Answering Exam Questions - Simple Harmonic Motion

Damping in Oscillations
Mechanical Oscillations and the Mass-Spring System
Practical Techniques for Damping Oscillations
Simple Harmonic Motion Basic Principles
The Simple Pendulum
 Physics Factsheet
April 2003 Number 53
Mechanical Oscillations & the Mass-spring System
This Factsheet covers the applications of simple harmonic motion So to test for SHM, we must analyse the forces acting and check that
(SHM) to mechanical oscillations. Factsheet 20 covers the basics of these two conditions apply:
SHM. Factsheet 54 will cover the simple pendulum.
• You need to be able to recognise SHM from the forces acting.
The diagram below shows a mass m oscillating along the line AB (i) there must be a restoring force acting. (ie a force acting
about the centre of oscillation (CO) towards a fixed point so as to reduce x)
No matter which way the mass is moving, the force acting is always (ii) the restoring force must be directly proportional to the
directed to CO. Because it always tries to return the mass to the displacement.
centre, it is called a restoring force (RF).
CO When x is positive the force F is in the negative direction. The opposite
RF
A m B is true when x is negative. This means that the two conditions can be
stated mathematically as
x
F ∝ −x or F = − ( cons tant ) × x !
• For SHM, the restoring force is directly proportional to the
displacement (x) from the centre of oscillation. RF ∝ x
So the further away the mass is from CO, the larger the size of the
force on it.

Example : A simple mass-spring system

This is one of the key examples of SHM that you need to know. The next stage in looking at SHM is to use Newton’s 2nd law of motion
in the form F = m × a. Again this is a vector equation with both F and a
A mass m rests on a horizontal frictionless surface and is connected to a being in the same direction. So, looking in the direction of x increasing:
spring with stiffness constant k. k
The tension in the spring and the extension are linked by Hooke’s law: F=m×a −kx = m × a a = −  x
m
Tension = k × extension This last equation is now compared with a standard equation for SHM:
a = -ω2 x
k k
m Unextended spring, stiffness k. So ω 2 = ω=
Mass at CO (equilibrium position) m m
We use ω to find the period, frequency, velocity & displacement.
F
m For example:
Direction of 2π 2π 2π m
x x increasing Period T = , so here, T = ⇒T = ⇒ T = 2π
ω ω k k
CO
m
When the mass is displaced as shown, there is a force F acting on the The formula for the periodic time is usually given on a ‘formula sheet’
mass where F = tension in the spring. Hooke’s law tells us the size of so you need not commit it to memory. It is worth noting that the period
the force: F = k × x. increases with the mass. Mass is a measure of inertia, it tells you how
sluggish the body is to movement. A really massive body would take
But since F and x are vectors, we must measure them in the same longer to oscillate ie, its period would increase. Because k is in the
direction. bottom line, the opposite effect takes place. If a spring is ‘lazy’, it is
easy to extend and k is small. Think of a bungee rope. k is small but
The rule is that always work in the direction of x increasing ‘1 over a small number is large’ and so the period is ‘large’.

F is in the opposite direction to x increasing, so we have F = -kx. Summary. In dealing with SHM questions first draw a sketch
So the restoring force is directly proportional to x - so, when the mass is showing the forces acting together with a clear indication of x ( the
released, it will move with simple harmonic motion. displacement ) increasing. Then, (in this case) from a knowledge of
springs deduce that, F = −k × x. This demonstrates SHM. The next
Exam Hint: You must be careful with x. We often use x for the stage is to find the corresponding equation for acceleration and
extension of a spring and also for the displacement of a body. In compare it with a = -ω2x to deduce ω. Almost all SHM quantities
this example, they are the same but this is often not the case. As a are determined by ω. One important exception is the amplitude.
rule, use X for the displacement and then figure out the extension. This is determined by something specific to the question such as
initial displacement or velocity. It is not determined by the mass of
the body nor the stiffness of the spring.

1
Mechanical Oscillations and the Mass-spring System Physics Factsheet

Example: Vertical mass-spring system

In this example, we also have to consider the effect of the weight of the (b)

x m 0.4 m
mass, since the spring would not be unextended at equilibrium. The key
ideas are:
• First consider the mass at equilibrium – this allows you to find any CO
unknown quantities, such as the spring constant or the extension at
equilibrium T
• When showing the motion is SHM, take x to be the downwards m
displacement from the equilibrium position (CO). So in this case,
the extension of the spring will be x + extension at equilibrium
mg
k
• As with the horizontal mass-spring system, we have ω = ;
m We measure everything downwards (as x is measured downwards)
unless you are asked to show the motion is simple harmonic, you Resultant force on m is mg – T
can quote this relationship But T = k × extension
Extension = 0.1 + x (since it is x m below CO)
An elastic cord has an unstretched length of 30cm. A mass of 60 So T = k(0.1 + x) = 6(0.1 + x)
grams is attached to one end and hangs freely from the cord whose
other end is attached to a fixed support. The mass then extends the So resultant force F = (0.06)(10) – 6(0.1 + x)
length of the cord to 40cm.. Assume that the cord obeys Hooke’s law. = 0.6 – 0.6 – 6x = -6x
(a) Calculate the ‘spring constant’ for the cord. So motion is simple harmonic

The mass is then displaced 5.0cm vertically downwards and (c) (i) From (b) we have –6x = 0.06a, so a = -100x
released from rest Hence ω2 = 100 and ω = 10
(b) Show that its motion is simple harmonic 2π 2π
So T = = = 0.63s (2SF)
ω 10
(c) Find:
(Or you could have done this using the standard formula forω2)
(i) the periodic time for one oscillation
(ii) the amplitude of the motion A
(ii) We are told ‘the mass is displaced 5cm vertically downwards’.
(iii) the greatest speed of the mass and state where it occurs
This then is the amplitude in this case.
(iv) the greatest acceleration of the mass and state where it occurs
(take the acceleration of free fall, g to be = 10 ms -2 ) When released, the mass will rise 5cm to the equilibrium
position and then another 5cm above the equilibrium position.
(a) So A = 0.05m.
30cm

40cm

(iii) The maximum speed occurs at the centre of oscillation.

V = Aω
= 5 × 10−2 × 10
T
= 0.5 ms−1
m
(iv) Remember that the defining equation for SHM is:- a = -ω2 x
It follows that the greatest acceleration occurs when x has its
mg greatest value.
Condition for equilibrium is T = mg. This is clearly when x = ± A ie at the points when the mass is
Converting mass into kg and extension into m: at its highest and lowest points.
mg 60×10-3 ×10 In this case, ω = 10 s-1 and A = 0.05 m.
k= = = 6Nm -1 The numerical value of the greatest acceleration is
extension 10 ×10-2
(10)2 × (0.05 ) = 5ms-2

Displacement, Velocity, Acceleration and Phase

The graphs (right) show the relationship between the displacement (x),
the velocity (v) and the acceleration (a) for a mass-spring system. If
you start with the displacement, the gradient of its graph against time acceleration (= -ω2X)
(at any position) will give you the value of the velocity at that point. velocity
The gradient of the velocity against time graph will give you the value
of the acceleration.

The three graphs are all sinusoidal (take sine or cosine shapes) and as
such are often drawn with angles on the x-axis. 0 time
T¼ T½ T¾ T
The diagrams provide a good example to explain the term ‘phase
difference.’
Displacement (X)
Look at the displacement and velocity in fig 1. They are ‘out of step’
with each other or, out of phase. The phase difference here is 90° (or
π/2) .The phase difference between displacement and acceleration is
180° (or π).
2
Mechanical Oscillations and the Mass-spring System Physics Factsheet

Exam Workshop (vi) the maximum acceleration, and label any two points (R & S)
This is a typical weak student’s answer to an exam question. The where this occurs. [3]
comments explain what is wrong with the answers and how they can R ! S
be improved. The examiner’s answer is given below.

A small machine part is vibrating with simple harmonic motion; its

displacement / time graph is shown below.
Maximum acceleration is at extreme of motion when x = amplitude.
displacement

a = -ω2x = - 2092×10 =436810 ms-1 2/3

! "

Points R & S correct The student has used the correct formula,
20 mm (allowance for incorrect ω =209 ) but has incorrect units.
time

If the machine part has a mass of 20 grams, calculate

(vii) the greatest restoring force acting on it. [1]
F = ma
F = 20×10-3 × 436810 Newtons. !ecf 1/1
30 ms
For the machine part, The student has quoted the correct formula and substituted what
(i) state the amplitude of the motion [1] s/he thinks the correct values.
The amplitude a = 10mm ! 1/1 This example illustrates the comparatively large forces acting on
the body (about 100 × weight ) when vibrating.
This is correct. The student has halved the peak to peak value and
got the units correct. Beware that at a later stage, this may need (viii) Without further calculation, comment on the resultant force
converting to metres. acting on the mass at the points you have labelled P & Q . [1]
0/1
(ii) calculate the periodic time [1]
The periodic time T = 30 ms = 30 × 10-3 s " 0/1 The student did not answer section v) but, the speed is a
maximum, ie not changing which means the acceleration at
This is incorrect; the student may have rushed the question. 30 points P & Q is zero. Hence the resultant force at these points is
ms is for three half cycles, not two An error like this will make zero
subsequent calculations wrong but the student is only penalised
once. The examiner will ‘re-work’ the rest of the question as if T Examiner’s Answers
were 30ms. For this reason you should explain your working as (i) The amplitude a = 10mm (= 0.01 m) !
fully as possible and write equations before substituting values.
(ii) 3 half cycles in 30 ms
2 half cycles in 20 ms
(iii) the frequency (f) [1] period = 20 ms !
1 1 10 3 1
frequency = = −3
= = 33.3Hz !ecf 1/1 (iii) f = = 50Hz !
time period 30 × 10 30 20 × 10 −3
(iv) ω = 2πf
The student has shown the work required. The error carried ω = 2π × 50 = 100 π = 314 s-1!
forward has been allowed for.
(v) The maximum speed is when:
(iv) the angular speed (ω) [1] x = 0 ⇒ Vmax = ωA = 314 × 10 × 10-3 = 3.14 ms-1 !
ω = 2πf One mark for correct units.
ω = 2π×33.3 = 209 s-1 !ecf 1/1 Points for P and Q shown
occurs when gradient of
displacement/time graph P Q
Again the student has shown working (incorrect f carried
forward). Correct answer. Note, ω measured in s-1 greatest(+ or -). Any two!
(vi) 1 mark for correct position of R and S
(v) the maximum speed, and label any two points (P &Q) on the
Other possible points for R and S shown dotted
graph where this occurs. [3]
v = ω A2 − x 2 "" 0/3 R ! S

The student has reproduced a formula but cannot apply it. No

points shown on graph.

! !
amax = ω2 × A = 3142 × 10 × 10-3 = 986 ms-2
(vii)F = ma = 20×10-3× 986 =19.7 N

(viii) the speed is a maximum , ie, not changing which means the
acceleration at points P & Q is zero. Hence the resultant force at
these points is zero!

3
Mechanical Oscillations and the Mass-spring System Physics Factsheet

Questions
Exam Hint: In answering SHM questions, focus on ω. Without this, 1. The diagram shows a mass m resting on a platform that can oscillate
not much else can be found. It is often supplied via periodic time with SHM in a vertical direction. The mass is shown above the
2π centre of oscillation with the forces acting on it. R is the reaction
T= or the frequency f = 2πω. Remember that ω is constant for a
ω between the platform and the mass.
 k  mass m
particular mass and spring  ω =  but the amplitude can be set
 m 
 R
at any sensible value. For example in the question above, the
amplitude could have been set at 2cm. (so changing greatest speed
and acceleration) but ω would not change. platform mg
• Do not confuse a for acceleration and A for amplitude. x
• A lot of numerical questions involve grams and centimetres,
kilograms and metres. Be ready to convert!
CO

Typical Exam Question

The diagram represents the cylinder and piston of a car engine.
As the piston moves from position T
down to B and back to T again, the Top T cylinder
crankshaft rotates through 360°. The (i) Write down the equation of motion for the mass whilst it
piston
distance BT is 10 cm. and the crank remains on the platform.
Bottom B
speed is 3000 revolutions per minute. (ii) What is the acceleration of mass/platform when the mass loses
crankshaft
For the piston, find: contact with the platform?
(i) its frequency;
(iii)Hence deduce the relationship between frequency and
(ii) its amplitude. displacement when contact is lost.
You may assume that the motion of the piston is simple harmonic. (iv) If the platform has a fixed frequency of 10 Hz, find the
Find amplitude when contact is just lost.
(iii) the maximum speed of the piston, stating its position where (v) If the platform has a fixed amplitude of 1.0 mm, find the
this occurs, frequency when contact is just lost.
(iv) the maximum acceleration of the piston, stating the position 2. A loudspeaker produces a musical note by the oscillation of a
where this occurs.
diaphragm. If the amplitude is 1.0 × 10-3 mm, what frequencies are
available if the diaphragm’s acceleration must not exceed 10 ms -2 ?
Answers:
3000
(i) f = = 50 Hz
60
BT
(ii) amplitude =
2
10 cm
=
2
= 5cm= 5 × 10-2m
3. A block of mass of 600 grams is hung from a spring. It performs
(iii) Vmax = A ×ω vertical oscillations with a periodic time of 0.5s.
= A×2πf (i) Calculate the spring constant.
= 15.7 m s-1 . Midpoint of BT. An additional mass of 200 grams is now added to the block.
(iv) a = -ω2x (ii) Calculate the new periodic time.
amax = ω2 × A Answers
= 4930ms-2. At both points B and T 1. (i) (R – mg ) = m×a. ⇒ R = m(g+a)
(ii) contact lost when R=0, ie a= -g
Or the acceleration is g toward the CO
(iii) a =-ω2x ⇒ -g = -ω2x. (g = ω2x ⇒ g = 4π2f 2x )
(iv) 9.8 = 4π2 102 x ⇒ x = 2.5mm
(v) 9.8 = 4π2f2 (1.0×10-3) ⇒ f = 16 Hz (2SF)
2. For SHM a = -ω2x
Acknowledgements: This Physics Factsheet was researched and written by Keith Cooper.
Numerical values only:
The Curriculum Press,Unit 305B, The Big Peg,120 Vyse Street, Birmingham, B18 6NF. ω2x ≤ 10 or 4π2f 2x ≤ 10 ⇒ f ≤ 500Hz for x = 10-6m
Physics Factsheets may be copied free of charge by teaching staff or students, provided that m 0.6
their school is a registered subscriber. 3. (i) T = 2π ⇒ 0.5 = 2π ⇒ k = 95 Nm-1
k k
No part of these Factsheets may be reproduced, stored in a retrieval system, or transmitted, 0.6 + 0.2
in any other form or by any other means, without the prior permission of the publisher. (ii) T = 2π ⇒ T = 0.58 s (2s.f.)
ISSN 1351-5136 94.7

4
 Physics Factsheet
September 2001 Number 20

Simple Harmonic Motion – Basic Concepts

This Factsheet will: Equations for SHM
! explain what is meant by simple harmonic motion The definition of SHM above can be expressed in the form of an
! explain how to use the equations for simple harmonic equation:
motion F = −kx
! describe the energy transfers in simple harmonic motion
Note that the minus sign appears because the force is directed back
! show how to represent simple harmonic motion
towards the fixed point.
graphically
! explain the link between simple harmonic motion and For the sake of convenience, this is more usually written:
circular motion F = −mω2x
Later Factsheets will deal with specific examples of simple
harmonic motion, such as the simple pendulum and mass- The reason why this form is more convenient will become apparent
spring system, and damped oscillatory motion. shortly.

What is simple harmonic motion? By using “F = ma”, this leads to the equation
Simple harmonic motion (SHM) is one form of oscillatory motion. a = −ω2x
SHM occurs when the resultant force acting on a body has particular
properties: This equation leads to the following equations (but you do not need to
know how unless you are studying SHM in A-level Maths!):

A body performs SHM if it is acted upon by a force v2 = ω2(r2 – x2) where v = speed
! the magnitude of which is proportional to the distance of the x = displacement
body from a fixed point r = amplitude = maximum displacement
! the direction of which is always towards that fixed point.

x = rcosωt where t = time

A simple example of SHM can be observed by attaching a mass to a This equation assumes that the particle starts at a point of maximum
spring, then pulling the mass down and releasing it. It will bob up and displacement – in the case of the spring example, this would mean it
down – this motion is SHM. starts from the “pulled down” position
In this case, the “fixed point” in the above definition is the equilibrium If we started timing from the equilibrium position then the
position of the mass – where it was before it was pulled down.
displacement would be given by x = rsinωt
The resultant force acting on the mass is composed of its weight and the
v = −r ω sin ω t
tension in the spring. This will always be “trying” to pull the mass back
Again, this assumes the particle starts at the point of maximum
to the equilibrium position, and the further away from equilibrium the
displacement. – if it starts from equilibrium, we’d have v = rωcosωt
mass is, the stronger this force will be.

Some
Some helpful
helpful maths
maths

1. cosω
1. cos ωtt means cos(ω
means cos( ω ×× t)t) –– so
so you
you have
have to
to work out ω
work out ωtt first,
first, then
then find
find its
its cosine.
cosine.

2.
2. Radians
Radians are are another
another wayway ofof measuring
measuring angles,
angles, rather
rather than
than degrees.
degrees. Angular
Angular velocities (see also Factsheet 19 Circular Motion) are
measured
measured in in radians per second (rad s-1). When you are using functions like cosωt, you will need to work in radians. The best way to
do
do this
this is
is to
to put
put your
your calculator
calculator into
into radians
radians mode,
mode, enter
enter the
the value
value you
you have
have and
and work
work out
out its
its cosine
cosine as
as normal.
normal.
!! OnOn aa standard
standardscientific
scientificcalculator,
calculator,you
youcan
canchange
changeinin and
andoutout
of of
radians mode
radians modeusing thethe
using button labelled
button DRG
labelled (where
DRG D =D
(where degrees and
= degrees
R
and= radians). SomeSome
R = radians). graphical calculators
graphical workwork
calculators in this
in way too; too;
this way on others you need
on others to gotothrough
you need the setup
go through menu.
the setup menu.
!! IfIf you
you have
have to
to convert
convert between
between degrees
degrees and
and radians
radians (which
(which youyou usually
usually won’t):
won’t):
#
# ToTo change
change degrees
degrees to
to radians,
radians, multiply by π/180.
multiply by π/180.
#
# ToTo change
change radians
radians to
to degrees,
degrees, multiply
multiply by
by 180/π
180/π

3.
3. sine
sine and
and cosine
cosine repeat
repeat themselves
themselves every 360oo (=
every 360 (= 2π radians). This is because 360o is a full circle, so if you add 360o to an angle in
degrees,
degrees, you
you get
get back
back to
to where
where you
you started.
started. We
We can
can say
say that
that sine
sine and
and cosine
cosine have
have aa period
period of
of repetition
repetition of 360oo or
of 360 or 2π
2π radians
radians

1
Simple Harmonic Motion – Basic Concepts Physics Factsheet

We can deduce some more results from these equations: • You won’t need to use the equations for the maximum value of
displacement, velocity or acceleration unless maximum values are
Maximum speed specifically mentioned.
If we look at the speed equation v2 = ω2(r2 – x2), we can see that the • If time is not mentioned anywhere, you are probably going to be
bigger x2 is, the smaller v2 is and vice versa. The smallest possible value using v2 = ω2(r2 – x2).
of x2 is 0. Putting x2 = 0 in gives us v2 = ω2r2, hence:
You may also find the following useful:
vmax = ωr; this occurs when x = 0 • The time required for the body to go between a maximum value of
displacement to the equilibrium position is one quarter of the period
• The time required for the body to go between one maximum of
Maximum acceleration displacement to the other (i.e. the two “ends” of the motion) is half
Since a = −ω2x, the acceleration will be greatest in magnitude when x is of the period
greatest in magnitude, hence
The way to approach problems is best seen from worked examples:
maximum magnitude of a = ω2r
Example 1. An object is oscillating with simple harmonic motion.
Its maximum displacement from its equilibrium position is 0.2m.
Acceleration at any time The period of the motion is 0.1 s. Find its speed when it is 0.06m
Using a = −ω2x together with x = rcosωt, we have from its equilibrium position

a = −ω2cosωt Since we are given T, our first step is to calculate ω.

2π 2π
T= ⇒ω= = 62.8 rads-1
ω Τ
Period and frequency
We know x = rcosωt. We now know: r (= max displacement) = 0.2m
Since cosine repeats every 2π radians (see maths box on page 1), the ω = 62.8 rad s-1
displacement will first return to its intial value when ωt = 2π. x =0.06 m
We want: v =?
The time required for this is the period of the motion, T. From above,
we have Since r, ω, x and v are involved (and not t), we use
T = 2π v2 = ω2(r2− x2)
ω
v2 = 62.82(0.22 – 0.062)
v2 = 143.5
The frequency of the motion is the number of oscillations per second. v = 12.0 ms-1
1 ω
f = =
T 2π
So ω = 2πf Example 2. A mass is moving with simple harmonic motion; its
displacement was at a maximum of 1.1m when t = 0. Its maximum
It is worth noting that the period and frequency do not depend on the speed is 0.33ms-1. Find:
amplitude of the motion. a) its frequency;
b) its speed after 2.0 seconds.
Exam Hint: You may need to know how to derive the expressions
for maximum speed and maximum acceleration. Other equations a) Since we are asked for frequency, we need to find ω first
can be learnt – but check your formula sheet, to make sure you do So we have: r = 1.1 m
not waste time learning equations you will be given. vmax = 0.33 ms-1
ω=?
So we need to use the equation with these three letters in it:
The equations are summarised in Table 1 below vmax = ωr
0.33 = ω1.1
Table 1. SHM equations sorted by type ω = 0.33÷ 1.1 = 0.30 rads -1
Now we have ω, we find f using
acceleration speed displacement period/frequency
equations equations equations equations ω = 2πf
0.3 = 2πf
a = −ω2x v2 = ω2(r2 – x2) Τ = 2π f = 0.30 ÷ (2π) ≈ 4.8 × 10-2 Hz
ω
a = -ω2cosωt v = -rωsinωt x = rcosωt ω = 2πf b) We have t = 2.0 s; ω = 0.30 rads -1; r = 1.1 m; v = ?
amax = ω2r vmax = ωr xmax = r So we must use:
v = -rωsinωt
Using the equations = -1.1 × 0.30 × sin(0.30 × 2.0)
At first sight, there may seem to be a bewildering number of equations = -1.1 × 0.30 ×0.565
to choose from. Here are some strategies to ensure you use the correct ≈ -0.19 ms-1
one(s):

• Focus on ω. If you are given the period or frequency, use these to Exam Hint:
find ω before doing anything else. Similarly, if you are asked to find 1. Make sure you do 0.30 × 2.0 before finding the sine
the period or frequency, find ω first. 2. Ensure your calculator is in radians mode!
• Write down what you know and what you want. Then choose the
equation with just these symbols in it.
2
Simple Harmonic Motion – Basic Concepts Physics Factsheet

Example 3. A heavy body is performing simple harmonic motion. Its Typical Exam Question
displacement is at its maximum value of 0.40 m when t = 0. It first A body performs SHM with a period 3 seconds. Timing starts at
reaches a point 0.20 m from its equilibrium point after 3.0 s. one of the extremes of displacement of the body.
a) Find the period of the motion. Determine the next three times when:
b) Explain why the words “first reaches” are important for your (a) displacement is at an extreme of the motion; [3]
calculation in a). (b) velocity is zero; [3]
c) Find the body’s displacement when its speed is 5.0 × 10-2 ms-1 (c) acceleration is zero. [3]

a) Since we are asked for the period, we first need to find ω (a) it will be at the other extreme half a cycle later, so the times are
We have: r = 0.40 m, x = 0.20 m, t = 3.0 s, ω = ? 1.5 s$, 3.0 s $, 4.5 s $
So we use (b) Velocity is zero when ω2(r2 – x2) = 0 – so when x = ± r$
x = rcos ωt So times are 1.5 s, 3.0 s and 4.5 s as in (a). $$
0.20= 0.40cos( ω × 3.0) (c) Acceleration = -ω2x. So acceleration is zero when body is at
equilibrium position.$ This is midway between the times it is at the
To solve this sort of equation, we must get the part with the cos on extremes. So we have 0.75 s, 2.25 s, 4.75 s $$
its own first:
0.20÷ 0.40 = cos(3ω)
0.50 = cos(3ω)
Typical Exam Question
Now we must use cos-1 (using INV COS on the calculator) to find the The body in the diagram performs simple harmonic motion
angle – in radians – whose cos is 0.5: between the points shown as dotted lines. The period of the motion
is 2s.
3ω = cos-1(0.50) = 1.05 20cm
ω = 0.35 rad s-1

2π
Now we can find T = =18 s
ω

b) Since SHM is repetitive, there will be other times when the body
reaches this displacement.
identical light helical springs
c) We have ω = 0.35 rad s , r = 0.40 m, v = 5.0 × 10 ms , x = ?
-1 -2 -1

So we use Calculate the following quantities:

v2 = ω2(r2 – x2) (a) amplitude; [1]
0.052 = 0.352(0.42 – x2) (b) maximum acceleration; [2]
0.0025 = 0.122(0.16 – x2) (c) maximum speed. [2]
0.0025÷ 0.122 = 0.16 – x2
0.0205 = 0.16 – x2 (a) amplitude = distance from equilibrium to max displacement pt
x2 = 0.16 – 0.0205 = 0.1395 The distance shown is twice that – so amplitude = 10cm = 0.10m$
x = 0.37 m 2π
(b) We need ω = =π
Τ
amax = ω r =π (0.10)$ = 0.99 ms-2$
2 2
Example 4. A body performs simple harmonic motion. Its motion is
(c) vmax = ωr = 0.10π $= 0.31 ms-1$
timed from a point of maximum displacement. Two seconds later, it
reaches the equilibrium position for the first time. Its maximum
acceleration is 0.60 ms-2. Find:
a) its period; Exam Hint: - Many candidates lose marks through omitting to
b) its maximum speed. change their answers to SI units – in the above example, if the
amplitude had not been changed into metres, the values for
a) Initially, we do not seem to have enough information to use any maximum acceleration and speed could have been inconsistent.
equation. But we do know that it takes two seconds to move from a
point of maximum displacement to the equilibrium position – which
corresponds to a quarter of the period.
Energy Transfers in SHM
So the period is 4 × 2 = 8 seconds
In SHM, with no external forces (so no “damping” or
“forcing”), the total energy of the oscillating system remains
2π 2π
b) For any further calculations, we will need ω = = constant.
Τ 8
=0.785 rad s-1
we know: amax = 0.60 ms-2 Although total energy remains constant, energy is transferred between
So use amax =ω2r kinetic energy and potential energy – when the body is at the
0.6 = 0.7852r equilibrium position, it is moving at its fastest, so its kinetic energy is
r = 0.6÷ 0.7852 = 0.973 m maximum, and when it is at the points of maximum displacement, its
speed is zero so its kinetic energy is also zero. These energy transfers
We need vmax = ωr = 0.785 × 0.973 =0.76 ms-1 are best represented graphically (see next section).

Note: we could have obtained this by dividing amax by ω.

3
Simple Harmonic Motion – Basic Concepts Physics Factsheet

Graphical Representation of SHM Energy Graphs

Displacement, speed and acceleration against time Energy against displacement

The graphs of these graphs, as might be expected from their respective We will consider total energy, kinetic energy and potential energy.
equations, produce a standard “sine wave” shape:
Since total energy is constant, this graph is simply a horizontal line.
displacement Kinetic energy, however, is more interesting:
x We know that v2 = ω2(r2 – x2)
So kinetic energy = ½ mv2 = ½ mω2(r2 – x2)
r period, T amplitude, r This leads to the graph below:

k.e.
t ½ mω2r2

-r

speed
x
v -r 0 r

ωr
We can combine kinetic energy, potential energy and total energy on
one graph: energy
t total energy
½ mr2ω2
potential
-ωr energy

kinetic
acceleration energy

a displacement
ωr 2 0

Note that the kinetic energy and potential energy at any time add up to
the total energy, which is constant.
t

Energy against time

−ω2r Since we know v = -rωsinωt, we can deduce
k.e. = ½mv2 = ½ mr2ω2sin2ωt
This produces the following graph:
Points to note k.e.
T
• The speed graph is the gradient of the displacement graph, and they ½ mr2ω2
have a phase difference of π/2 (= 90o or a quarter of a period).
• The acceleration graph is the gradient of the speed graph, and is π/2
out of phase with the speed, and π out of phase with displacement.
• Each of these graphs has the same period. t
• The acceleration is always opposite in sign to the displacement
Note that the kinetic energy goes through two cycles during one period
of the oscillation.
Acceleration against Displacement
The total energy is, again, a constant, producing a horizontal straight
a line graph. The potential energy graph is an “upside down” version of
ω2r the kinetic energy:

p.e.

-r 0 r
x ½ mr ω2 2

-ω2r t

4
Simple Harmonic Motion – Basic Concepts Physics Factsheet

Circular Motion and SHM

To see the link between circular motion and SHM, consider an object Typical Exam Question
performing horizontal circular motion at constant speed in front of a A large fairground wheel, which rotates at a constant rate, casts a
screen, with the plane of the motion perpendicular to the screen. Light is shadow on to a nearby building. At a time when the sun's rays
then directed onto the object, so that the object’s shadow falls onto the strike the building horizontally, a boy measures the speed of the
screen. The diagram below shows a view looking down on the shadow of one of the cars on the wheel as it passes different floors
apparatus. of the building.

Shadows
A A* screen
LIGHT

shadow
Big wheel

B*
B
Rays of sunlight
At a floor which is level with the centre of the wheel, the speed of
the shadow is 0.17 ms-1. At a floor 10m higher, the speed is 0.16 ms-1.
The shadow’s motion will be in a straight line, between points A* and Calculate:
B*. In fact, this motion is simple harmonic; to show this we will need to (a) the time it takes to complete one rotation; [6]
introduce some angles and simple trigonometry. (b) the diameter of the wheel. [2]

(a) Shadow is projection of circular motion ⇒ SHM

A Level with centre of wheel ⇒ speed is maximum.
A*
So 0.17 = ωr$
P
θ x When x = 10 m, v = 0.16ms -1. So using v2 = ω2(r2 – x2):
r
C
O 0.162 = ω2(r2 – 102)$
0.162 = ω2r2 – ω2102$
B* But ω2r2 = (ωr)2 = 0.172
So 0.162 = 0.172 – 100ω2 $
100ω2 = 0.172 – 0.162 = 0.0033
ω2 = 3.3 × 10-5
• We will measure the displacement of the shadow from the point O,
ω = 5.7 × 10-3 rad s-1 $
which is midway between points A and B. It is level with the centre
of the circle, C. This corresponds to the equilibrium point in SHM. T = 2π/ω = 1.1 × 102 s$

• We will assume the object starts its circular motion at point A. This (b) This is 2r.
means its shadow will start at point A*. This corresponds to our 0.17 = ωr ⇒ r = 0.17/ω = 29.8 m$
assumption that SHM starts at a point of maximum displacement. Diameter = 60 m (2SF)$

• The angle the line CP makes an angle θ with the line CA at time t. Questions
1. Explain what is meant by simple harmonic motion.
• The particle is moving with a constant angular velocity ω.
2. Write down expressions for the maximum speed and acceleration of
a particle carrying out SHM.
We now need to find an expression for the displacement, x, of the
shadow at time t. 3. Sketch graphs to illustrate displacement, speed and acceleration
against time for a particle carrying out SHM. State the relationship
By trigonometry, x = rcosθ, where r = radius of the circle. between these graphs.
Since the particle has a constant angular velocity, θ = ωt
So at time t, displacement of shadow is given by 4. Sketch a graph to show how kinetic energy varies with displacement
x = rcosωt or a particle carrying out SHM. Include on your sketch the
This is therefore SHM. maximum value of the kinetic energy.
5. Explain why a projection of circular motion will not produce simple
Any SHM can have circular motion linked with it in this way – it is harmonic motion if the angular velocity is not constant.
known as associated circular motion. The SHM is sometimes
described as a projection of the circular motion. Note that ω in the 6. A particle carrying out SHM has a period of 2.0 s and a maximum
SHM corresponds to the angular velocity in the associated circular speed of 9.4 ms-1. Given that timing starts when the particle’s
motion displacement is at a maximum, find the first 3 times when it is at a
distance of 1.5 m from its equilibrium position.
Exam Hint: - This only works if the circular motion is at constant
7. A body carries out SHM. When its displacement from its equilibrium
angular velocity – so in many cases it will not apply to vertical
position is 0.10 m, its speed is 5.0 ms -1. When its displacement from
circular motion.
equilibrium is 0.30 m, its speed is 2.0 ms-1.
Calculate its maximum displacement.

5
Simple Harmonic Motion – Basic Concepts Physics Factsheet

Exam Workshop Answers

This is a typical poor student’s answer to an exam question. The 1 – 4 can be found in the text
comments explain what is wrong with the answers and how they
can be improved. The examiner’s answer is given below. 5. If the angular velocity of the circular motion is not constant, then
the angle through which the body turns will not be directly
The spherical object shown in the diagram below is known to proportional to time. Accordingly, the displacement of the
perform simple harmonic motion between points A and B. The projection will not be proportional to the cosωt, when ω is
object appears stationary when viewed with a strobe light at 21 Hz constant.
and 28 Hz but at no frequencies in between.
6. T = 2 s ⇒ ω = 2π/T = π rad s-1
A 20cm B vmax = ωr = 3π ⇒ r = 3.0 m
x = 3.0cosπt
Distance of 1.5m from equilibrium ⇒ x = ±1.5 m
10cm x = 1.5 m ⇒ 1.5 = 3.0cosπt
0.50 = cosπt
1.047 = πt
x y
t = 0.33 s
4cm 9cm
Determine:
x = -1.5 m ⇒ -1.5 = 3cosπt
(a) the frequency of the motion. [1] -0.50 = cosπ t
28Hz % 0/1 2.094 = πt
t = 0.67 s
The student clearly did not understand what was happening with
the strobe frequencies, but giving some answer, rather than no We know the body takes half a period (= 1s) to travel from one
answer, was sensible, since it allows the student to continue with extreme displacement to the other.
the question and hence gain some marks.
So to travel from x = -1.5 m to the negative extreme takes
(1.0 – 0.67) = 0.33 s
(b) the maximum speed of the object. [2] It then takes a further 0.33s to return to x = -1.5 m
ω = 2πf = 56π $
v=ωr = 56π × 20 = 3520 ms-1 (3SF) % 1/2 So the first 3 times are: 0.33s; 0.67s; 1.33s

The student gains the mark for using his/her value of frequency 7. v2 = ω2(r2 – x2)
to find ω, but then uses both the wrong value for r (remember 52 = ω2(r2 – 0.12) &
amplitude is half the “peak-to-peak” value – this is commonly 22 = ω2(r2 – 0.32) '
examined!) and the wrong units (cm rather than m)
& ÷': 5 = ω (r - 0.1 ) = (r - 0.1 )
2 2 2 2 2 2

(c) the acceleration at position x. [2] 2 2

ω ( r - 0.3 ) (r - 0.32 )
2 2 2 2

ω2x = 56π2 × 0.04 $= 31000ms-2 (3SF) 1/2

25 ( r 2 − 0.01)
== 2
Again the student gains credit for using his/her own value of 4 (r − 0.09)
frequency – and here, s/he has remembered to work in SI units. 25(r2 – 0.09) = 4(r2 – 0.01)
But the student has forgotten that acceleration is a vector – it 25r2 – 2.25 = 4r2 – 0.04
must have a direction! The words “towards the centre” or the 21r2 = 2.21
use of a minus sign would have completed the answer. r2 = 0.105
Also, although the student did not make this error in calculation, r = 0.32 m (2SF)
it was unwise to write 56π2 when (56π)2 is meant, since it may
lead to making the mistake of squaring just the π. When you write down your final answer you are expected
to use the same number of significant figures as the data
(d) the speed at point y. [2] that you were given. In mid calculation it doesn’t really
v2 = ω2(r2 – x2) $ matter if you use one more significant figure because it is
v2 = 56π2(202 – 92) your method rather than your mid-way result that is being
v = 9870000 ms-1 % 1/2 marked.

Since the student has shown a suitable method, one mark can be The same thing applies to units. You may choose to leave
awarded. However, s/he has forgotten to take the square root! out units in the middle of calculations but you must include
The size of the answer should have alerted him/her to something them with your final answer.
wrong.

This Factsheet was researched and written by Cath Brown.

Examiner’s Answer Curriculum Press, Unit 305B The Big Peg, 120 Vyse Street, Birmingham B18 6NF. Physics
(a) The frequency must be a common factor of 21 and 28Hz i.e. 7Hz. $ Factsheets may be copied free of charge by teaching staff or students, provided that their school
(b) ω = 2πf = 14π. $ vmax = rω = 0.1 × 14π = 4.4ms-1 $ is a registered subscriber. They may be networked for use within the school. No part of these

(c) a = ω2x = 142π2 × 0.04 = 77ms-2$ towards the centre $

Factsheets may be reproduced, stored in a retrieval system or transmitted in any other form or by
any other means without the prior permission of the publisher. ISSN 1351 -5136
(d) v2 = ω2 (r2 − x2)$
v2 =142π2 ((0.1)2 – (0.09)2)
v = 1.9ms-1 $

6
 Physics Factsheet
April 2003 Number 54

The Simple Pendulum

This Factsheet follows on from Factsheet 20 (Simple Harmonic Motion)
Radians
and Factsheet 53 (Mechanical Oscillations).
Radians are an another way of measuring angles.
π 180
The simple pendulum is an example of a system that oscillates with radians = degrees × degrees = radians ×
simple harmonic motion(SHM). It consists of a small mass (a bob) 180 π
attached to a thin thread and allowed to oscillate. The centre of Approximations: If θ is a small angle, measured in radians, then:
oscillation is the position of the bob when the thread is vertical. sinθ ≈ θ cosθ ≈ 1 tanθ ≈ θ

There are two possible ways of measuring the displacement of the mass Arcs: The length of an arc of a circle is given by rθ, where r is the
m from the centre of oscillation: x (distance) or θ (angle) as shown radius of the circle and θ the angle subtended by the arc.
below. θ must be measured in radians (see box right). θ and x are
x r
related by: θ =
! rθ
θ

θ θ Period of a pendulum
2π g 2π !
Since T = and ω = ,T= = 2π
T ω ! g g
l l
!
The most striking aspect of this equation is that the time period is
m m m independent of the bob’s mass m. Changing the mass will not change
mg sinθ
θ the period. The only way to alter the period of a simple pendulum is to
m x m alter its length l.
mg cosθ
Fig 1a Fig 1b !
mg Another point worth noting is the units in the equation T = 2π
g
Fig 1b shows how to analyse the forces acting on the bob. The units of the left hand side must be equivalent to those on the right.
The two forces involved are the weight mg (acting vertically) and the
Writing m for metres and s for seconds, the units for acceleration are ms-2.
tension T acting along the thread.
We resolve the forces along and perpendicular to the thread. ! m
So the units of 2π are = s2 = s
g m
Along the thread, no motion takes place, so we have  2
s 
T = mgcosθ
The unit for T is s, so the equation is consistent in terms of units.
At right-angles to the thread, in the direction of increasing θ, we have:
Resultant force = -mgsinθ Worked Example: A long case (grandfather) clock has a pendulum.
The angle θ is small and measured in radians, so sin θ ≈ θ. To a good approximation, we may consider it to be a ‘simple
So resultant force = − mgsinθ pendulum’. The bob may be moved up or down by means of an
= − mgθ (using the above approximation) adjusting nut.
(a) If the clock runs too fast, which way should the bob be moved, up
x
= − mg or down? Explain your answer.
! (b) The clock ‘beats seconds.’ That is, each half swing, a tick or a
We now have the requirement for SHM i.e. the restoring force is tock, is one second. Find the length of the simple pendulum
proportional to the displacement x required if the acceleration due to gravity is 9.80 ms-2.
To complete the proof, we use Newton’s second law (F = ma) in the (a) If the clock is running too fast (gaining), then the pendulum is
direction of x increasing: oscillating too quickly - its period (T) is too small. So T must be
F = ma !
x increased and since T = 2π we must increase the length l -
−mg = ma g
! the bob should be moved down
g (b)Each half swing is one second ⇒ full oscillation takes 2 seconds
a = − x This is of the form a = -ω2x, - the standard equation for SHM.
!
! 1 !
g g 2 = 2π so =
So we have ω2 = , and so ω = 9.80 π 9.80
! !
9.80
! = 2 = 0.993m
π

1
The Simple Pendulum Physics Factsheet

Typical exam question Using a pendulum to find the acceleration of gravity

The diagram shows a simple pendulum with a length of 0.90m. The acceleration of gravity can be found with a simple pendulum. We
The pendulum bob is drawn aside as shown through a distance X. !
know that the period is given by T = 2π .
g
So, by measuring T and ! we should be able to figure out a value for g.
!
Squaring the above equation gives T 2 = 4π 2 .
g
l = 0.9m
In this equation, T and ! are the variables whilst π and g are constants.
X  4π 2 
Re-writing the equation with the constants in a bracket: T 2 =  × ! .
 g 
(a) If the distance X is 10 cm, calculate, for the bob, This is the same form as the equation of a straight line. So, a graph of T2
(i) the periodic time 4π 2
(ii) the initial acceleration of the bob against ! will be a straight line with gradient
g
(iii) the subsequent maximum speed of the bob.
(b) A second mass is now held on the vertical passing through the Table 1 shows in the first two columns the length and period of a simple
pendulum’s point of suspension. Where must this second mass pendulum. You would be expected to figure out the third column and
be placed so that when both masses are released simultaneously, complete it.
they will collide with each other?
!/m T/s T2/s2
(c) If the distance X is increased to 15cm, will the two masses again
collide? Explain your answer. You may take the acceleration of 0.4 1.27 1.61
gravity to be 9.8 ms-2. 1.0 2.00 4.00
1.6 2.55 6.50
Answer 2.0 2.82 7.95
! 0.9
(a) (i) T = 2π = 2π = 1.9s The graph has been drawn for you and a suitable point (1.8, 7.2)
g 9.8 identified for gradient calculation.
2π 2π
Now use this to find ω. ω = = ω = 3.3s-1.
T 1.9 T2/s2
(ii) for SHM a = - ω2x. Released when x = 0.1m. 8.0 ×
initial acceleration = (3.3)2× 0.1 = 1.1 ms-2 (2s.f.)
(iii) Vmax = Aω = 0.1 × 3.3 = 0.33ms-1 × (1.8 , 7.2)
T 1.9 6.0
(b) Time for bob to reach vertical is seconds = = 0.475 s.
4 4
During this time, second mass falls vertically under gravity.
Use s = ut + ½ g t2 4.0 ×
s = 0 + ½ × 9.8 × (0.475)2 = 1.1 m
(c) The second mass must be placed 1.1 m above lowest position of bob
ie 0.2 m above point of suspension. If the distance X is increased to 2.0
15 cm, the period of the pendulum will not change. It will still take ×
0.475 s to cover the increased distance. Hence the masses will still
collide.
0 1.0 2.0
l /m
Exam Hint:
1 Most SHM questions require a value for ω. A common way of
supplying ω is through the periodic time T. Remember from your From the graph, the units for the gradient are s2/m. This is ‘the upside
SHM work that ω is radians per second, 2π is radians and T is down’ of acceleration units.
seconds so it is easy to quote: 4π 2
This is correct because the gradient is .
2π g
ω=
T 7.2 4π 2 1.8
2 This question supplies numerical values to two significant figures. To find g we solve the equation: = ⇒ g = 4π 2 × = 9.87ms -2
This means that final answers should be to two significant figures; 1.8 g 7.2
however, it always a good idea to work to more than this until
stating the final answer. In this case, as an intermediate step, we
find a quarter period to 3 sig fig ( 0.475 s). Writing 0.48 may give
an inaccurate answer in later parts of the question.

2
The Simple Pendulum Physics Factsheet

Exam Workshop
This is a typical weak student’s answer to an exam question. The comments explain what is wrong with the answers and how they can be
improved. The examiner’s answer is given below.

The diagram shows a simple pendulum. The length of the pendulum, (v) As the bob continues to oscillate, find the maximum kinetic
l is 80cm and the mass of the bob is 100 grams. energy of the bob and the position where this occurs. [3]
The maximum kinetic energy = ½ × mu2 !
!
= ½ × 0.1 × 0.0352 ecf
= 0.000061 J ^ 2/3

Again one mark for carrying an error forward, one for k.e. but
l omitted position

x (vi) Hence find the maximum change in potential energy and deduce
the vertical displacement h through which the bob moves. [3]
h Potential energy mgh ^ ^ ^ 0/3

In this question, take the acceleration of free fall to be g = 9.8 ms-2

For the pendulum, calculate: The candidate cannot get started. Not even one mark for p.e.; the
(i) the periodic time T [2] change has not been equated to change of k.e
!
T = 2π !
g
80 (vii)The bob is now drawn aside horizontally through a distance X = 15 cm
T = 2π "
9.8 and released from rest. Without further calculation, state and
T = 2π × 2.86 explain what will happen to the periodic time T and the
T = 18 s 1/2 maximum amounts of kinetic and potential energies. [4]
The period increases " ^ 1/4
The maximum energies increase !^
The candidate has the correct formula but not changed
centimetres into metres
First part wrong, & no explanation. Second correct but lacks
explanation.
(ii) the angular velocity ω [2]
angle turned 2π
ω= = !
time taken T Examiner’s Answers
2π !! 80 !
ω= = 0.35 rad s-1 ecf 2/2 (i) T = 2π = T = 2π = 1.8 s ( s.f.)
18 g 9.8

! !
The candidate has used the correct formula and brought down an g 9.8
(ii) ω = = = 3.5 s-1
error; no penalty here. ecf=error carried forward. l 0.8
(iii) Amplitude = 10 cm = 0.10 m !
! ! !
The bob is now drawn aside horizontally through a distance X = 10cm (iv) Vmax = A x ω = 0.1 × 3.5 = 0.35 ms-1
and released from rest. For the subsequent motion of the bob, find
(iii) the amplitude [1]  m  !  0.1  ! 2 !
 × (0.35) = 6.1 × 10 J (at lowest point)
-3
The amplitude = 10 cm = 0.1 m ! 1/1 (v) kemax =   u 2 = 
2  2 
! ! !
Correct, the candidate has changed units, (vi) ∆(k.e)2 = ∆(p.e) ⇒ 6.1 × 10-3J = mgh ⇒ h = 6.3 mm (2s.f)
!
The period is independent of amplitude (determined only by l and g)
and remains constant.The maximum k.e. and p.e. both increase.
(iv) the maximum velocity of the bob [3] !
max velocity = A × ω !
= 0.1 × 0.35 ecf
= 0.035 ms-1 ! 3/3
Full marks here, again an allowance for carrying an error
forward.

3
The Simple Pendulum Physics Factsheet

Questions Answers
1. Write down the equation for the period of a simple pendulum. !
1. T = 2π
g
2. Write down the equation for (a) the frequency and (b) the angular
frequency (ω) in terms of the acceleration of gravity and the length 1 g
of the pendulum. 2. (a) f =
2π !
g
3. Explain how you would accurately measure the period of a simple (b) ω (= 2πf ) =
pendulum. !
3. Choose the centre of the oscillation and place a fiducial mark, such
4. In an experiment to measure the acceleration of free fall using a as a pin held in a cork, at that place. Pull the pendulum aside and
simple pendulum count down from 3 to zero each time the bob passes the fiducial
(a) explain what measurements you would take; mark, moving in the same direction. As the bob passes this mark on
(b) state what quantities you would plot along the y and x axes of “zero” start the timer and time 20 oscillations. Repeat this
the graph that you would plot; procedure and average your value. Divide the mean time for 20
(c) explain how you would use your graph to calculate the oscillations by 20 to get a value for the period.
acceleration of free fall.
4. See text.
5. (a) A simple pendulum has a frequency of 1.5 Hz. Calculate
(i) its periodic time 5) (a) (i) 0.67 s
(ii) the length of the pendulum (take g = 9.8 ms-2). (ii) 11.0 cm

(b) The motion of the pendulum is recorded by a moving strip as (b) (i) 100 mm
shown. (ii) 50 mm

(c)
displacement /mm

10 mm
speed u ms-1

The strip moves uniformly with a speed u of 150 mm s-1.By time/s

considering the number of oscillations in 1 second,
(i) Find the “wavelength” of the trace obtained.
(ii) The pendulum is now replaced by one with a frequency of
3.0Hz. Find the new “wavelength” obtained.
0.67 s
(c) On the graph label and number the axes so as to give an accurate
description of the original pendulum, given that it has an
amplitude of 10mm.

maximum speed maximum acceleration

in opposite directions in opposite directions

(d) (i) 9.4 × 10-2 ms-1

(ii) 0.89 ms-2

Acknowledgements: This Physics Factsheet was researched and written by Keith Cooper.
(d) Calculate: The Curriculum Press,Unit 305B, The Big Peg,120 Vyse Street, Birmingham, B18 6NF.
(i) the maximum speed of the pendulum bob and mark two Physics Factsheets may be copied free of charge by teaching staff or students, provided that their
points where these speeds are in opposite directions. school is a registered subscriber.
(ii) the maximum acceleration of the bob and mark two points No part of these Factsheets may be reproduced, stored in a retrieval system, or transmitted, in
where these accelerations are in opposite directions. any other form or by any other means, without the prior permission of the publisher.
ISSN 1351-5136