Calculate Intensity of Sound Wave at Cliff from Radio Source

Click For Summary

Homework Help Overview

The problem involves calculating the intensity of a sound wave from a radio source located across a valley from a cliff, where an echo is heard. The context includes the speed of sound in air and the intensity of sound at a specific distance from the source.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between distance and intensity, noting that intensity decreases with the square of the distance from the source. There are attempts to calculate the intensity at the cliff based on the distance derived from the echo time.

Discussion Status

Some participants are exploring the calculations related to intensity, with one providing a specific numerical attempt. There is a request for clarification on the calculations used, indicating a lack of consensus on the correct approach or result.

Contextual Notes

Participants are working with the assumption that the sound source is isotropic and are considering the implications of distance on intensity. There is a specific intensity value given for a distance of 1.0 m, which is central to the calculations being discussed.

kokenwa
Messages
10
Reaction score
0

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?
 
Physics news on Phys.org
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.
 

Similar threads

Waves: Calculate the sound intensity from two speakers
  • · Replies 3 ·
Replies
3
Views
3K
Max velocity of a vibrating loud speaker membrane given sound intensity
  • · Replies 4 ·
Replies
4
Views
2K
Calculating τ by knowing I is proportional to A^2
  • · Replies 8 ·
Replies
8
Views
3K
Intensity of sound wave and energy
  • · Replies 1 ·
Replies
1
Views
2K
Audio interference of sound waves from two speakers
  • · Replies 19 ·
Replies
19
Views
5K
Calculating Intensities of Out-of-Phase Sound Waves
  • · Replies 1 ·
Replies
1
Views
2K
Total sound intensity from sources
  • · Replies 11 ·
Replies
11
Views
3K
Sound Wave Interference Problem
  • · Replies 5 ·
Replies
5
Views
2K
Loudspeaker Question (Sound and Intensity)
  • · Replies 1 ·
Replies
1
Views
3K
Calculate Intensity of Sound Waves at 3.0 m from Source
  • · Replies 1 ·
Replies
1
Views
2K