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anonymous

  • one year ago

The amplitude of a wave is A and intensity is I. Which amplitude is neccessary for the intensity to be doubled to 2I? a. A^2 b. √A c.√2A d. 2A

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  1. Michele_Laino
    • one year ago
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    the intensity is proportional to the square of the amplitude

  2. anonymous
    • one year ago
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    I=kA^2 so i multiplied both sides by two to get 2I=2A^2 am i right?

  3. Michele_Laino
    • one year ago
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    yes! correct! So, the new amplitude is: sqrt(2)*A

  4. anonymous
    • one year ago
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    But wouldn't it give √2I too? Or it doesn't matter ?

  5. Michele_Laino
    • one year ago
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    no the intensity is 2*I, so we can write: \[2I = K{A^2} + K{A^2} = 2K{A^2} = K{\left( {\sqrt 2 A} \right)^2}\]

  6. Michele_Laino
    • one year ago
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    so the new amplitude is: \[{\sqrt 2 A}\]

  7. anonymous
    • one year ago
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    But if we square root one side of the equation won't we do the same to the other side? So it becomes like √2I=k√2A ?

  8. anonymous
    • one year ago
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    @Michele_Laino

  9. Michele_Laino
    • one year ago
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    I haven't squared root the sides of that equation. The new intensity is 2*I, and as all intensities, it has to be proportional to the square of some amplitude, so I can write: \[\Large 2I = K{B^2}\] where B is the new amplitude. Now comparing that equation with the previous equation: \[\Large 2I = 2K{A^2}\] we can write: \[\Large {B^2} = 2{A^2}\] and taking the square root, we get: \[\Large B = \sqrt 2 A\]

  10. Michele_Laino
    • one year ago
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    @rosestella

  11. anonymous
    • one year ago
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    Ohh! I get it if you put √2A in 2I=kB^2 you get 2A^2 THANKYOU SO MUCH !!!!

  12. Michele_Laino
    • one year ago
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    :):)

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