How Does Distance Affect Decibel Levels in Sound Propagation?

In summary, the difference in decibel levels of a sound source, B1 and B2, is related to the ratio of its distances, r1 and r2, from the receivers by B2 - B1 = 20 log (r1/r2). However, in the scenario presented in part (b), there is not enough information given to determine the distance of the speaker from each observer. Without knowing the power output of the source, it is impossible to solve for both the intensity level and distance. Therefore, the problem cannot be solved.
  • #1
gh0st
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Question

15. (a) Show that the difference in decibel levels, B1 and B2, of a sound source is related to the ratio of its distances, r1 and r2, from the receivers by

B2 - B1 = 20 log (r1/r2)

(b) A speaker is placed between two observers who are 110 m apart, along the line connecting them. If one observer records an intensity level of 80 dB, how far is the speaker from each observer ?

Answer given by lecturer : [ 10 m and 100 m ]

====================================================

Before i start, i would like to apologize for the messy post as i have no idea how to type out the actual symbols, therefore I'm only using those basic ones on my keyboard ..

Okay, this is my problem ... I managed to solve 15(a) but I'm stuck at 15(b)

====================================================

According to 15(b), i came up with this diagram :

http://img246.imageshack.us/img246/606/untitledmn4.jpg

My worksteps

Formulas used :

i. Intensity Level in Decibels, B = 10 log ( I / Io ) dB ... ( Io is intensity reference, 1.00 x 10^-12 Wm^-2 )

ii. Intensity , I = Power / Area ... ( Area is 4*pi*r^2 )

iii. Assumed that Observer 1 was the person who recorded 80dB mentioned. Therefore :

80dB = 10 log ( I / Io ) dB

Intensity 1 = [ anti-log ( 80/10) ]*[Io]
= ( 1 x 10^-8 )*(1x10^-12)
= 1 x 10^-4 Wm^-2

iv. Used Intensity formula :

I = P/A
I = P/4*pi*r^2
1 x 10^-4 = P/4*pi*r^2

P = ( 1 x 10^-4 )*4*pi*r^-2

v. Okay, so P, Power is not given, so i thought of deriving it from Observer 2, using the Intensity formula ...

Intensity 2 = P/4*pi*r^-2 , however this r is ( 110-r ) , as it's the distance between speaker and obersver 2, as shown in the diagram ..

P = Intesity2*4*pi*[(110-r)^-2]

vi. Since the sound is coming from 1 speaker therefore the value for P should be the same ...

P = ( 1 x 10^-4 )*4*pi*r^-2 and P = Intesity2*4*pi*[(110-r)^-2]vii. I concluded that

( 1 x 10^-4 )*4*pi*r^-2 = Intesity2*4*pi*[(110-r)^-2][/b]

viii. However, the problem is there are 2 unknown variables here ... Intensity 2 and r ( which is sought by the question ... ) And i got stuck here ... Tried many ways in substituting either r or intensity but i got back the same equation ... Also, the reason i posted along 15(a) is because i was wondering if (b) is related to (a) , but i can't find any connection ..

===================================================

please help me , i can't think of where i got it wrong .. and thanks in advance.

edit : sorry for the topic title, i forgot to add it in after typing in this post .
 
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  • #2
I think you are stuck on part (b) because there is not enough information. You don't know the power output of the source, and you only know one observed level. It can't be done.
 
  • #3


Last edited by a moderator: May 3, 2017

Answers and Replies

- Apr 2, 2007

- #2

Astronuc
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Science Advisor

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I think the problem is in the use of the intensity formula. The intensity should be the same at both observers.

So one should be able to equate the two intensities and determine the value of r.

You have the correct equation, but one must remember that the intensity is a measure of power per unit area, and the power must be the same at both observers.

One can also use the inverse-square law for sound, which is essentially the same as the intensity formula.

P/4πr^2 = P/4π(110-r)^2

The power (P) should be the same at both observers.

- Apr 3, 2007

- #3

HallsofIvy
Science Advisor
Homework Helper

41,833

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Astronuc is right. Your equation, ( 1 x 10^-4 )*4*pi*r^-2 = Intesity2*4*pi*[(110-r)^-2], is correct but you need to solve for r.

( 1 x 10^-4 )*4*pi*r^-2 = Intesity2*4*pi*[(110-r)^-2] can be rewritten as
( 1 x 10^-4 )= Intesity2*[(110-r)^-2]/r^-2. Since 110-r= (110-r)^-1, that can be rewritten as
( 1 x 10^-4 )= Intesity2*[(110-r)/r]^2.

Now, (110-r)/r= 110/r- 1 and so [(110-r)/r]^2= 110^2/r^2- 2(110/r)+ 1. Replace [(110-r)/r]^2 by that in the equation above and you have a quadratic equation in r^2.

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Related to How Does Distance Affect Decibel Levels in Sound Propagation?

1. What are decibels and how are they measured?

Decibels are a unit of measurement for sound level. They are measured using a logarithmic scale, with each increase of 10 decibels representing a tenfold increase in sound intensity. This means that a sound at 40 decibels is ten times louder than a sound at 30 decibels.

2. How do decibel levels affect our hearing?

Exposure to high decibel levels can cause damage to our hearing. Sounds at or above 85 decibels can cause permanent hearing loss if exposed for long periods of time. It is important to protect our ears from loud noises to prevent hearing damage.

3. What are some common examples of decibel levels in everyday life?

A quiet library or bedroom is typically around 30 decibels, a normal conversation is around 60 decibels, and a rock concert can reach up to 110 decibels. An increase of 10 decibels represents a sound that is twice as loud.

4. How do you calculate decibel levels?

Decibel levels are calculated using a logarithmic formula. The formula is 10 log (I/I0), where I is the sound intensity being measured and I0 is the reference intensity of 10^-12 watts per square meter.

5. What are some ways to reduce decibel levels?

There are several ways to reduce decibel levels, including using earplugs or earmuffs in loud environments, turning down the volume on electronic devices, and limiting exposure to loud noises. Building materials such as soundproofing insulation can also help reduce decibel levels in a room.

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