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Posted

Well since my physics teacher doesn't...well...teach everyhing, I cant figure out this homework problem. If anyone could help or point me in the right direction it would be greatly appriciated. Here's the problem:

 

 

A loudspeaker is placed between 2 observers who are 110 meters apart along a line connecting them. If one observer hears a sound level of 60dB, and the other records 80dB, how far is the speaker from each observer?

 

Thanks in advance

-tw

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Posted

Just sayin', instead of posting it here expecting us to help you, is there a way for you to look for online lessons? I mean, I'd be willing to help you if I weren't only in trigonometry, but I don't think the moderators would agree with me.

Posted

A loudspeaker is placed between 2 observers who are 110 meters apart along a line connecting them. If one observer hears a sound level of 60dB, and the other records 80dB, how far is the speaker from each observer?

 

Acoustics is probably the branch of physics I'm worst at, but here is what I would do.

 

1. Use the definition of sound intensity level: Intensity level=(10 dB) log [intensity/10^-12 W/m^2] to find the intensity of each sound wave.

 

2. Use the inverse-square law for intensity: [intensity 1/Intensity 2]=[((distance 1)^2)/((distance 2)^2)] to find the two distances. You will also need to use this simple algebraic fact: distance 1+distance 2=110 m.

 

I'm not 100% sure if this will work, but it makes sense and is the only way I can think of doing it. Also, you might want to look up these equations in your book as they are a bit difficult to read on here without an equation writing program.

Posted

A loudspeaker is placed between 2 observers who are 110 meters apart along a line connecting them. If one observer hears a sound level of 60dB, and the other records 80dB, how far is the speaker from each observer?

 

Acoustics is probably the branch of physics I'm worst at, but here is what I would do.

 

1. Use the definition of sound intensity level: Intensity level=(10 dB) log [intensity/10^-12 W/m^2] to find the intensity of each sound wave.

 

2. Use the inverse-square law for intensity: [intensity 1/Intensity 2]=[((distance 1)^2)/((distance 2)^2)] to find the two distances. You will also need to use this simple algebraic fact: distance 1+distance 2=110 m.

 

I'm not 100% sure if this will work, but it makes sense and is the only way I can think of doing it. Also, you might want to look up these equations in your book as they are a bit difficult to read on here without an equation writing program.

 

...I am now afraid to go into Physics next year, in 11th grade. Thank you, my good man, for scaring the living hell out of me.

Posted

^Don't worry, this isn't high school physics. I'm posting this solution as a third semester college physics student. If you are good with math, high school physics is no problem. I don't know whether or not you will be in a calculus based class, but if you aren't physics is as easy as solving a right triangle. The calculus class isn't too bad either since there are many constants and resulting derivatives/integrals are rarely more difficult than 1/x.

Posted

This is 11th grade physics I am currently in. Its not hard, its just our teacher doesn't always teach. By the way, the answers are 108.9m and 1.1m away from the speaker.

 

Onewheeled, dont be scared. Its just a math class with labs and themed (I guess you could say) math lessons. Its not hard, you just need some sort of Pre-calc or regular calculus before going into it.

Posted

I took three semesters of college physics and never had anything about acoustics.

So apparently what is covered can vary greatly.

 

 

Jon *I once failed a physics test for using the left hand rule* Azevedo

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