Problem 12 - College Physics

Question

A sound wave from a siren has an intensity of \(100.0 \mathrm{~W} / \mathrm{m}^{2}\) at a certain point, and a second sound wave from a nearby ambulance has an intensity level \(10 \mathrm{~dB}\) greater than the siren's sound wave at the same point. What is the intensity level of the sound wave due to the ambulance?

Step-by-Step Solution

Answer

The intensity level of the sound due to the ambulance is significantly higher, and can be computed using the given formula after getting the dB level for the siren and adding 10dB to it. After that, the actual intensity can be computed by rearrangement of the formula.

Step 1: Understand the Definition of Decibel

We know that the decibel level \(L\) of a sound with intensity \(I\) can be found using the formula \(L = 10 \cdot \log_{10}(I/I_0)\) where \(I_0\) is the reference intensity, usually the quietest sound that the average human ear can hear, \(I_0 = 10^{-12} W/m^2\). The log here is the logarithm base 10.

Step 2: Find the Decibel Level of the Siren's Sound

Plug the given intensity of the siren's sound, 100.0 W/m^2, into the formula to find \(L_{siren} = 10 \cdot \log_{10}(100/I_0)\). Simplify this to find the decibel level of the siren's sound.

Step 3: Find the Decibel Level of the Ambulance's Sound

It's given that the intensity level of the ambulance's sound is 10 dB greater than the siren's sound. So, \(L_{ambulance} = L_{siren} + 10\). By substituting the value of \(L_{siren}\) found in the previous step, the value of \(L_{ambulance}\) can be determined.

Step 4: Convert Decibel Level to Intensity

Now we need to find the intensity of the ambulance's sound. The formula can be rearranged as \(I = I_0 \times 10^{(L/10)}\). Substitute the value of \(L_{ambulance}\) into this formula to find the intensity of the ambulance's sound.