EDEXCEL NATIONAL CERTIFICATE/DIPLOMA
ADVANCED MECHANICAL PRINCIPLES AND APPLICATIONS
UNIT 18
NQF LEVEL 3
OUTCOME 4
BE ABLE TO DETERMINE THE CHARACTERISTICS OF SIMPLE
HARMONIC MOTION IN ENGINEERING SYSTEMS
Simple harmonic motion generation: general equations for simple harmonic motion derived
from a consideration of uniform circular motion e.g. expressions for circular frequency,
displacement with time, velocity with time, velocity with displacement, acceleration with time,
acceleration with displacement, periodic time, frequency of vibration; application to
mechanical systems where output simple harmonic motion is generated by input uniform
circular motion e.g. scotch yoke mechanism; parameters to be determined e.g. frequency of
vibration, periodic time, displacement, velocity and acceleration at a given instant
Vibrating mechanical systems: systems (mass-spring, simple pendulum); expressions for
circular frequency in terms of system parameters; application of general equations for simple
harmonic motion e.g. natural frequency of vibration, periodic time, velocity and acceleration at
a given instant
© D.J.Dunn www.freestudy.co.uk
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1.
SIMPLE HARMONIC MOTION
The best way to explain this is with a mechanism called a Scotch Yoke shown below.
When the wheel rotates, the pin makes the yoke move up and down repeatedly (reciprocating
motion). The pin is set at radius R. If we start off with the pin in the horizontal position and rotate
the wheel anti-clockwise an angle θ, the pin moves a vertical distance x = R sin(θ). If we project
this point onto the graph we see that as the wheel rotates the point traces out a sinusoidal graph.
We don't have to choose the horizontal position for the starting point, we could choose any starting
point say at angle φ before the horizontal. This gives an initial starting value xo for the
displacement. φ is called the phase angle. In this example both φ and xo are negative.
In general we define SHM by the equation x = R sin(θ + φ)
ANGULAR VELOCITY
Let us suppose that the wheel of the Scotch Yoke rotates at a constant angular velocity ω rad/s.
Remember the conversions from rotational speed are:
ω = 2πN when N is in rev/s or 2πN/60 when N is in rev/min.
The angle turned in time t seconds is θ = ωt so the equation for SHM is usually written as:
x = R sin(ωt + φ)
Note that when φ = 90o or π/2 radian the graph becomes a cosine curve.
© D.J.Dunn www.freestudy.co.uk
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Another mechanism that produces the same motion is an
eccentric cam. The cam is circular but the centre of
rotation is distance e from the centre. When the cam
rotates at uniform speed the follower will move up and
down with SHM.
For this cam x = e sin (θ) = e sin (ωt)
You might think of some other ways of producing SHM
but it should be noted that a crank and piston
mechanism does not produce SHM but it does produce
a reciprocating motion that is similar to SHM. You
should have studied this in outcome 3.
DISPLACEMENT, VELOCITY AND ACCELERATION
We defined SHM as motion that produces a displacement of
x = R sin(ωt + φ)
You should know that velocity is the rate of change of displacement so the velocity of any object
performing SHM is given by:
v = dx/dt = ωR cos(ωt + φ)
Acceleration is the rate of change of velocity with time and this is found by differentiating the
velocity.
a = dv/dt = - ω2R sin(ωt + φ).
Notice that this reduces to
a = -ω2x
This is the usual definition of S.H.M. The
equation tells us that any body that performs
SHM must have an acceleration that is
directly proportional to the displacement and
is always directed to the point of zero
displacement.
Consider the plots of x, v and t
The plots on the left of the diagram are
shown for φ = 0. Note that the velocity is a
maximum when x is zero and acceleration is
a minimum when x is zero. Each time we
differentiate, the plots are shifted 90o to the
left and acceleration is 180o out of phase
with displacement.
© D.J.Dunn www.freestudy.co.uk
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WORKED EXAMPLE No. 1
An eccentric cam has an eccentricity e = 20 mm and rotates at 50 rad/s. Calculate the time and
angle when the follower has risen 10 mm from the mean position.
The mean position is when θ = 0
Calculate the velocity and acceleration at this point.
Calculate the maximum and minimum displacement, velocity and acceleration.
SOLUTION
It should be clear that φ = 0 if time starts at the mean position and note the rotation is
clockwise. Clearly the maximum and minimum displacement is ± 20 mm
x = e sin(θ) put x = 10 mm and
10 = 20 sin(θ)
sin(θ) = 0.5
Remember to put your calculator in radian mode when using radians.
θ = sin-1(0.5) = 0.524 radian. (30o)
ωt = θ
t = 0.524/50 = 0.01047 seconds
v = ω e cos(ωt) = (50)(20) cos(0.524) = 866 mm/s
a = -ω2 e sin(ωt) = -(50)2(20) sin(0.524) = -25 000 mm/s2
Check a = -ω2x = -(50)2(10) = -25000 mm/s2
The maximum and minimum values are found as follows.
For displacement this is clearly ± 20 mm
v = ω e cos(ωt) This is a maximum when cos(ωt) = 1 (The maximum value of a cosine)
This is when θ = 0 radian. Hence
v = 50 x 20 x 1 = 100 mm/s
This is a minimum when cos(ωt) = -1 This is when θ = 180o (π radian)
v = 50 x 20 x (-1) = -100 mm/s (passing through the mean on the way down).
a = -ω2 e sin(ωt) This is a maximum when sin(ωt) = 1 and this is when θ = 270o (3π/2 radian)
a = - 502 x 20 x (-1) = 50 000 mm/s2 at the bottom position.
It is a minimum when sin(ωt) = -1 when θ = 90o (π/2 radian).
a = - 502 x 20 x (1) = -50 000 mm/s2 at the top position.
We could use a = -ω2x At the top x = 20 mm so a = - 502 x 20 = -50 000 mm/s2
At the bottom x = -20 mm so a = -50 000 mm/s2
© D.J.Dunn www.freestudy.co.uk
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WORKED EXAMPLE No. 2
A mechanism performs SHM and an accelerometer fitted to it measures the maximum
accelerations as 45000 mm/s2 and amplitude of 50 mm. Determine the speed of rotation in
rev/s. The phase angle φ is zero. Write down the equations that describe the displacement,
velocity and acceleration.
SOLUTION
x = A sin (ωt) and clearly A is 50 mm so x = 50 sin (ωt)
v = ωAcos(ωt) = v = 50 ωcos(ωt)
a = - ω2A sin(ωt) = - 50ω2 sin(ωt)
a = - ω2x
45000 = - ω2 (-50) hence ω = 30 rad/s N = ω/2π = 4.775 rev/s
The full equations are now:
x = 50 sin (30t)
v = 1500 cos(30t)
a = - 45000 cos(30t)
SELF ASSESSMENT EXERCISE No.1
1.
A mechanism has a motion described by the equation x = A sin(ωt + φ)
Given A = 8 mm, ω = 25rad/s and φ = 0.2 radian, find the displacement, velocity and
acceleration at t = 0.01 seconds. (3.48 mm, 180 mm/s and -2175 mm/s2)
2.
For the same problem what is the maximum and minimum displacement, velocity and
acceleration? (±8 mm, ±200 mm/s and ±5000 mm/s2)
© D.J.Dunn www.freestudy.co.uk
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2
FREE VIBRATIONS
Many things that oscillate naturally have an identical motion to that described as SHM. Examples
are a simple pendulum and a mass oscillating on the end of a spring. These are all called free
vibrations. In the case of the pendulum, the displacement is the angle from the vertical position.
In all these examples it can be shown from first principles that the displacement changes with time
sinusoidally and that the acceleration of the mass is directly proportional to the displacement and
directed towards the rest position. In other words
a = -C x
Since this is the requirement of SHM it follows that C = ω2
A free vibration is one that occurs naturally with no energy being added to the vibrating system.
The vibration is started by some input of energy. If there was no friction they would oscillate
indefinitely but friction always makes the oscillations die away with time as the energy is
dissipated. In each case, when the body is moved away from the rest position, there is a natural
force that tries to return it to its rest position.
We will examine the pendulum and mass-spring system in detail but the same basic principles may
be applied to any free vibration. You will discover that the equations for displacement, velocity and
acceleration of the mass are the same as for the mechanisms described earlier.
ANGULAR FREQUENCY, FREQUENCY AND PERIODIC TIME
It may be difficult to appreciate that there is no wheel rotating and driving the system but clearly
they have a frequency of oscillation f. This is the number of cycles or oscillations per second
measured in Hertz and correspond to the speed of rotation of the wheel in the mechanisms.
Every oscillation corresponds to one revolution (2π radian) of the wheel in the earlier mechanism so
we define ω as the angular frequency and ω = 2πf
If the body oscillates at 2 cycles/s (2 Hz) it takes 1/2 second to complete one complete cycle. If it
oscillates at 5 Hz it takes 1/5 second. If it oscillates at f Hz it takes 1/f seconds. This formula is
important and gives the periodic time T.
T = time needed to perform one cycle.
It follows that
T = 1/f and f = 1/T
Since f = ω/2π then
T = 2π/ω and ω = 2π/T
© D.J.Dunn www.freestudy.co.uk
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NATURAL VIBRATIONS - EXPLANATION
In the following work we will show how some simple cases of natural vibrations are examples of
simple harmonic motion. Remember that one important point common to all of them is that there
must be a natural force that makes the body move to the rest position. Also the body has mass
(inertia) and in order to accelerate, there must be an inertia force or torque present.
A free vibration has no external energy added after it starts moving so it follows that all the forces
and all the moment of force acting on the body must add up to zero. This is the basis of the analysis.
MASS ON SPRING
Most natural oscillations occur because the restoring force is due to a spring. A spring is any elastic
body which when stretched, compressed, bent or twisted, will produce a force or torque directly
proportional to displacement. Examples range from the oscillation of a mass on the end of a spring
to the motion of a tree swaying in the wind. We will only consider the mass on a spring as shown.
Consider the mass is pulled down with a force F as shown.
The spring is stretched a distance xo. This is called the
initial displacement.When the mass is the released, it
oscillates up and down with simple harmonic motion.
Let’s analyse the forces involved. F is the applied force in
Newtons.
x is the displacement from the rest position at any time
and k is the spring stiffness.
Fs = Spring force that tries to return the mass to the rest
position.
From spring theory we know that Fs = k x
Since the motion of the mass clearly has acceleration then there is an inertia force Fi.
From Newton’s second law of motion we know that Fi = mass x acceleration = M a
Balancing forces gives
F = Fi + Fs = M a + k x
If the mass is disturbed and released so that it is oscillating, the applied force must be zero and this
is the requirement for it to be a free natural oscillation.
0=Ma+kx
Rearrange to make a the subject
a = -(k/m) x
In other words acceleration is directly proportional to displacement and directed towards the rest
position. If x was plotted against time, a sinusoidal graph would result. The constant of
k
k
proportionality is
and this must be the angular frequency squared so ω =
M
M
ω
1 k
=
2π 2π M
Because this is a natural oscillation the frequencies are often denoted as ωn and fn. This equation is
true for all elastic oscillations.
The frequency of oscillation is f =
NOTE - in reality the spring has a mass and this is also moving. It can be shown that to correct for
this 1/3 of the spring mass should be added to the other mass.
© D.J.Dunn www.freestudy.co.uk
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WORKED EXAMPLE No. 3
A spring of stiffness 20 kN/m supports a mass of 4 kg. The mass is pulled down 8 mm and
released to produce linear oscillations. Calculate the frequency and periodic time. Sketch the
graphs of displacement, velocity and acceleration. Calculate the displacement, velocity and
acceleration 0.05 s after being released.
SOLUTION
k
20000
ω=
=
= 70.71 rad/s
M
4
ω
f=
= 11.25 Hz
2π
1
T = = 0.0899 s
f
The oscillation starts at the bottom of the cycle so xo = -8 mm. The resulting graph of x against
time will be a negative cosine curve with an amplitude of 8 mm.
The equations describing the motion are as follows.
x = xocosωt
When t = 0.05 seconds x = -8 cos(70.71 x 0.05)
x = 7.387 mm. (Note angles are in radian)
This is confirmed by the graph.
If we differentiate once we get the equation for velocity.
v = -ωxosin ωt
v = -ωxosin ωt = -70.71 (-8)sin(70.71 x 0.05)
v = -217 mm/s
This is confirmed by the graph.
Differentiate again to get the acceleration.
a = -ω2xocosωt and since x = xocosωt a = -ω2x
a = -70.712 x 7.387 = -36 934 mm/s2
This is confirmed by the graph.
© D.J.Dunn www.freestudy.co.uk
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SIMPLE PENDULUM
The restoring force in this case is gravity. When the
pendulum is displaced through an angle θ, the
weight tries to restore it to the rest position. This
analysis is based on moment of force (torque).
RESTORING TORQUE
Weight = mg
To find any moment we must always use the
distance to the centre of rotation measured at 90o
(normal) to the direction of the force. In this case
the distance is L sinθ
Denote the torque as Tg .This tries to return the
mass to the rest position.
Tg = mg (Lsinθ)
INERTIA TORQUE
Since the pendulum has angular acceleration α as it slows down and speeds up, it requires an inertia
torque to produce it. Denote this torque as Ti.
From Newton’s second law for angular motion Ti = Iα
α is the angular acceleration and I is the moment of inertia.
The mass is assumed to be concentrated at radius L. (If it was not, the problem would be more
complicated).The moment of inertia is then simply given as I = mL2
BALANCE OF MOMENTS
If there is no applied torque from any external source, then the total torque on the body must be
zero.
Tg + Ti = 0
mg Lsinθ + mL2α = 0
mg Lsinθ = - mL2α
g Lsinθ = -L2α
The sin of small angles is very similar to the angle itself in radians (try it on your calculator). The
smaller the angle, the truer this becomes. In such cases sin(θ) = θ radians and so we may simplify
the equation to
gθ = -Lα
α = -(g/L)θ
This meets the requirements for SHM since the acceleration α is directly proportional to the
displacement θ and the minus sign indicates that it is always accelerating towards the rest point. It
follows that the constant of proportionality is so (g/L).
ω2 = g/L
ω = (g/L)½
If the displacement θ was plotted against time we would get a sinusoidal plot. Note that the
displacement in this example is angle and should not be confused with the angle on the Scotch
Yoke. The frequency of oscillation is obtained from f = ω/2π.
1 g
Hence
f =
2π L
© D.J.Dunn www.freestudy.co.uk
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Note that the mass makes no difference to the frequency. On earth we can only change the
frequency by altering the length L. If we took the pendulum to the moon, it would oscillate more
slowly since gravity is smaller. In outer space where g is very close to zero, the pendulum would
have no weight and would not swing at all if moved to the side.
Remember also that the above equation is only true if the pendulum swings through a small angle.
If the angle is large, the motion is not perfect S.H.M.
WORKED EXAMPLE No. 4
A mass is suspended from a string 60 mm long. It is nudged so that it makes a small swinging
oscillation. Determine the frequency and periodic time. If the amplitude of oscillation is 4o how
long does it take to move 2o from the vertical?
SOLUTION
f = (1/2π) (g/L)½
f = (1/2π) (9.81/0.06) ½ = 2.035 Hz
T = 1/f = 0.491 s
Equation for displacement - θ = A sin(ωt) where A = 4 o
ω = 2πf = 12.786 rad/s
2 = 4 sin(12.786 t) Note no need to convert to radian.
0.5 = sin(12.786 t)
sin-1(0.5) = 12.786 t You do need radian mode for this.
0.5236 = 12.786 t
t = 0.041 second
© D.J.Dunn www.freestudy.co.uk
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SELF ASSESSMENT EXERCISE No. 2
1.
A simple pendulum oscillates with an amplitude of 3o at 0.7 Hz. Determine the length of the
cord. How long does it take to move 1o from the vertical? (0.507 m and 0.0773 s)
2.
Calculate the frequency and periodic time for the oscillation produced by a mass – spring
system given that the mass is 0.5 kg and the spring stiffness is 3 N/mm. (12.3 Hz, 0.081 s).
3.
A mass of 4 kg is suspended from a spring and oscillates up and down at 2 Hz. Determine the
stiffness of the spring. (631.6 N/m).
The amplitude of the oscillation is 5 mm. Determine the displacement, velocity and
acceleration 0.02 s after the mass passes through the mean or rest position in an upwards
direction. (1.243 mm, 60.86 mm/s and -196.4 mm/s2)
4.
From recordings made of a simple harmonic motion, it is found that at a certain point in the
motion the velocity is 0.3 m/s and the displacement is 20 mm, both being positive downwards
in direction. Determine the amplitude of the motion and the maximum velocity and
acceleration. Write down the equations of motion.
Note that the data given is at time t = 0. You will have to assume that
x = xocos(ωt + φ) at time t=0
Ans.
x= 0.0311 cos(ωt - 50o)
v = -0.3914 sin(ωt - 50o)
a = -157.9 x
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