An Ultrasound Test was run under the following conditions Power: 40 v Sampling R

Question

An Ultrasound Test was run under the following conditions Power: 40 v Sampling Rate: 50 MHz Gain: 3 Determine: a) The speed of sound in this material if the sample thickness was 12.5 mm and Frequency: 5000 kHz Cycles: 2 the results as shown in Figure I b) The thickness of the material in Figure 2, if same material was used. (5 points) 1226,221) (1400,202) 1000 1500 2000 Figure 1. Ultrasonic results for 12.5 mm thickness 250 200 150 100 (189,81) 50 -(104,56) 200 400 600 Figure 2. Ultrasonic results for unknown thickness

Explanation / Answer

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a. The required formula is l = ct/2, where l = thickness of material, c = velocity of sound in material, t = traverse time. For the first sample, l = 12.5 x 10^(-3) m. Now time taken = 1400 - 1226 samples = 174 samples. Since sampling rate = 50MHz, time taken between 2 samples = 174/50M = 3.48 x 10^-6 seconds. Therefore

c = 2l/t = 2 x 12.5 x 10^-3 / 3.48 x 10^-6 = 7.18 x 10^3 m/s

b. l = ct/2. time taken = t = 189 - 104 = 85 samples = 85/50M = 1.7 x 10^-6 s. c = 7.18 x 10^3 m/s. Thus,

l = 7.18 x 10^3 x 1.7 x 10^-6 /2 = 12.2 mm

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