anonymous
  • anonymous
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
Physics
schrodinger
  • schrodinger
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Michele_Laino
  • Michele_Laino
the intensity is proportional to the square of the amplitude
anonymous
  • anonymous
I=kA^2 so i multiplied both sides by two to get 2I=2A^2 am i right?
Michele_Laino
  • Michele_Laino
yes! correct! So, the new amplitude is: sqrt(2)*A

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anonymous
  • anonymous
But wouldn't it give √2I too? Or it doesn't matter ?
Michele_Laino
  • Michele_Laino
no the intensity is 2*I, so we can write: \[2I = K{A^2} + K{A^2} = 2K{A^2} = K{\left( {\sqrt 2 A} \right)^2}\]
Michele_Laino
  • Michele_Laino
so the new amplitude is: \[{\sqrt 2 A}\]
anonymous
  • anonymous
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 ?
anonymous
  • anonymous
Michele_Laino
  • Michele_Laino
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: \[\Large B = \sqrt 2 A\]
Michele_Laino
  • Michele_Laino
anonymous
  • anonymous
Ohh! I get it if you put √2A in 2I=kB^2 you get 2A^2 THANKYOU SO MUCH !!!!
Michele_Laino
  • Michele_Laino
:):)

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