This document discusses oscillations and simple harmonic motion. It begins by listing learning outcomes related to oscillations, including describing examples of free and forced oscillations. It then provides definitions and equations related to simple harmonic motion, such as the defining equation a=-ω2x. Graphs of displacement, velocity, and acceleration over time for simple harmonic motion are shown. Examples of the simple pendulum and a mass on a spring are provided to illustrate simple harmonic motion. The document also discusses the kinetic and potential energy changes that occur during simple harmonic motion.

1. Oscillations
• 13. Oscillations
• Content
• 13.1 Simple harmonic motion
• 13.2 Energy in simple harmonic motion
• 13.3 Damped and forced oscillations: resonance
• Learning Outcomes
• (a) describe simple examples of free oscillations.
• * (b) investigate the motion of an oscillator using experimental
and graphical methods.
• (c) understand and use the terms amplitude, period, frequency,
angular frequency and phase difference and express the
period in terms of both frequency and angular frequency.
• (d) recognise and use the equation a = – ω2x as the defining
equation of simple harmonic motion.
1

2. • (e) recall and use x = xo sin ωt as a solution to the equation a = –
ω2x.
• (f) recognise and use v = vo cos ωt, v = ± ω√ (x2
o − x2)
• * (g) describe with graphical illustrations, the changes in
displacement, velocity and acceleration during simple harmonic
motion.
• (h) describe the interchange between kinetic and potential
energy during simple harmonic motion.
• * (i) describe practical examples of damped oscillations with
particular reference to the effects of the degree of damping and
the importance of critical damping in cases such as a car
suspension system.
• (j) describe practical examples of forced oscillations and
resonance.
• * (k) describe graphically how the amplitude of a forced
oscillation changes with frequency near to the natural frequency
of the system, and understand qualitatively the factors which
determine the frequency response and sharpness of the
resonance.
• (l) show an appreciation that there are some circumstances in
which resonance is useful and other circumstances in which
resonance should be avoided. 2

3. Oscillations and vibrations
 Vibrations and oscillations occur all the time and are everywhere.
 Vibrations are physical evidence of waves, such as a loud stereo shaking a
table, i.e. sound waves cause vibrations
 One complete movement from the starting point or rest point or equilibrium
position and back to the starting point or rest position or equilibrium position
is known as an oscillation
 The time taken for one complete oscillation is referred to as the period T of
the oscillation
 The number of oscillations per unit time is the frequency f
 Frequency f = 1/T , may be measured in hertz (1 Hertz = 1 s-1) or in min-1,
hour-1 etc
 The distance from the equilibrium position is known as the displacement and
it is a vector quantity since the displacement may be on either side of the
equilibrium position
 The amplitude (a scalar quantity) is the maximum displacement
3

4. Examples of oscillatory motion
 beating of a heart
 a simple pendulum
 a vibrating guitar string
 vibrating tuning fork
 atoms in solids
 air molecules oscillate when sound waves travel
through air.
 oscillations in electromagnetic waves such as light and
radio waves
 oscillations in alternating current and voltage.
4

5. Recap from study of waves
 Some oscillations maintain a constant period
even when the amplitude of the oscillation
changes. This is known as isochronous and
has been made use of in timing devices
 Galelli Galileo discovered this for a
pendulum. A pendulum swinging with a
large amplitude is not isochronous
5

6. Displacement-time graphs
 It is possible to plot displacement-time graphs for
oscillators
 The graph describing the variation of displacement
with time may have different shapes depending on
the oscillating system
 For many oscillators the displacement-time graph
of a free oscillation is approximately a sine or
cosine curve
6

7. Simple harmonic motion (shm)
 A sinusoidal displacement time graph is a
characteristic of an important type of oscillation
called simple harmonic motion(shm)
 In harmonic oscillators the amplitude is constant
with time
 SHM is defined as the motion of a particle about
a fixed point such that its force ‘F’ or
acceleration ‘a’ is proportional to its
displacement ‘x’ from the fixed point, and is
directed towards the point
 F is known as the restoring force
 Mathematically it is defined as a = - ω2x where ω
is the angular frequency and is equal to 2πf 7

8. cont…
 The defining equation is represented in a graph of a against x as a
straight line of negative gradient through the origin.
 Gradient is negative because of the minus sign in the equation which
represents that acceleration is always directed towards the fixed point
from which the displacement is measured
 This means that in shm, acceleration is directly proportional to the
displacement/distance from the fixed point and is always directed to
that point
 Acceleration is always opposite to the displacement since the force is
also opposite to the displacement
8
a
x0

9. Comparisons
 In linear motion, acceleration is
constant in magnitude and direction
 In circular motion acceleration is
constant in magnitude but not
direction
 In simple harmonic motion the
acceleration changes periodically in
magnitude and direction
9

10. Solution of equation for shm
• In order to find the displacement time relation for a particle
moving in shm, we need to solve the equation a = - ω2x
which requires mathematics beyond the requirements of
A/AS
• However we need to know the form of the solution
x = x0 sin ωt or x = x0 cos ωt
where x0 is the amplitude (maximum displacement) of
the oscillation
• The solution x = x0 sin ωt is used when at time t = 0, the
particle is at its equilibrium position where x = 0, and
conversely if at time t = 0 the particle is at its maximum
displacement, x = x0 the solution is x = x0 cos ωt
10

11. Velocity & acceleration for shm
• The velocity v of the particle is given by the expressions
v = x0ω cos ωt when x = x0 sin ωt
v = -x0ω sin ωt when x = x0 cos ωt
• The maximum speed is given by v0 = x0ω
• An alternate expression for the velocity is v = ±ω√(x0
2 –
x2)
(which will be derived next)
• The acceleration a of the particle is given by the
expressions
a = -x0ω2 sin ωt when x = x0 sin ωt
a = -x0ω2 cos ωt when x = x0 cos ωt
11

12. Displacement, velocity and acceleration graphs
x
v
a
t
t
t
Displacement (x), velocity (v) & acceleration time graph

13. Alternate expression for velocity
• Recall that x = x0 sin ωt and v = x0ω cos ωt
• So sin ωt = x/x0 and cos ωt = v/(x0ω)
• Trigonometric relationship between sine and cosine is
sin2θ + cos2θ = 1
• Applying the above relationship, we have
x2/x0
2 + v2/(x0
2ω2) = 1 which gives
v2 = x0
2ω2 - x2 ω2 , hence
13
v = ± ω√(x0
2 - x2)

14. Example
The displacement x at time t of a particle moving in shm is given by x =
0.25 cos 7.5t where x is in metres and t is in seconds.
a) use the equation to find the amplitude, frequency and period for the
motion
b) find the displacement when t = 0.50 s
Solution
a) Compare the equation with x = x0 cos ωt
The amplitude x0 = 0.25 m, ω = 2πf = 7.5 rad/s, therefore f = 1.2 Hz
and period T = 1/f = 0.84 s
b) Substitute t = 0.50 s in the equation
ωt = 7.5 x 0.50 = 3.75 rad = 215°
so x = 0.25 cos 215° = -0.20 m
14

15. Worked examples of shm
• Mass on a helical spring
• Simple pendulum
15

16. Mass on a helical spring – Hooke’s law
• Consider a mass m suspended from a spring
• The weight mg is balanced by the tension T in the spring
• When the spring is extended downwards by an amount x away from
the equilibrium position, there is an additional upward force called
the restoring force in the spring given by F = - kx
• When the mass is released the restoring force F pulls the mass
upwards towards the equilibrium position. The minus sign shows
the direction of this force.
• As the force is proportional to the displacement, the acceleration is
also proportional to the displacement and is directed towards the
equilibrium position meeting the condition for shm
• The full theory shows that the period of oscillation T = 2π√(m/k)
• since F = ma, then ma = - kx
hence a = - (k/m)x = dv/dt = d2x/dt2
16

17. The simple pendulum
• A simple pendulum is a point mass m on a light inelastic string although in real
experiments we use a finite pendulum bob of finite mass
• When the bob is pulled aside through an angle and released, there will be a restoring
force acting in the direction of the equilibrium position
• Because the pendulum moves in an arc of a circle, the displacement will be an angular
displacement rather than a linear displacement
• The 2 forces on the bob are its weight mg and the tension T in the string
• The component of the weight along the direction of the string mg cos θ, is equal to the
tension T in the string
• The component of the weight at right angles to the direction of the string, mg sin θ , is
the restoring force F. This makes the bob accelerate towards the equilibrium position
• The restoring force depends on θ . As θ increases the restoring force is not proportional
to the displacement and so the motion is oscillatory but not shm, but if the angle is kept
small (less than 5°), θ is proportional to sin θ and exhibits shm (check using your calc)
• The full theory shows period of oscillation T = 2π√(l/g) where g is the acceleration of
free fall
• A simple pendulum can be used to experimentally determine g by repeating the
experiment with different lengths of pendulum and plotting a graph of T2 against 4π2/g
17

18. Hooke’s law & shm
 Any system which obeys Hooke's Law exhibits shm
but i) extensions must not exceed the limit of proportionality
ii) the spring must have small oscillations as large
amplitude oscillations may cause the spring to
become slack
iii) the spring should have no mass; if the mass is > 20x
the mass of the spring, the error is 1%
 This example of shm is a particularly useful model for
interatomic forces and vibration of molecules containing atoms
oscillating as if connected by tiny springs
 The frequency of oscillation can be measured using
spectroscopy which gives direct information about the bonding
18

19. Example
A light spring of spring constant k hangs vertically from a fixed point and a mass m
is attached to its free end.
a) State 2 conditions that must be met before the subsequent motion may be
considered to be simple harmonic.
b) Derive an expression for the period T of the resulting motion.
Solution
a) 2 conditions for shm are:
a) The equilibrium position due to the mass is within the Hooke’s law limit of
the spring
b) the mass is given a small vertical displacement such that the spring’s Hooke’s
law limit is not exceeded
b) Let x = displacement of mass m, a = acceleration of mass m, F = ma = -kx
Force in a spring is, - kx = ma , hence a = - (k/m)xa
As a is proportional to - x , so resulting motion is shm
i.e. a = - ω2x
a = - (k/m)x = - ω2x, so angular frequency ω = √(k/m)
Therefore period T = 2π/ω = 2π√(m/k) 19

20. Example
A light string of length l hangs vertically from a fixed support and a mass m is attached to its
free end. The mass is given a horizontal displacement and released to swing freely.
a) State a condition which must be satisfied before the resulting oscillation may be
considered shm.
b) Derive an expression for the period T of the resulting motion.
Solution
θ
a) A required condition is that the angular displacement θ is small l
b) Let x = displacement of mass m, a = acceleration of mass m
In the direction perpendicular to string, F = ma
- mg sin θ = ma, so - g sin θ = a
For small θ , sin θ ≈ x/l, so - gx/l ≈ a x
As a is proportional to - x , so resulting motion is shm
i.e. a = - ω2x
Hence, - (g/l)x = - ω2x, so ω = √(g/l) mg
Therefore period T = 2π/ω = 2π√(l/g)
20

21. Example
A helical spring is clamped at one end and hangs
vertically. It extends by 10 cm when a mass of 50 g is
hung from its free end.
Calculate: a) the spring constant of the spring
b) the period of small amplitude
oscillations of the mass
Solution
a) k = F/x, k = 4.9 Nm-1
b) T = 2π√(m/k) T = 0.63 s
21

22. Energy changes in shm
• A system exhibiting simple harmonic motion would possess a
constant total energy at all points of time
• The total energy normally comprises a portion of potential
energy and another balanced portion of kinetic energy.
• There is thus a continuous interchange of the two energies
during oscillations.
• For example, a weighted helical spring has a total energy that
is the sum of the kinetic energy of the moving mass and the
stored elastic potential energy of the spring.
• Plotting on the same graph for energy versus
time/displacement, the two sinusoidal curves are completely
out of phase.
• It can be proven that the total energy of a weighted spring is ½
mω2xo
2 which is a constant.
.
22

23. Energy vs time graph
energy
Energy versus time graph
0
0 T/4 T/2 3T/4 T time
total energyK.E. P.E.

24. Displacement, velocity and acceleration graphs
x
v
a
t
t
t
Displacement (x), velocity (v) & acceleration time graph

25. Energy changes in shm
• The kinetic energy of a particle of mass m oscillating with shm is
½mv2 and from the earlier derivation v2 = x0
2ω2 - x2 ω2
• So k.e Ek at displacement x is ½mω2( x0
2- x2)
• To find the potential energy Ep we need to find the work done against
the restoring force;
since F = ma , Fres = - mω2x but average restoring force = ½mω2x
• Hence work done = average restoring force x displacement
= ½mω2x2
• The total energy Etot of the oscillating system is given by
• Etot = Ek + Ep = ½mω2( x0
2- x2) + ½mω2x2
= ½mω2x0
2
• This total energy is constant as it merely expresses the law of
conservation of energy
• Pg 272 Chris Mee figs 10.22, 10.23,10.24
25

26. Example
A particle of mass 60 g oscillates with shm with angular frequency of
6.3 rad/s and amplitude 15 mm.
Calculate a) the total energy
b) the k.e and p.e at half amplitude (i.e. at x = 7.5 mm)
Solution
Etot = Ek + Ep = ½mω2( x0
2- x2) + ½mω2x2 = ½mω2x0
2
a) Etot = ½mω2x0
2 = 2.7 x 10-4 J
b) Ek = ½mω2( x0
2- x2) = 2.0 x 10-4 J
Ep = ½mω2x2 = 6.7 x 10-5 J
26

27. Natural frequency & resonance
 A particle is said to be undergoing free oscillations when
the only external force acting on it is the restoring force
 The total energy remains constant at all points of time
 A free oscillation is one where an object or system
oscillates in the absence of any damping forces, and it is
said to oscillating in its natural frequency
 In real situations, frictional and other resistive forces cause
the oscillator’s energy to be dissipated, and this energy is
converted eventually into heat energy. The oscillations are
said to be damped
 When one object vibrates at the same frequency as
another it is said to be in resonance
 The ‘swing’ of a frictionless pendulum is an example of a
free oscillation.
27

28. Resonance
 In the absence of external forces to an oscillating system, the system oscillates at its
natural frequency f0
. The only forces acting are the internal forces of the oscillating
system
 When an external force is applied to an oscillating system, the system is under forced
oscillations and will vibrate at the frequency of the applied force rather than at the
natural frequency of the system
 Whether or not the forcing frequency equals the natural frequency, the oscillations are
said to be forced when a periodic force acts.
 When the forcing frequency is equal to the natural frequency, net energy is taken in
and the amplitude of oscillation builds up further and the applied periodic force is said to
have set the system in resonance. Under such condition, further resonance will result in
more energy being taken in to build up the amplitude further.
 Resonance occurs when a system is forced to oscillate at its natural frequency by the
driving frequency
 When resonance occurs, the amplitude of the resulting oscillations is a maximum as
maximum energy is transferred from the forcing system
E.g. Barton's pendulum – only the pendulum with the same length as the original will
oscillate with the biggest amplitude
 Applications – wind instruments, excessive noise from a moving bus, radio & tv tuning
 The Tacoma Narrows suspension bridge in Washington State, USA in 1940 collapsed due
to a moderate gale (of same frequency as natural frequency of bridge) setting the bridge
into resonance until the main span broke up 28

29. Damped oscillations
• A damped oscillation is one where frictional forces present
gradually slow down the oscillation and the amplitude decreases
with time i.e. decreasing energy
• Damped oscillations are divided into under-damped, critically
damped and over-damped oscillations
• An under-damped(lightly damped) oscillation is one where the
amplitude of oscillation or displacement of the system decreases
with time. Example: oscillation of a simple pendulum with the
damping or dissipative force as air resistance
• In a critically damped system, oscillations are reduced to zero in the
shortest possible time. Examples: moving coil ammeter or volt
meter, shock absorber, door closer
• In an over-damped(heavy damping) system, a displacement from its
equilibrium position takes a long time for the displacement to be
reduced to zero. Example: door dampers
29

30. Damped oscillations
Displacement vs Time Graph
x
t
under-damped critically-damped
over-damped

31. Effects of damping on forced oscillations
• Pg 277 Chris Mee fig 10.29 & 10.30
• As the degree of damping increases:
– The amplitude of oscillation at all frequencies is reduced
– The frequency at max amplitude shifts gradually towards lower
frequencies
– The peak becomes flatter
31

32. Electrical resonance
• Electrical oscillators made from combinations of capacitors
and inductors(coils) can also be forced into oscillations or be
made to resonate
• This is the basis of tuning in electronic circuits which pick
out the required transmission in a receiver
• The natural frequency of an electrical oscillator depends on the
capacitor and inductance of the coil used. By varying the
capacitance, we can tune in to different ‘channels’
• The range of frequencies selected depends on the damping
which in turn depends on the resistance in the circuit
32

33. 14.6 Production and use of ultrasound
(a) explain the principles of the generation and detection of
ultrasonic waves using piezo-electric transducers
(b) explain the main principles behind the use of ultrasound to
obtain diagnostic information about internal structures
(c) show an understanding of the meaning of acoustic
impedance and its importance to the intensity reflection
coefficient at a boundary
(d) recall and solve problems by using the equation I = I0e–μx for
theattenuation of X-rays and of ultrasound in matter
33

34. Ultrasound
Incident
wave Reflected wave
Transmitted wave
Boundary between media
34

35. Recap – piezo-electric effect
• Piezo-electric devices contain a crystal which can expand and compress when external
pressure is varied e.g. quartz
• The crystal’s structure is such that the centre of positive charges coincides with the centre
of negative charges when not stressed
• When expanded, both centres will not coincide.
• When compressed, the centres will be in the opposite direction as compared to under
expansion.
• The separation results in a voltage across the crystal surface and this effect is known as
the piezo-electric effect
35

36. Piezo-electric transducer
• A piezo-electric device is a sensor that detects differences in pressure
(sound wave)
• Variation in pressure will result in an ac voltage
• The magnitude of the voltage generated depends on the magnitude of the
pressure on the crystal and the polarity depends on whether the crystal is
compressed or expanded i.e. whether the pressure is greater than or less
than the ambient pressure
• A transducer is any device that converts energy from one form to another
• The piezoelectric transducer converts mechanical energy (vibration) into
electrical energy in the form of ac voltages
• It also can convert electrical voltages back to vibration.
• Hence it acts as a receiver as well as an emitter.
• To detect the voltages, opposite faces of the crystal are coated with a
metal (silver) and electrical connections are made to these metal films and
since the voltages are very small they are amplified
• The crystal and its amplifier may be used as a simple microphone for
converting sound signals into electrical signals
• Ultrasound waves may be generated using a piezo-electric crystal such
as quartz, as it can convert electrical voltages to vibrations
36

37. Ultrasound
• When a potential difference is applied between the electrodes of the crystal, an
electric field is set up in the crystal which causes forces to act on the ions
• Quartz has a tetrahedral silicate structure with the oxygen ion negatively charged
and the silicon ion positively charged, and as these ions are not held rigidly in
position, they will be displaced slightly when an electric field is applied
• The positive ions will be attracted to the negative electrode, and the negative ions
will be attracted to the positive electrode and depending on the direction of the
electric field, the crystal will become slightly thinner or thicker
• An ac voltage applied across the electrodes will cause the crystal to vibrate with a
frequency equal to that of the applied voltage with a small amplitude
• If the frequency of the applied voltage is equal to the natural frequency of vibration
of the crystal, resonance will occur and the amplitude of vibration will be a
maximum
• The dimensions of the crystal can be such that the oscillations are in the
ultrasound region(> 20 kHz) and this will give rise to ultrasound waves in any
medium surrounding the crystal
• In the medical field the ultrasound frequency is in the megahertz region
• Human range of hearing is from 20 Hz to 20,000 Hz
37

38. The reflection and absorption of
ultrasound
• Ultrasound is typical of many types of waves in that , when it is incident on a boundary
between 2 media, some of the wave power is reflected and some transmitted
• For a wave of incident intensity I, reflected intensity IR and transmitted intensity IT, by
conservation of energy,
I = IR + IT
• Although for a beam of constant intensity, the sum of the reflected and transmitted
intensities is constant, their relative magnitudes depends not only on the angle of
incidence of the beam on the boundary but also on the media themselves
• Hence the relative magnitudes of IR and IT are quantified by reference to the specific
acoustic impedance, Z of each media
• The specific acoustic impedance Z is defined as the product of the density ρ of the
medium and the speed c of the wave in the medium i.e. Z = ρc
• For a wave incident normally on a boundary between 2 media having specific acoustic
impedances of Z1 and Z2, the ratio of the reflected intensity IR to the incident intensity I
known as the intensity reflection coefficient symbol α
i.e. intensity reflection coefficient α = IR/I = (Z2-Z1)2/(Z2+Z1)2
38

39. Typical values of specific acoustic impedance and speed of ultrasound
Medium Speed/m s-1 specific acoustic
impedance/kg m-2 s-1
Air 330 430
Water 1500 1.5 x 106
Blood 1600 1.6 x 106
Fat 1500 1.4 x 106
Muscle 1600 1.7 x 106
Soft tissue 1600 1.6 x 106
Bone 4100 5.6 – 7.8 x 106
39

40. Example
Calculate the intensity reflection coefficient for a
parallel beam of ultrasound incident normally on the
boundary between:
(1) air and soft tissue, specific acoustic impedances of
430 kg m-2 s-1 and 1.6 x 106 kg m-2 s-1 respectively
(2) muscle and bone, specific acoustic impedance of
1.7 x 106 kg m-2 s-1
and 6.5 x 106 kg m-2 s-1 respectively
Solution
Using α = IR/I = (Z2-Z1)2/(Z2+Z1)2
(1) α = 0.999 almost 1!
(2) α = 0.34
40

41. Linear absorption coefficient
• The intensity reflection coefficient for a boundary between air and soft tissue
from the last example is almost unity which means that when ultrasound is
incident on the body, very little ultrasound is transmitted into the body
• In order to overcome this, it is important that there is no air between the
transducer and the soft tissue(skin) and this is achieved by using a water based
jelly whose specific acoustic coefficient is approximately 1.5 x 106 kg m-2 s-1
• Once the ultrasound is within the medium, the intensity of the wave is reduced
by absorption of energy as it passes through the medium
• This causes heating and appropriate frequencies of ultrasound are actually used
in physiotherapy to assist in sprains and similar injuries
• For a parallel beam, the absorption is approximately exponential and for a beam
of ultrasound that is incident normally on a medium of thickness x, the
transmitted intensity I is related to the incident intensity I0 by the expression
I = I0 e-kx or I = I0 exp(-kx)
where k is a constant depending on the medium known as the linear absorption
coefficient. The unit is cm-1
• The coefficient k depends not only on the medium but also the frequency of the
ultrasound 41

42. Typical values of linear absorption coefficient, k of ultrasound
Medium Linear absorption
coefficient/cm-1
Water 0.0002
Bone 0.13
Muscle 0.23
Air 1.2
42

43. Example
A parallel beam of ultrasound is incident on the surface of a muscle and
passes through a thickness of 3.5 cm of the muscle. It is then reflected at
the surface of a bone and returns through the muscle to its surface.
Calculate the fraction of the incident intensity that arrives back at the
surface of the muscle given that the linear absorption coefficient for
muscle is 0.23 cm-1 and the fraction reflected at bone-muscle interface is
0.34(from the last example)
Solution
The beam passes through a total thickness of 7.0 cm of muscle
For the attenuation in the muscle,
using I = I0 exp(-kx) = I0 exp(-0.23 x 7.0) = 0.20I0
Given that the fraction reflected at the bone-muscle interface i.e. α is
0.34,
therefore the fraction received back at surface = 0.34 x 0.20 = 0.068 =
1/15 43

44. Obtaining diagnostic information using ultrasound
• The ultrasound transducer is placed on the skin with the water
based jelly excluding any air between the transducer and the skin
• Short pulses of ultrasound are transmitted into the body where
they are partly reflected and partly transmitted at the boundaries
between media in the body
• The reflected pulses or echoes return to the transducer where
they are detected and converted into voltage pulses which are
amplified and processed by electronic circuits to be displayed on
a screen or oscilloscope
• The time between the transmitted and reflected pulses gives
information as to the distance of the boundary from the
transducer
• The intensity of the reflected beam gives information as to the
nature of the boundary
• 2 techniques are in common use for the display of an ultrasound
scan:
– A-scan
– B-scan
44

45. A-scan
• A short pulse of ultrasound is transmitted into the body through the coupling
medium
• At each boundary some of the energy of the pulse is transmitted and some is
reflected
• The transducer(generator/detector) detects the reflected pulses as it now acts as a
receiver
• The signal is amplified and displayed on a c.r.o.
• Reflected pulses received at the transducer from deeper in the body tend to have
lower intensity than those reflected from boundaries near the skin
• This is due not only to the energy being absorbed by the various media but also on
the return of the reflected pulse to the transducer, some of the energy of the pulse
will again be reflected at intervening boundaries
• To allow for this, echoes received later at the transducer are amplified more than
those received earlier
• A vertical line is observed on the c.r.o. corresponding to the detection of each
reflected pulse
• The time-base of the c.r.o. is calibrated so that knowing the speed of the ultrasound
wave in each medium, the distance between boundaries can be determined 45

46. B-scan
• This consists of a series of A-scans all taken from
different angles so that a 2-D image can be formed
• The ultrasound probe consisting of a
generator/detector, for a B-scan does not consist of a
single crystal, but rather an array of smaller crystals
each one at a different angle to its neighbours
• The separate signals received from each of the crystals
in the probe is processed and a pattern of spots is built
up to create a 2-D image for immediate viewing,
photographing or to be stored in computer memory
46

47. Advantages of ultrasound scanning
• Health risk to patient and operator is very much less
compared to use of X-rays
• The equipment is much more portable and easy to use
• Using higher frequency ultrasound enables greater
resolution to be obtained i.e. greater details to be seen
• Modern techniques allow for the detection of very low
intensity reflected pulses, hence boundaries between
tissues where there is little change in acoustic
impedance can be detected
47