# Homework Help: Calculating decibel - Urgent

1. Jan 26, 2005

### mayaa

Can someone help me calculate the following problem:

I want to find out how many times more intense is 44 decibels compared to 32 db.

Example: If in a house the maximal level of sound disturbance allowed is 32db, and the actuall measured sound level is 44db, then how much more intense is the disturbance in comparision to the allowed sound level?

2. Jan 26, 2005

### Integral

Staff Emeritus
We are not real good at giving "quick" answers around here.

However, If you will show that you have put some thought into the problem, Or can ask a specific question about the problem, we are very good at giving quick help. Help which will help you do your own homework. Not us do your homework for you.

What do you know about the realationship between sound intensity and Db?

3. Jan 26, 2005

### mayaa

It is some time back since I studied physics and sound and waves were not my favourite subject back then either.

I just know that Db is a measure for sound intensity and is related to the energy and the sound 'pressure' level. (my native language is not english so you have to excuse me). I have read a little today and did some calculations. To save some time and make it short for the question I asked I calculated as follows:

I1 = 32db
I2 = 44db
What I need to find is I ratio
I ratio = I2 / I1
I ratio = 10 1.2
That is 16.

Is this correct?

4. Jan 26, 2005

### da_willem

The amount of Bells depend logatitmically (with base 10) on the intensity (I) of the wave: (I'll unaesthetically write down B for the intensity in amount of Bells and dB the intensity in amount of deciBells)

$$B=log(\frac{I}{I_0})$$

Where $I_0$ is a reference intensity of $10^{-12} W/m^2$. The amount of deciBells is ten times the amount of Bells (like a meter is 10 decimeters). So:

$$dB=10 log(\frac{I}{I_0})$$

[as the intensity of a sound wave is proportional to the pressure squared you might also sometimes see: $$dB=10 log(( \frac{P}{P_0})^2)=20log(\frac{P}{P_0})$$ ]

Now for the difference in dB's you don't even need the reference intensity, because of the nice feature of logaritms:

$$log(a)-log(b)=log(a/b)$$

5. Jan 26, 2005

### da_willem

And yes that looks correct...

6. Jan 26, 2005