calculate values of x for

plotting. Values of t 5 1, 2, 3, etc. are too large and represent points
which are thousands of oscillations apart.
The frequency of the oscillation is 5000 Hz, so the oscillation has a
period of 1/f 5 1/5000 5 0.0002 s or 0.2 ms.
The oscillation will have a maximum displacement at 0, 0.2, 0.4,
0.6 ms etc.
The oscillation will have a minimum (negative) displacement at 0.1,
0.3, 0.5 ms etc.
The oscillation will have zero displacement in between these points, at
0.05, 0.15, 0.25, 0.35 ms, etc.

+6
x/mm

t/ms
0.1 0.2 0.3 0.4 0.5 0.6

–6

Fig 14
Notes
(b) x 5 0.006 cos (10 000 pt) Remember to set your
x 5 0.006 cos (10 000  p  0.24  1023) calculator to radians when
you calculate this formula.
x 5 1.9  1023 m 5 1.9 mm

Velocity and time Essential Notes

The velocity of a particle that is moving with SHM can be found from the
Remember (from AS/A-Level
gradient of the displacement vs time graph.
Year 1, Section 4) that velocity
5 x/t.

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 AQA Physics A-level Year 2

Fig 15
Displacement and velocity vs time
for a simple harmonic oscillator

gradient = 0 gradient = 0

displacement/m
gradient is
maximum time/s
gradient is
(negative) maximum
(positive)
gradient = 0

velocity/m s–1
time/s

The velocity of a simple harmonic oscillator reaches a maximum when the

displacement is zero. The velocity is zero when the displacement is at a
maximum. Think of a pendulum where the velocity is zero for an instant at
each end of the pendulum’s swing, and the velocity of the pendulum bob
is at its greatest when it swings through the equilibrium position.

Fig 16
Velocity and displacement for a
pendulum bob

x=–A x=+A
v=0 x=0 v=0
maximum v = ±2πfA

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 Section 6

This pattern of changing velocity can be linked to the displacement by the

equation

v  ± 2
ff A2  x 2

so that the maximum velocity, vmax occurs at x = 0:

vmax 5   A = 2pfA

Example
A mass suspended from a spring oscillates with a period of 1.5 s. The
amplitude of the oscillations is 5 cm. Find
(a) the maximum velocity
(b) the minimum velocity
(c) the velocity when the displacement is half the amplitude
(d) the displacement when the velocity is half of its maximum value.
Answer
The period of the oscillation is 1.5 s, so the frequency is
1
f   0.67
67 H
Hzz
1.5
(a) The maximum velocity occurs when there is no displacement, x 5 0:

v   2 f A 2  x 2   2 f A
 2  0.67  0.05  0.21 m s1
(b) The minimum velocity is when x 5 A, and then v 5 0 m s21.
(c) The velocity at x 5 0.025 m is
Essential Notes
v  2  0.67  0.052  0.0252  0 18 ms1
0.1
Velocity is not proportional to
1
(d) When the velocity is 2
of its maximum value, v 5 0.105 m s21. displacement. So, for example,
1
v2 the velocity at x = A does not
2
v 2  42 f 2( A 2  x 2 ) so x 2  A2  1
42 f 2 equal vmax.
2
This give
gi s x  0.043 m from
fr equili
equi
uilibriu
librium.

Essential Notes
Acceleration Remember that the
Acceleration is the rate of change of velocity, so an oscillating object will instantaneous velocity is the
have its greatest acceleration when the velocity is changing most quickly, gradient of a displacement–
at the ends of the oscillation. time graph, x/t, and that the
instantaneous acceleration is
the gradient of a velocity-time
maximum acceleration 5 2 A
graph, v/t.

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 AQA Physics A-level Year 2

Fig 17
Displacement, velocity and
acceleration vs time graph for a
simple harmonic oscillator

displacement/m
t/s

gradient = 0

maximum
velocity/m s–1
gradient
(positive)
t/s
maximum
gradient
(negative)
gradient = 0
acceleration/m s–2

t/s

The magnitude of the acceleration reaches its maximum value at the same
time as the displacement, though they are in opposite directions.

Example
A loudspeaker is vibrating at 250 Hz with an amplitude of 1 mm. If the
vibrations change to a frequency of 500 Hz and an amplitude of 2 mm;
(a) What would happen to the maximum velocity?
(b) What would happen to the maximum acceleration?
Answer
(a) vmax = 2πfA. Since A and f are both doubled vmax → 4 its original
value.
(b) amax = (2πf )2A. Since A is doubled and f 2 goes to 4 its original
value, the maximum acceleration increases by a factor of 8.

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 Section 6

3.6.1.3 Simple harmonic systems

Mass on a spring
A mass bouncing on a spring is another example of simple harmonic
motion.
To show that this is SHM we need to consider the forces acting on the
mass.

Fig 18
(a) (b) (c) The forces acting on a mass
Unloaded spring Mass in equilibrium Mass pulled down a
on a spring
distance, x, and released

spring
constant

length, l k

extension, e T = ke

T = k(e+x)
x

mg

mg

When the mass is in equilibrium, the sum of the forces must be zero. So,
taking downward as the positive direction:

W −T = 0 or mg − ke = 0

When the mass is pulled down by a further displacement, x, the sum

of the forces is equal to the mass  acceleration (from Newton’s
Second Law):
Essential Notes
mg  k (e  x ) = ma , The negative sign shows that
the net force is upwards,
but mg 5 ke, so: whereas the displacement x is
measured from the equilibrium
position in the downwards
k
2 kx 5 ma or a    x direction.
 m

For any given mass and spring, m and k are constants, so the acceleration
is proportional to the displacement and directed in the opposite direction.
The motion is therefore simple harmonic.

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 AQA Physics A-level Year 2

If we compare the equation for acceleration with the general SHM

equation, we can find an expression for the period of the oscillations.
k
Mass–spring system a    x
 m

General SHM equation a  ( 2 f )2 x

k
This gives  ( 2 f )2
m

1 k
So f 
2 m
As T 5 1/f,

m
T  2
k

The period of a mass–spring system depends on the mass and the spring
constant.

Example
A family car has a mass of 1000 kg when it is not loaded. This mass is
supported equally by four springs. When the car is fully loaded its mass
goes up to 1250 kg and the springs compress by a further 2 cm. When
the car goes over a bump in the road, it bounces on its springs. Find the
period of these oscillations.
Answer
m
The formula for the period of a mass–spring system is T  2 .
k
We know the mass of the system but we need to
calculate the spring constant, k.

The extra weight of 250 kg  9.81 N kg21 5 2450 N, will depress the four
springs by 0.02 m.

If each spring carries 14 of the weight, we can use Hooke’s Law,

F 5 ke, to find the spring constant of one spring:

F 613
Fig 19 k   3.0 6 × 10 4 Nm1
Simple pendulum e 0.0
02

Then we can find the period of oscillation of a spring. The effective

mass oscillating on each spring is 1250/4 5 313 kg.

m 313
T  2  2  0.6
64 s
k 3.0
06 10 4
6 ×1
suspension
has negligible
mass The simple pendulum
A ‘simple’ pendulum has all its mass concentrated at the free end. This
bob mass is called the pendulum ‘bob’. The mass of a simple pendulum’s
suspension is small compared to the bob’s mass and so we can neglect

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 Section 6

it. When this is not true, the pendulum is referred to as a compound

pendulum.
We can show that an oscillating simple pendulum is an example of simple
harmonic motion. To do this we need to prove that the acceleration is
proportional to the displacement from equilibrium, and is directed in
the opposite direction to the displacement. First we need to consider
the forces acting on the pendulum. There are only two forces acting, the
weight of the bob, W, and the tension in the string, T. We resolve the
weight into two components, acting parallel and perpendicular to the
tension (Fig 20).

Fig 20
Forces acting on the bob of a simple
θ pendulum

T length, l
mass, m

W cos θ θ W sin θ

W = mg
Essential Notes
If the angle u is small enough,
The displacement x is
around an arc of a circle. we can assume that u in
–x radians is approximately equal
to sin u. (This is valid up to
about 10°, sin 10° 5 0.17364
negative positive and u (radians) 5 0.17453, a
direction direction difference of only about 0.5%.)

 sin u is the only force acting perpendicular to the string. It causes an

W
acceleration towards the equilibrium position.
W sin u 5 ma or mg sin u 5 ma
a  g sin

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 AQA Physics A-level Year 2

Essential Notes If the angle u is small, we can say that sin u  u 52x/l.

The negative sign appears So

because we have defined
displacement to the left as g
a    x
negative, and a negative l
displacement produces an
acceleration to the right.
The definition of SHM states that the acceleration must be proportional
to the displacement, and is directed in the opposite direction. Since g/l is
constant for this pendulum, we can say that acceleration is proportional
to displacement and the motion is therefore simple harmonic. This is only
true for small angles of swing.
We can find an expression for the period of the motion by comparing the
acceleration with the general SHM equation.

g
Simple pendulum a    x
l

General SHM equation a   2 f( )2 x

We can see that
g
l
(
 2 f )2
1 g
So f 
2 l

As T  1/f ,

l
T  2
g

The period of a simple pendulum just depends on its length and on the
acceleration due to gravity. The mass of a pendulum does not affect its
period.

Example
A child on a playground swing oscillates with a motion that is
approximately simple harmonic.
(a) If the child has a mass of 30 kg and the swing is 2 m long, calculate
the period of oscillation.
(b) The child’s younger sister, mass 20 kg, now gets onto the swing
instead. What difference would you expect to the period of the
pendulum?
Answer
l
(a) We use the equation T  2
g
2
T  2  2.8 s (to 2 s.f.)
9.8
81

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 Section 6

(b) The period will be almost unchanged. The period of a pendulum

does not depend on its mass. The effective length of the pendulum
would be altered if its centre of mass was different, and this would
affect the period slightly.

Energy in oscillations
Objects that oscillate are continually transferring energy from potential
energy to kinetic energy and back again. A pendulum uses its kinetic
energy to do work against gravity, storing gravitational potential energy.
This energy is then transferred as kinetic energy as the pendulum
accelerates back towards the centre of the oscillation.
Fig 21
Energy transfer during the oscillation
of a pendulum

gravitational potential gravitational potential

energy energy

kinetic energy

If the oscillating system does not transfer energy to the surroundings, the
total energy remains constant:

Total energy 5 kinetic energy 1 potential energy

The variation of kinetic and potential energy with displacement is shown

in Fig 22.
Fig 22
energy Energy vs displacement for a
total energy pendulum oscillation

potential
energy Ep

kinetic
energy Ek

–A 0 A
displacement

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 AQA Physics A-level Year 2

Fig 23 Horizontal mass–spring system

A horizontal mass–spring system We can investigate energy changes in a simple harmonic oscillator by
considering a mass on a horizontal, frictionless surface. The mass is
spring connected to a spring which extends and contracts. We assume that the
constant k mass m
spring obeys Hooke’s law in compression as well as extension (Fig 23).
The mass is pulled to the right, to a maximum value of displacement, A, and
then released. The mass oscillates in simple harmonic motion about the
extension x equilibrium position, compressing and extending the spring. At each end
natural of the oscillation, when x 5 ±A, the system has potential energy, stored as
length
elastic strain energy in the spring. As the mass passes through the equilibrium
position, when the spring is at its natural length, all the energy is kinetic.
The kinetic energy of the mass is given by
1
Ek 5 2 mv 2

Since this is SHM, v2 5 4p2 f 2(A2 2 x 2), so

Ek 5 2mp2f 2(A2 2 x2)

5 2mp2f 2 (A2 – A2cos22pft)
5 2mp2f 2A2 (1 2 cos22pft)
Notes Since 1 2 cos2u 5 sin2u, the kinetic energy is given by:
Remember from AS/A-Level
Year 1, Section 4 that the Ek 5 2mp2 f 2 A2 sin2 2pft
energy stored in a spring is 12
FL, where L is the The potential energy is stored in the spring as elastic strain energy:
extension, and since Hooke’s
1
Law gives F = kL, the energy Ep 5 2 kx2
stored is equal to 12 kL2.
where k is the spring constant. For SHM, x 5 A cos 2pft, so

E p  12 kA 2 cos2 2 fftt
kA

The potential energy is therefore proportional to cos22pft.

The total energy, ETotal, of the mass–spring system is:

ETotal = Ek + Ep
1
= 2mp2f 2A2 sin2(2pft) + 2 kA2 cos2 (2pft)

Since the period, T, of a mass–spring system is given by:

T = 2p√(m/k), the frequency, f, is given by:

f = ( 12 p)√(k/m). So 4p2f 2m = k.

So we can write the equation for total energy in terms of the mass, the
amplitude and the frequency of oscillations:
1
ETotal = Ek + Ep = 2mp2f 2A2 sin2(2pft) + 2 4π2 f 2mA2 cos2 (2pft)
= 2mp2f 2A2{sin2 (2pft)+ cos2 (2pft)}
= 2mp2f 2A2

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 Section 6

Fig 24
kinetic The variation of kinetic and potential
energy/J energy Ek energy with time in simple harmonic
oscillations

potential
energy Ep

time/s

Example
A mass of 500 g is hung from a spring which extends by 6 cm. The mass
is pulled down a further 4 cm and then released. Calculate the maximum
kinetic energy of the mass, and hence find its maximum velocity.
Answer
The energy stored in a stretched spring is 12 kx 2 , where k is the spring
constant.

force 4.9
91N
k   81.8 Nm1 (taking
ki g as 9.81N kg1 )
king
ex nsion
exte 0.0
05

The elastic potential energy in the spring when it is extended by a

further 4 cm (total extension = 10 cm) is:

E  1
2
 81.8  0.10 2  0.409 J

This is transferred as kinetic energy and gravitational potential energy

when the spring is released. The maximum kinetic energy of the mass
will occur when it passes through the equilibrium position. When the
mass passes back through the equilibrium position, the gravitational
potential energy gained is

 E p  mg h  0.5  9.81  0.04  0.196 J

The energy still stored in the spring is E  1

2
 81.8  0.06 2  0.147 J
Therefore the kinetic energy is the remainder:

Ek 5 0.409 J 2 0.196 J 2 0.147 J 5 0.066 J

mv 2 , the velocity is
Because kinetic energy is E k  12 m

2E k
v = 51 m s1
 0.5
m
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