# DB difference, intensity

1. Apr 30, 2009

### Oijl

1. The problem statement, all variables and given/known data
Two sounds differ in sound level by 2.80 dB. By what factor is the one intensity greater than the other?

2. Relevant equations

3. The attempt at a solution

Is the equation

$$B$$ = 10dB log(I/Io)

relevant?

2. Apr 30, 2009

### rl.bhat

The equation is
Bdb = 10log(I/Io)
Now let B1 = 10log(I1/Io) and
B2 = 10log(I2/Io)
Take the difference and find the ratio of I1/I2.

3. Apr 30, 2009

### danago

Lets say sound 1 has an intensity of I1 and intensity level B, and sound 2 has an intensity of I2 with intensity level B+2.8

B = 10 log (I1 / I0)
B+2.8 = 10 log (I2 / I0)

See if you can use those

EDIT: rl.bhat beat me to it :P

4. Apr 30, 2009

### Oijl

I'm sorry, but the algebra is confusing me. Could you show the first few steps, please?

5. Apr 30, 2009

### rl.bhat

B1 = 10log(I1/Io)
B2 = 10log(I2/Io)
B1 -B2 = 10[log(I1/Io) - log(I2/Io)]
Use laws of logarithm to further simplification.