Calculating Sound Level from One Firecracker

  • Thread starter Thread starter physics1987
  • Start date Start date
  • Tags Tags
    Sound
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
6 replies · 12K views
physics1987
Messages
4
Reaction score
0
In my homework, I was assigned a question,
If two firecrackers produce a sound level of 95dB when fired simultaneously at a certain plance, what will be the sound level if only one is exploded ( hint: add intensities, not dB?

My immediate guess is to divide 95 in two, but it seems too simple, plus if it is only one fire cracker, would the sound level be 1/2 ? I don't think I am right, but can someone help me?
thanks
 
on Phys.org
physics1987 said:
In my homework, I was assigned a question,
If two firecrackers produce a sound level of 95dB when fired simultaneously at a certain plance, what will be the sound level if only one is exploded ( hint: add intensities, not dB?

My immediate guess is to divide 95 in two, but it seems too simple, plus if it is only one fire cracker, would the sound level be 1/2 ? I don't think I am right, but can someone help me?
thanks

Welcome to PF

What is the definition of a db?
 
LowlyPion said:
Welcome to PF

What is the definition of a db?

a decible (measured in hertz)
 
physics1987 said:
a decible (measured in hertz)

sorry, I got mixed up between frequency and dB...I guess my question is, is it possible to just divide dB by the amount of objects a noise is coming from?
ie. 3 sirens = 30dB
therefore 2 sirens =20dB?
 
physics1987 said:
sorry, I got mixed up between frequency and dB...I guess my question is, is it possible to just divide dB by the amount of objects a noise is coming from?
ie. 3 sirens = 30dB
therefore 2 sirens =20dB?

The sound intensity may be half in your example but that doesn't change the db's by half, because what is the a db?

It's a Log base 10 of the ratio of sound intensity to some reference.

So if Log10(P1/Po) = 95,
then Log10(1/2*P1/Po) = your answer.
 
LowlyPion said:
The sound intensity may be half in your example but that doesn't change the db's by half, because what is the a db?

It's a Log base 10 of the ratio of sound intensity to some reference.

So if Log10(P1/Po) = 95,
then Log10(1/2*P1/Po) = your answer.

thank you for the help,
I was able to solve for Intensity (P1) as .0031622777w/m squared
I then inputted this into the formula that you gave me to come up with the answer of 92, and I also double checked the formula and got the original answer of 95 for both firecrackers. Thank you so much!