University Physics: Ultrasonic Testing and Sound Physics of Gold Bars

This assignment focuses on sound physics and its application in detecting adulteration in gold bars using ultrasonic probes. The solution explores the relative sizes of reflected longitudinal beams from tungsten and the back wall, considering acoustic impedance and attenuation. It also calculates the thickness of tungsten inserted in a gold bar using longitudinal wave velocity and time-of-flight measurements, including necessary assumptions. Furthermore, the assignment derives a line equation relating the measured thickness with the thickness of tungsten, providing a comprehensive analysis of the ultrasonic testing method. The solution provides detailed calculations, explanations, and assumptions made throughout the process, offering a complete understanding of the physics behind the detection of adulterated gold.

SOUND PHYSICS
Question 1
It was recently discovered that gold bars (density 19.32 g/cm3) have been adulterated with a
cheaper metal of similar density such as Tungsten (density 19.25 g/cm3). It has been proposed
that ultrasonic probes can be used to detect the presence of a different metal in the gold. For
more information please visit the following websites:
http://www.olympus-ims.com/en/applications/ut-testing-gold-bars/
http://testyourgold.com/how-ultrasound-tests-gold-and-silver/
Estimate the relative sizes of the reflected longitudinal beams that would be received by an
ultrasonic probe i.e. the relative size of the beam reflected off the tungsten layer and the beam
reflected off the back wall (20 marks)
SOLUTION
Sensitivity is the capability of an ultrasonic system to detect defects at a certain depth in a test
material. The transducer system is more sensitive when the received signal is greater.
Adulteration of a gold bar by the method of inserts will result in predictable variations in the
way ultrasonic waves traverse the metal. The pattern of wave reflections is changed by inserts of
material which is not gold in a bar, a response that can also result due to internal voids. Huge
inserts that cover much of the volume of the bar can also be noticed by means of changes in
sound velocity.

i. Relative size of Tungsten
Tungsten longitudinal velocity is calculated as shown;
Where:
VL =
Longitudinal
Wave Velocity
E =
Modulus of
Elasticity which is
= 411
 =
Density which is =
19.25 =
Poisson's Ratio
which is 0.284
VL= 5.25 M/S
Z= pc
Z= acoustic impedance

C= material sound velocity
P= material density
Z= 19.25×c
= 19.25 ×c (longitudinal wave velocity)
= 19.25 × 5.25 m/s
=101.0625
ii. Beam of backwall
Z= pc
Z= acoustic impedance
C= material sound velocity
P= material density
Z= 19.32×c
= 19.32 ×c (longitudinal wave velocity)
= 19.32 × 3240 m/s
=62,596.8
Db loss (amplitude loss)= 10 log10(4 z1z2/(z1±z2)2)

= 10 log10(4 ×101.0625×62596.8/(101.0625±62596.8)2)
=21.913
As ultrasound passes via a medium it attenuates. There are 3 causes of attenuation: diffraction,
scattering and absorption, assuming no major reflections.
ASSUMPTIONS
a. Sound velocity is dependent on the elasticity and density of the solid.
b. On a material denser than the external medium, reflected light undergoes a 180° phase
reversal.
State any assumptions made. Please note that Tungsten has a Young's modulus is 411 GN/m2,
and a poisson's ratio is 0.284 and Gold has longitudinal wave velocity of 3240 m/s.
A standard gold bar which has a standard thickness of 8.3 mm, has 6 mm of tungsten inserted in
the middle. If we use an ultrasonic probe to measure the thickness using longitudinal waves,

what thickness would be found?
(20 marks)
SOLUTION
Ultrasonic vibrations travel just the same as light in the form of a wave. The difference is
ultrasound requires an elastic medium such as a solid or liquid while light can travel even in a
vacuum.
Shear modulus G= VT2× density
= 32402(10±8) ×19.32
=2.02
Young Modulus = FL,
AL , ,
L,,= YM A
FL,
convert the given values to units of time assuming you are using pulse-echo. So, the thickness of
tungsten is calculated as follows
(8.3 × 2)÷3.24 = 5.123 time through Gold
(6 × 2) ÷ 3.24 = 3.7 time through both Gold and Tungsten
5.123- 3.7= 1.423 the difference in time
1.423 × 5.25 = 7.47mm the thickness of Tungsten

ASSUMPTIONS
a. The longitudinal waves are identical with waves in air.
b. It is possible to measure the thickness of gold sheet within sub-micron accuracy
Using all the above information, find the values of a and b in the following line equation that
relates the measured thickness, d, with the thickness of tungsten, dt: dt = a(b -d) (10 marks)
Measured thickness = d
Thickness of Tungsten dt: dt = a(b -d)
L,,= d = YM A
FL,
dt: dt = a(b -d)
L,,= d = YM A
F ( a ( b−d ) + 8.3)
a = YM A+d +Fad+8.3 Fd
Fdb

b= YM A+d + Fad+ 8.3 Fd
Fda