What are the Decibel Levels of Two Combined 70 dB Sound Sources?

  • Context: Undergrad 
  • Thread starter Thread starter MathematicalPhysicist
  • Start date Start date
  • Tags Tags
    Sound
Click For Summary

Discussion Overview

The discussion revolves around the calculation of decibel levels when combining two independent sound sources, specifically two 70 dB sources. Participants explore the relationship between amplitude and intensity in sound, as well as the implications of these relationships for calculating resultant sound levels.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants propose that doubling the amplitude of a sound signal results in an increase of approximately 4 dB, while others suggest it may be closer to 6 dB based on their interpretations of the relationship between intensity and amplitude.
  • One participant calculates the resultant sound level of two 70 dB sources using the formula for decibels, concluding that the combined sound level is approximately 73 dB.
  • Another participant questions the clarity of the text regarding the formulas needed to solve the problems presented, expressing frustration over the lack of foundational explanations.
  • There is a discussion about the interpretation of "power intensity" and its equivalence to intensity, with some participants agreeing on the mathematical relationships involved.
  • Participants engage in clarifying the mathematical steps involved in deriving the resultant sound level and the implications of their calculations.

Areas of Agreement / Disagreement

Participants express differing views on the correct increase in decibels when amplitude is doubled, with some suggesting 4 dB and others suggesting 6 dB. The calculation of the resultant sound level from two 70 dB sources also remains a point of contention, with no consensus reached on the final answer.

Contextual Notes

Some participants note that the original text does not adequately cover the necessary formulas or concepts, leading to confusion in solving the problems. There are also unresolved assumptions regarding the definitions of terms like "power intensity."

MathematicalPhysicist
Science Advisor
Gold Member
Messages
4,662
Reaction score
372
im reading the text of dave benson about music and maths and i have 2 questions from the text which i need an answer to.
1. power intensity is proportional to the square of the amplitude. how many decibles represent a doubling of the amplitude of a signal?
(my answer is 4 but I am not sure).
2. (multiple choices) two independents 70 dB sound sources are heard together. how loud is the resultant sound to the nearest dB?

a. 140 dB
b. 76 dB
c. 73 dB
d. 70 dB
e. none of above

now I am stuck at it because i think it is either a or d.

thanks for any help.
 
Physics news on Phys.org
Originally posted by loop quantum gravity
im reading the text of dave benson about music and maths and i have 2 questions from the text which i need an answer to.
1. power intensity is proportional to the square of the amplitude. how many decibles represent a doubling of the amplitude of a signal?
(my answer is 4 but I am not sure).
2. (multiple choices) two independents 70 dB sound sources are heard together. how loud is the resultant sound to the nearest dB?

a. 140 dB
b. 76 dB
c. 73 dB
d. 70 dB
e. none of above

now I am stuck at it because i think it is either a or d.

thanks for any help.
I'm not sure about the first question but I will happily answer the second. The formula for decibles is

B = 10*log(I/A)

where B is the number of decibles, A is the reference level which is 10^-12 watts per metre^2 and I is the intensity of the sound. Now a 70dB source has an intensity as follows

70 = 10*log(I/A)
10^7 = I/A
I = A 10^7

there are two 70dB sources in total so the above intensity will be doubled. Therefore the total intensity is

I = 2*10^7 A

putting this back into the decible formula

B = 10*log(I/A)
B = 10*log(2*10^7 A / A)
B = 10*[log(2)+log(10^7)]
B = 10*[log(2)+7]
B ≈ 73.0103dB

So (c) would be the correct answer!
 
Last edited by a moderator:
thanks, surprsingly i haven't met this formula in the text.
 
Originally posted by loop quantum gravity
thanks, surprsingly i haven't met this formula in the text.
It is quite stupid of the author then to set problems without teaching the machinary to solve them!
 
indeed.
you can find the text I am talking about in here http://www.math.uga.edu/~djb/html/music-hq.pdf
the problems are at page number 11.
 
Last edited by a moderator:
I haven't taken a look at the book but I will take a shot at the first question you asked. I'm not quite sure what "power intensity" is but I will make the assumption that power intensity = intensity = I. The question is restated below

"power intensity is proportional to the square of the amplitude. how many decibles represent a doubling of the amplitude of a signal?
(my answer is 4 but I am not sure)."

therefore I = k C^2, where C is the amplitude and k is the constant of proportionality.

it is easily shown that 4*I = k (C*2)^2, therefore doubling the amplitude quadruples the intensity. So using the formula I used in my last post

B = 10*log(I/A)

now doubling the amplitude quadruples the intensity so

B'= 10*log(4*I/A)
B'= 10*log(4)+10*log(I/A)
B'= 10*log(4)+B

therefore if my assumption is correct i.e. power intensity is just another word for intensity, then the number of decibles corrosponding to a doubling of the amplitude is 10*log(4)≈6.0206dB.
 
Last edited by a moderator:
Originally posted by MathNerd
I haven't taken a look at the book but I will take a shot at the first question you asked. I'm not quite sure how amplitude fits in with dB but I will make the assumption that power intensity = intensity = I. The question is restated below

"power intensity is proportional to the square of the amplitude. how many decibles represent a doubling of the amplitude of a signal?
(my answer is 4 but I am not sure)."

therefore I = k C^2, where C is the amplitude and k is the constant of proportionality.

it is easily shown that 4*I = k (C*2)^2, therefore doubling the amplitude quadruples the
tell me if I am wrong but when you multiply I=k*C^2 by 4 it should be 4I=4kC^2.
 
Originally posted by loop quantum gravity
tell me if I am wrong but when you multiply I=k*C^2 by 4 it should be 4I=4kC^2.
Notice that in my post I had the following
4*I = k (C*2)^2

notice that the C*2 is all squared

So, k (C*2)^2 = k C^2 2^2 = 4k C^2
So, 4*I = 4k C^2

therefore your expression is equivalent to mine.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 9 ·
Replies
9
Views
5K
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
7K
  • · Replies 5 ·
Replies
5
Views
6K