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Sound Assignment - Physics

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Added on  2020-05-28

Sound Assignment - Physics

   Added on 2020-05-28

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SOUND PHYSICSQuestion 1It was recently discovered that gold bars (density 19.32 g/cm3) have been adulterated with a cheaper metal ofsimilar density such as Tungsten (density 19.25 g/cm3). It has been proposed that ultrasonic probes can be usedto 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)SOLUTIONSensitivity 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.
Sound Assignment - Physics_1
i.Relative size of TungstenTungsten longitudinal velocity is calculated as shown;Where:VL=LongitudinalWave Velocity E=Modulus of Elasticity which is = 411=Density which is = 19.25=Poisson's Ratio which is 0.284VL= 5.25 M/SZ= pcZ= acoustic impedanceC= material sound velocityP= material density
Sound Assignment - Physics_2
Z= 19.25×c= 19.25 ×c (longitudinal wave velocity)= 19.25 × 5.25 m/s=101.0625ii.Beam of backwallZ= pcZ= acoustic impedanceC= material sound velocityP= material densityZ= 19.32×c= 19.32 ×c (longitudinal wave velocity)= 19.32 × 3240 m/s=62,596.8Db loss (amplitude loss)= 10 log10(4 z1z2/(z1±z2)2)= 10 log10(4 ×101.0625×62596.8/(101.0625±62596.8)2)
Sound Assignment - Physics_3

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