What Is the Reflection Distance and SPL of Sound in an Auditorium?

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Homework Help Overview

The discussion revolves around calculating the reflection distance and sound pressure level (SPL) of sound in an auditorium setting. The original poster presents a scenario involving a speaker positioned 90 feet from a chair, with specific measurements related to sound pressure level and reflection time from the roof.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the calculations related to the distance of the first reflection, the speed of sound at a given temperature, and the resulting SPL. There are attempts to clarify unit conversions and the implications of the calculated distances.

Discussion Status

The discussion is ongoing, with participants providing feedback on calculations and unit usage. Some participants have pointed out potential errors in the original calculations, particularly regarding the interpretation of units and the resulting distances.

Contextual Notes

There are concerns about the accuracy of the calculations, especially regarding the reflection distance being significantly larger than expected compared to the direct path. Participants are also addressing the importance of unit consistency in their calculations.

Jese Reyes
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Homework Statement


Spaker located at 90 feet from a auditorium chair. The sound pressure level at the chair is 70 dB. The first reflection from the roof takes 200ms to get to the chair.

What distance does the first reflection travel to get to the chair if the room temperature is 303.15 K?
What is the SPL of the first reflection assuming a perfect reflection from the roof?
What would the delay be of the first reflection after the direct sound has reached the chair first?

Homework Equations


[/B]
Reflection delay = (reflected path - direct path)/ speed of sound
Reflection Level at listening position = 20 log (Direct sound/Reflection distance)

The Attempt at a Solution



90ft = 27.432 mts
331 + .6(30C) = 349 ms This is the speed of sound at 303.15K

Solve for Rflection path 200ms= (rpath - 27.432 mts)/349 ms rpath=69800 mts does this make sense?

SPL = 20 log (27.432/6980) = -68.11 dB

Im not sure my math is making sense. I am having trouble understanding how to integrate these type of problems when calculating delays, reflection distance and SPL.
 
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Jese Reyes said:

Homework Statement


Spaker located at 90 feet from a auditorium chair. The sound pressure level at the chair is 70 dB. The first reflection from the roof takes 200ms to get to the chair.

What distance does the first reflection travel to get to the chair if the room temperature is 303.15 K?
What is the SPL of the first reflection assuming a perfect reflection from the roof?
What would the delay be of the first reflection after the direct sound has reached the chair first?

Homework Equations


[/B]
Reflection delay = (reflected path - direct path)/ speed of sound
Reflection Level at listening position = 20 log (Direct sound/Reflection distance)

The Attempt at a Solution



90ft = 27.432 mts
331 + .6(30C) = 349 ms This is the speed of sound at 303.15K

Solve for Rflection path 200ms= (rpath - 27.432 mts)/349 ms rpath=69800 mts does this make sense?

SPL = 20 log (27.432/6980) = -68.11 dB

Im not sure my math is making sense. I am having trouble understanding how to integrate these type of problems when calculating delays, reflection distance and SPL.
Welcome to the PF. :smile:

First, it's important to use standard abbreviations for units, and to be careful to write them correctly. So this:
90ft = 27.432 mts
331 + .6(30C) = 349 ms

Should be this:
90ft = 27.432 m
331 + .6(30C) = 349 m/s

And this:
Solve for Rflection path 200ms= (rpath - 27.432 mts)/349 ms rpath=69800 mts does this make sense?

Should be this:
Solve for Reflection path: distance traveled in 200ms is D(200ms) = 349 m/s * 200ms = _____
 
berkeman said:
Welcome to the PF. :smile:

First, it's important to use standard abbreviations for units, and to be careful to write them correctly. So this:
90ft = 27.432 mts
331 + .6(30C) = 349 ms

Should be this:
90ft = 27.432 m
331 + .6(30C) = 349 m/s

And this:
Solve for Rflection path 200ms= (rpath - 27.432 mts)/349 ms rpath=69800 mts does this make sense?

Should be this:
Solve for Reflection path: distance traveled in 200ms is D(200ms) = 349 m/s * 200ms = _____
Thank you, i will make my future posts more accurate.
 
You're welcome. So what do you get for the path length, and what does that give you for the path length difference?
 
berkeman said:
You're welcome. So what do you get for the path length, and what does that give you for the path length difference?
I get the result of 69,800 m. The difference would be 69,772.68 m. This number in my opinion is too high compared to the 90 ft distance of the speaker and chair.
 
Jese Reyes said:
I get the result of 69,800 m.
That is longer than the 27.4m direct path, so that seems correct.
Jese Reyes said:
The difference would be 69,772.68 m.
What does that mean? Oh wait, I thought your were using the "," character above as the decimal point (European style). If not, then 69,800m is incorrect. That calculation should give 69.8m.

Oh, I see you multiplied by 200s instead of 200ms. Note that 200ms = 0.2s (be sure to keep track of your units)
 
berkeman said:
That is longer than the 27.4m direct path, so that seems correct.

What does that mean? Oh wait, I thought your were using the "," character above as the decimal point (European style). If not, then 69,800m is incorrect. That calculation should give 69.8m.

Oh, I see you multiplied by 200s instead of 200ms. Note that 200ms = 0.2s (be sure to keep track of your units)
Makes complete sense! Now the numbers fit with the context of the problem. Such a small detail, i need to be more careful with my units. Thank you so much!
 
You're welcome, good job. And thanks for showing so much of your work -- that makes it a lot easier to catch small math errors or mistakes in unit conversions.
 

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