Sound Intensity (Level) - Hearing Aid

[Const@ntine]

Homework Statement

A man is wearing a hearing aid device that increases the frequency of the sounds the human ear can conceive, by 30 db. The device "catches" a sound of frequency f = 50 Hz, and intensity I = 3.00 * 10^-11 W/m². What's the intensity that the man's ear hears?

Homework Equations

b = 10log(I/I₀)
ΔPmax = sqrt(2pvI)
ω = 2πf
I = 0.5pv(ωSmax)²

p = 1.20 km/m3
v = 343 m/s
I₀ = 10^-12 W/m²

The Attempt at a Solution

Uh, I don't really have anything. From the problem's info I figured I'd put the original I into the formula for b, find it, then add 30 db to b, and then use the formula again to find the new I. But apparently that's not true. I'm not sure where the frequency is even used.

I know this is very bare-bones, but I coulf really use some help/guidance here.

Any kind of help is appreciated!

 

[TSny]

Hello.

A man is wearing a hearing aid device that increases the frequency of the sounds the human ear can conceive, by 30 db.

Is the word "frequency" correct here?

From the problem's info I figured I'd put the original I into the formula for b, find it, then add 30 db to b, and then use the formula again to find the new I.

That "sounds" reasonable to me.

But apparently that's not true.

Why do you say that?

I'm not sure where the frequency is even used.

Maybe the frequency is not relevant.

 

[Const@ntine]

Hello.
Is the word "frequency" correct here?

Hi!

Sorry, meant to write the sound intensity level of the frequency (the b). It's a translation so I missed that.

That "sounds" reasonable to me.

That's what I figured.

Why do you say that?

With that method, I get a very different result from the book's. Here:

Let's assume that the man has no hearing device:

b = 10log(I/I₀) = 10log(3 * 10^-11/10^-12) = ... = 14,77 db

So, the new b, b', is b + 30 db, so b' = 44,77 db

Now, let's take the hearing aid into account:

b' = 10log(I'/I₀) <=> 4,477 = log(I'/I₀) <=> e^4,477 = I'/10^-12 W/m² <=> I' = 8.8 * 10^-12 W/m²

The book's answer on the other hand is 3.00 * 10^-8 W/m²

Maybe the frequency is not relevant.

Why give it then? I'm thinking that there might be a problem in the wording or something along those lines. Or I'm just missing something.

 

[lightgrav]

They probably meant to say that it amplifies ALL frequencies by 30 dB.
"adding" a Bell means multiply the Intensity by 10.
+30 dB is +1+1+1 = 3 Bels, so you multiply the Intensity by 10, 3 times.

 

[TSny]

Hi!

Let's assume that the man has no hearing device:

b = 10log(I/I₀) = 10log(3 * 10^-11/10^-12) = ... = 14,77 db

So, the new b, b', is b + 30 db, so b' = 44,77 db

Now, let's take the hearing aid into account:

b' = 10log(I'/I₀) <=> 4,477 = log(I'/I₀) <=> e^4,477 = I'/10^-12 W/m² <=> I' = 8.8 * 10^-12 W/m²

The book's answer on the other hand is 3.00 * 10^-8 W/m²

Your work looks good except for one thing. What is the base of the logarithm function that is used in the formula for decibles?

Why give it then? I'm thinking that there might be a problem in the wording or something along those lines. Or I'm just missing something.

I think you'll get the right answer if you make the correction hinted at above. So, apparently the 50 Hz is not needed. It is not too uncommon for extra info to be given in a problem. It tests your confidence in solving the problem.

However, I did wonder if somehow you were supposed to take into account the fact that the human ear is less sensitive at 50 Hz compared to, say, 1000 Hz. But that would have required knowing how to take this into account (which I'm not familiar with). Anyway, it looks like you get the right answer without worrying about the frequency.

@lightgrav has a nice way to get the answer in a flash!

 

[Const@ntine]

They probably meant to say that it amplifies ALL frequencies by 30 dB.
"adding" a Bell means multiply the Intensity by 10.
+30 dB is +1+1+1 = 3 Bels, so you multiply the Intensity by 10, 3 times.

So theoretically, for every x * 10 dBs, I can just go to my initial Intensity (I), and multiply it x times by 10. Kinda like a "cheat code" or "experienced problem solving", right? It gets me out of using the formula. But even if I do, I get the same result and the theory is proven. Thanks for this tidbit!

PS: Is there a link where I could read more about this? My book only has the b = 10log(I/I0) formula and not much else about Bell and the like.

Your work looks good except for one thing. What is the base of the logarithm function that is used in the formula for decibles?

Oh darn it, yeah. The base is 10, not e. I got carried away and put it in wrong.

I think you'll get the right answer if you make the correction hinted at above. So, apparently the 50 Hz is not needed. It is not too uncommon for extra info to be given in a problem. It tests your confidence in solving the problem.

Yeah, if I change the e for a 10 I get the book's answer. Rookie mistake, really. As for the extra info, all of our professors (and my teachers in the past) always said that we should use every bit of data, so I assumed I was just missing something.

However, I did wonder if somehow you were supposed to take into account the fact that the human ear is less sensitive at 50 Hz compared to, say, 1000 Hz. But that would have required knowing how to take this into account (which I'm not familiar with). Anyway, it looks like you get the right answer without worrying about the frequency.

Eh, I doubt it. The book doesn't go into detail in this section. It just has a couple of "put the numbers in the formulas" exercises.

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Thanks a lot both of you, I really appreciate it!