Q19E - University Physics with Modern Physics

Question

BIO For a person with normal hearing, the faintest sound that can be heard at a frequency of 400 Hz has a pressure amplitude of about $\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{Pa}$. Calculate the (a) intensity; (b) sound intensity level; (c) displacement amplitude of this sound wave at 20°C.

Step-by-Step Solution

Answer

A) $\mathrm{I}=4.36\times {10}^{-12} \mathrm{W}/{\mathrm{m}}^2$  b) $\mathrm{\beta }=6.4\mathrm{dB}$  c)  $\mathrm{A}=5.8\times {10}^{-11} \mathrm{m}$ 

STEP 1 Concept of the intensity of the sound wave in terms of the pressure amplitude

Formula is $\mathbf{I}\mathbf{=}\frac{{{\mathbf{vp}}^{\mathbf{2}}}_{\mathbf{max}}}{\mathbf{2}\mathbf{D}}$ is the relation that describes the intensity of the sound wave in terms of the pressure amplitude

Step 2 Calculate the intensity of he sounds

Substitute the values in the equation $\mathrm{I}=\frac{{{\mathrm{vp}}^2}_{\max }}{2\mathrm{D}}$ we get,

$$\begin{gathered} \mathrm{I}=\frac{344\times {\left(6\times {10}^{-5}\right)}^2}{2\times 1.42\times {10}^5} \\ =4.36\times {10}^{-12}\mathrm{W}/{\mathrm{m}}^2 \end{gathered}$$

The relation that describes the intensity level of a sound wave is $\mathrm{\beta }=10\log \left(\frac{\mathrm{l}}{{\mathrm{l}}_0}\right)$ .Substitute the values in the above equation we get,

$$\begin{gathered} \mathrm{\beta }=10\log \left(\frac{4.36\times {10}^{-12}\mathrm{W}/{\mathrm{m}}^2}{{10}^{-12}\mathrm{W}/{\mathrm{m}}^2}\right) \\ =6.4\mathrm{dB} \end{gathered}$$

Therefore, the intensity of the sound is 6.4dB

Step 3 The relation between the displacement amplitude and the pressure amplitude

The equation is $\mathrm{A}=\frac{{\mathrm{P}}_{\max }}{\mathrm{kB}}$

We know that the Bulk modulus for the air is $\mathrm{B}=1.42\times {10}^5\mathrm{Pa}$ and the pressure amplitude is $6\times {10}^{-5} \mathrm{Pa}$ but in order to make use of equation (3) we need to determine k, where k can be represented as follows

$$\begin{gathered} \mathrm{k}=\frac{2\mathrm{\pi }}{\mathrm{\lambda }} \\ \mathrm{\lambda }=\frac{\mathrm{v}}{\mathrm{f}}=\frac{344}{400}=0.86\mathrm{m} \\ \mathrm{k}=\frac{2\mathrm{\pi }}{0.86\mathrm{m}}=7.3\mathrm{rad}/\mathrm{m} \\ \mathrm{A}=\frac{6\times {10}^{-5}\mathrm{Pa}}{7.3\mathrm{rad}/\mathrm{m}\times 1.42\times {10}^5\mathrm{Pa}} \\ =5.8\times {10}^{-11}\mathrm{m} \end{gathered}$$

 

Therefore, the pressure amplitude is $5.8\times {10}^{-11} \mathrm{m}$