# dB problem

• Jan 9th 2007, 04:53 AM
dcfi6052
dB problem
How do you solve this? I don't even know where to start.

The sound of heavy equipment at a construction site is measured at 145 dB. The sound at a rock concert is measured at 125 dB. How many times louder is the noise at the construction site than that at the rock concert?

Can someone please explain how to do this.
• Jan 9th 2007, 08:47 AM
topsquark
Quote:

Originally Posted by dcfi6052
How do you solve this? I don't even know where to start.

The sound of heavy equipment at a construction site is measured at 145 dB. The sound at a rock concert is measured at 125 dB. How many times louder is the noise at the construction site than that at the rock concert?

Can someone please explain how to do this.

The "Bel" unit of sound intensity is on a logarithmic scale. (People usually use the unit deci-Bel, or dB.) The equation is:
$\beta = 10 log \left ( \frac{I}{I_0} \right )$ <-- For clarity's sake, this is $log_{10}$.

where $I_0$ is taken as the threshold of human hearing, about $10^{-12} \, W/m^2$, and $\beta$ is the sound intensity in dB.

We wish to determine how many times greater the intensity of the heavy equipment is versus the rock concert.

So for the construction equipment:
$145 = 10 log \left ( \frac{I}{10^{-12}} \right )$

$14.5 = log \left ( \frac{I}{10^{-12}} \right )$

$\frac{I}{10^{-12}} = 10^{14.5}$

$I = 10^{-12} \cdot 10^{14.5} \approx 316.228 \, W/m^2$

Similarly, for the rock concert at $\beta = 125 \, dB$:

$I = 10^{-12} \cdot 10^{12.5} \approx 3.16228 \, W/m^2$

So the construction equipment is $\frac{316.228}{3.16228} = 100$ times louder than the rock concert.

-Dan