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An ultrasonic sound wave is used to detect the position of defect in a steel bar of thickness 50 cm. If the echo times are 40 and 90 s, locate the position of defect.

Mumbai university > FE > SEM 1 > Applied Physics 1

Marks: 5M

Year: Dec 2013

1 Answer
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Given:-

Thickness, b = 50 cm = 0.5 m

Pulse arrival time for a defect, td=40μs=40×106s

Pulse arrival time for the other end of the bar, tb=90μs=90×106s

To Find: - The location of fault (x)

Solution:-

The thickness of the Steel bar can be expressed as,

b=vtb2

v=2btb

The fault position can be determined by,

x=vtd2

x=td2×2btb

x=40×1062×2×0.590×106

x=0.2222m

x=22.22cm

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