Calculate Intensity of Sound Wave at Cliff from Radio Source

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SUMMARY

The discussion revolves around calculating the intensity of sound waves from a radio source located 206 meters away from a cliff, where the speed of sound is 343 m/s. The intensity of the music at a distance of 1 meter from the radio is given as 1.0 x 10-5 W/m2. Participants clarify that intensity decreases with the square of the distance, leading to the formula I = I0 / r2. The correct intensity at the cliff, located 206 meters away, is calculated to be approximately 2.35 x 10-7 W/m2.

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  • Understanding of sound wave propagation and intensity
  • Knowledge of the inverse square law
  • Basic algebra for calculations
  • Familiarity with units of measurement in physics (W/m2)
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  • Study the inverse square law in detail
  • Learn about sound intensity levels and decibels
  • Explore the concept of isotropic sound sources
  • Review calculations involving distance and intensity in wave physics
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Students in physics, acoustics enthusiasts, and anyone interested in understanding sound wave behavior and intensity calculations.

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Homework Statement


Michelle is enjoying a picnic across the valley from a cliff. She is playing music on her radio (assume it to be an isotropic source) and notices an echo from the cliff. She claps her hands and the echo takes 1.2 s to return.
(a) Given that the speed of sound in air is 343 m/s on that day, how far away is the cliff?
-i got the correct answer for this one, and it was 206m


(b) If the intensity of the music 1.0 m from the radio is 1.0 10-5 W/m2, what is the intensity of the music arriving at the cliff?


i can't find the answer for b. can someone help me out?
 
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Intensity is proportional to 1/r^2.

That is, if the source is twice as far away, the intensity is 1/4; three times as far away, 1/9 etc...
So if the source is 206 times as far away...?
 
i put in 2.35 x 10^-5 and it was still wrong
 
how did you get 2.35 x 10^-5 ? - show your working.

this is not what I got.
 

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