Sound intensity as a function of distance from source

  • #1

Homework Statement

THe intensity of a sound wave at a fixed distance from a speaker vibrating at 1.24 kHz is .388 W/m^2.
Calculate the intensity if the frequency is reduced to 2.64 kHz and the displacement is doubled.

used I=.5p(w^2)(A^2)v for the initial and the new intensity. I ended up with
4I0(f/f0)^2=I plugging in the numbers didn't get me the right answwer. can someone tell me what im doing wrong. thanks.
 

Answers and Replies

  • #2
dynamicsolo
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Homework Statement

THe intensity of a sound wave at a fixed distance from a speaker vibrating at 1.24 kHz is .388 W/m^2.
Calculate the intensity if the frequency is reduced to 2.64 kHz and the displacement is doubled. /QUOTE]

Could you check the values you've posted here? You're saying that the frequency was reduced from 1.24 kHz to 2.64 Hz? Is this second frequency correct?
 
  • #3
yeah the problem is stated that way which is why i think i keep getting the wrong answer.
 
  • #4
dynamicsolo
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You still haven't shown how you actually did the calculation or what the answer is claimed to be. The equation you have is for the power of a wave (I presume the 'p' represents density of the medium); since the distance is kept fixed, the intensity will change in the same proportion as the power. With both the amplitude and the frequency doubled, the power of the wave and the intensity at the specific distance should quadruple.
 
  • #5
dynamicsolo
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With both the amplitude and the frequency doubled, the power of the wave and the intensity at the specific distance should quadruple.
Well, that's teach me to write some of these comments too late at night. Doubling each of those two factors will increase the power (and intensity at fixed distance) by a factor of 4, so doubling both will give an increase by a factor of 16.
 

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