Lowering Jet Sound Intensity Below Pain Threshold

Click For Summary

Homework Help Overview

The discussion revolves around calculating the distance required to reduce the sound intensity of a jet engine to below the pain threshold of 130 phon, given that the sound level at 1 meter from the jet during takeoff is approximately 150 phon.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the definition of intensity and its relation to power and area, with references to the formula for sound intensity from a point source. There are requests for further clarification on the calculations involved.

Discussion Status

Some participants have provided insights into the formulas related to sound intensity and the assumptions regarding the jet engine as a point source. There is an ongoing exploration of how to apply these concepts to find the required distance.

Contextual Notes

Participants are working within the constraints of needing to understand the relationship between sound intensity and distance, as well as the definitions of units involved, such as phon. There is an acknowledgment of the need for further explanation and guidance on the calculations.

teng125
Messages
416
Reaction score
0
the pain threshold of your ears is about 130 phon.The sound levelclose to a jet (r=1m) when it takes off is about 150 phon.How far do you have to go away in order to lower the intensity level below the pain threshold??

anyone pls tell me the formula to use so that i can calculate as i can't find and don't know which formula to use.

thanx very much
 
Physics news on Phys.org
Intensity is defined as power per unit area;

[tex]I = \frac{P}{A}[/tex]

Now, if we assume the jet engine is a point source, which emits sound equally in all directions thus forming a sphere of radius r and a point r distant from the source, then the surface area of a sphere is given by [itex]4\pi r^2}[/itex], thus the intensity at a point r from the source is given by;

[tex]I = \frac{P_{0}}{4\pi r^2}[/tex]

Where P is the power of the source. Can you go from here?
 
nono,pls explain a bit further
thanx...
 
The phon is basically just a unit of intensity, it is actually a unit of loudness but in this case can be treated as a unit of sound intensity. You have been given the intensity at 1 meter from the source, thus you can calculate the power of the source (which is constant). Now you just plug in your numbers for power and intensity and solve for r.

Can you go from here?
 
okok.i'll try
thanx
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
19K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 12 ·
Replies
12
Views
2K