# Sound Intensity (Level) - Hearing Aid

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1. Jul 13, 2017

### Techno_Knight

1. The problem statement, all variables and given/known data

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?

2. Relevant 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

3. 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!

2. Jul 13, 2017

### TSny

Hello.
Is the word "frequency" correct here?

That "sounds" reasonable to me.

Why do you say that?

Maybe the frequency is not relevant.

3. Jul 13, 2017

### Techno_Knight

Hi!

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

That's what I figured.

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) <=> e4,477 = 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

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.

4. Jul 13, 2017

### 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.

5. Jul 13, 2017

### TSny

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

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!

Last edited: Jul 13, 2017
6. Jul 15, 2017

### Techno_Knight

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!

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

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.

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!