10R Physical #50 (sound intensity)

[MikeMitchell]

Sorry, don't even know where to start.
Thanks!

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[Majik]

Is this a passage based question & did they give the threshold intensity for pain? Because if not, I'm assuming they'd expect you to use the threshold intensity for hearing (which doesn't make sense to me at all).

Anyways, the way I'd go about solving this:

1.) First write out an expression
120dB = 10log(I/Io)

2.) Now use what you know about logs. In order for intensity level to equal 120 dB, the intensity for pain should be 12 times greater than the threshold for pain (or hearing). In other words, we want the ratio log(I/Io) to be 12log(I/Io) and for that to happen, Intensity must be equal to: 1 or 10^0 because 1/10^-12 =1 x 10^(0 + 12) = 10^12. The log of 10^12 equals 12. 12 x 10 = 120 dB.

 

[MikeMitchell]

Thanks for the reply, its a standalone question. The answer key says...

Because the intensity level is 120 dB = 10 log (I/I0), log (I/I0) must equal 12 and I/I0 must
equal 1012. Therefore I = 100 W/m2, because I0 = 10-12 W/m2. Thus, B is the best answer.

Should have posted that from the beginning. I understand the right side of the equation set as 10log(10^12) equals 120 but I still don't understand how to arrive at that.

 

[MT Headed]

Well, it's kind of a English -> algebra -> solve type problem, yes?

First half of sentence 1: dB = 10 log (I/Io)
Second half of sentence 1: Io = 1 x 10^-12
Second sentence: dB = 120
Question: solve for I

So now the algebra is: 120 = 10 log (I/(1x10^-12))
Divide by 10: 12 = log(I/(1x10^-12))
Take 10 to both sides: 10^12 = I / (1x10^-12)
Multiply both sides by 10^-12: 1 = I = 1 x 10^0

I suspect you are tripped up by the English terms used when referring to the quantity of sound. Intensity is a measurement of power delivered over an area: W/m^2. IntensityLevel is a totally different concept, and it more closely resembles how humans perceive sound. IntensityLevel, measured in unitless dB, is related to the base 10 logarithm of the intensity.

Doubling the Intensity of a sound will double the power delivered to a given area, but it won't seem like it's twice as loud. Doubling the IntensityLevel will make a sound seem twice as loud. IntensityLevel has little to do with physical phenomena like watts and meters, and a lot to do with the biological design of the human hearing system.

 

[Majik]

Thanks for the reply, its a standalone question. The answer key says...

Because the intensity level is 120 dB = 10 log (I/I0), log (I/I0) must equal 12 and I/I0 must
equal 1012. Therefore I = 100 W/m2, because I0 = 10-12 W/m2. Thus, B is the best answer.

Should have posted that from the beginning. I understand the right side of the equation set as 10log(10^12) equals 120 but I still don't understand how to arrive at that.

It's a little weird since for this problem, you're working backwards and forwards. The question is asking what the intensity of pain must be equal to:

120 dB = 10log(intensity of pain/threshold hearing)

120 db is what we want the right side to equal to. We're told the threshold (Io) for hearing is 10^-12:

120 dB = 10log(intensity of pain/10^-12)

If you're good with anti-logs, then you can re-arrange this equation and solve for Intensity. I'm not, so I approached it differently. This is what I did: I realized I needed to multiply 10 (the 10 in the equation) by some number (12) to get 120 dB (intensity level).

I needed log(intensity of pain/10^-12) to equal 12.
Log(10^12) = 12 so therefore: intensity of pain/10^-12 must divide out to that value. I realized: (10^0 / 10^-12) = (10^12), therefore the intensity of pain must equal 10^0 (or 1).

It's really difficult to explain this. The way I'm describing it probably makes it seem more difficult than it really is, but all you really need is basic math understanding of how logs work and how to manipulate the equation. Hope this helps.

 

[paul411]

Well, it's kind of a English -> algebra -> solve type problem, yes?

First half of sentence 1: dB = 10 log (I/Io)
Second half of sentence 1: Io = 1 x 10^-12
Second sentence: dB = 120
Question: solve for I

So now the algebra is: 120 = 10 log (I/(1x10^-12))
Divide by 10: 12 = log(I/(1x10^-12))
Take 10 to both sides: 10^12 = I / (1x10^-12)
Multiply both sides by 10^-12: 1 = I = 1 x 10^0

Looks good to me.

 

[MikeMitchell]

Thanks everyone, very clear now. Main issue was with the language, and zero power equals 1 rule.