Calculating Sound Level from One Firecracker

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In summary, If two firecrackers produce a sound level of 95dB when fired simultaneously at a certain place, what will be the sound level if only one is exploded? 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?
  • #1
physics1987
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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
 
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  • #2
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?
 
  • #3
LowlyPion said:
Welcome to PF

What is the definition of a db?

a decible (measured in hertz)
 
  • #5
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?
 
  • #6
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.
 
  • #7
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!
 

1. How is sound level calculated from one firecracker?

Sound level is calculated by measuring the sound pressure level (SPL) in decibels (dB) at a given distance from the firecracker. A sound level meter is typically used to measure SPL.

2. What factors affect the sound level of a firecracker?

The sound level of a firecracker is affected by its size, type, and the amount of explosive material used. The environment in which the firecracker is ignited, such as an open field versus a closed room, can also impact the sound level.

3. How far away should I stand from a firecracker to protect my hearing?

The recommended safe distance to protect your hearing from a single firecracker is at least 50 feet (15 meters). However, this distance may vary depending on the size and type of the firecracker, so it is always best to err on the side of caution and stand farther away.

4. What is the maximum sound level a firecracker can produce?

The maximum sound level a firecracker can produce is around 150 dB, which is equivalent to a jet engine at takeoff. However, most firecrackers typically produce sound levels between 120-140 dB.

5. Can the sound level of a firecracker cause permanent hearing damage?

Yes, exposure to high sound levels, such as those produced by firecrackers, can cause permanent hearing damage. It is important to always wear hearing protection and stand at a safe distance when near firecrackers to prevent potential hearing loss.

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