study question re: sound intensity calculation

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4 comments

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    Adrielle Slaugh

    I didn't understand that calculation at all. I tried figuring it out and couldn't get it. Did you ever get an answer to that one?

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    Anton Gross

    No, not great assistance on this one. Sorry Adrielle.

     

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    Anson Nickel

    Heyo. I can't recall, but I think they give you the following equation in the exam:

     

    Delta(db) = 10 log (I/Io)

     

    Essentially, the db change equals 10 log (ratio of intensities). In this case, the problem states that intensity was tripled (ie - 3/1).

     

    Change in (db) = 10 log(3) = 4.77 db. 75 db + 4.77 db = 79.77 db (probably rounded to 80 on the exam).

     

    Sorry for the formatting - does this make more sense?

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    Luis Vargas

    Anson, if the question asked for the raise in intensity, wouldn't the answer be just 4.77? 

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