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Please find attached questions along with supplements. I recommend Jeetendra Gour as a tutor to work out this assignment

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Please find attached questions along with supplements. I recommend Jeetendra Gour as a tutor to work out this assignment

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Answered Same Day Dec 26, 2021

Solution

David answered on Dec 26 2021
104 Votes
a) from the given data -
largest trace amplitude Log10AN = 3.6075 µm
So, AN = 10
3.6075 = 4050.4194 µm or 4.0504 mm
Log10AE = 2.8762 µm
So, AE = 10
2.8762 = 751.9691 µm or 0.7519 mm
Now vector amplitude A = (AN
2 + AE
2)1/2
= (4.05042 + 0.75192)1/2
= (16.4057 + 0.5653)1/2...
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