Q60P - University Physics with Modern Physics

Question

The sound from a trumpet radiates uniformly in all directions in 20°C air. At a distance of 5.00 m from the trumpet the sound intensity level is 52.0 dB. The frequency is 587 Hz. (a) What is the pressure amplitude at this distance? (b) What is the displacement amplitude? (c) At what distance is the sound intensity level 30.0 dB?

Step-by-Step Solution

Answer

A) The pressure amplitude of the trumpet is 0.0114 Pa B) The displacement amplitude of the sound is $7.49\times {10}^{-7}\mathrm{m}$  C) The distance at which the sound intensity level is equal to 30.0 dB is 63.0 m.

STEP 1 The pressure amplitude of the sound

Formula is ${\mathrm{p}}_{\max }=\sqrt{2\mathrm{\rho Vl}}$ where, ${\mathrm{p}}_{\max }$  is the pressure amplitude of the sound. p is the density of the air, vis the speed of the sound in air, l is the intensity of the sound.

The displacement amplitude of the sound is given as $\mathrm{A}=\frac{\left({\mathrm{p}}_{\max }\right)\mathrm{v}}{\mathrm{B}\left(2\mathrm{\pi f}\right)}$ where, A is the displacement amplitude. $\left({\mathrm{p}}_{\max }\right)$ is the pressure amplitude, B is the bulk modulus of the air, f is the frequency of the sound, v is the speed of the sound in air.

STEP 2 Calculate the intensity of the sound

The pressure amplitude of the sound is given as ${\mathrm{p}}_{\max }=\sqrt{2\mathrm{\rho vl}}$ and the intensity of the level of sound is $\mathrm{\beta }=\left(10\mathrm{dB}\right)\log \left(\frac{\mathrm{l}}{{\mathrm{l}}_0}\right)$ where, $\mathrm{\beta }$ is the intensity level of the sound. I is is the intensity of the sound, ${\mathrm{l}}_0$ is the intensity of the reference sound. 

Substitute 52.0 dB for $\mathrm{\beta }$, $1\times {10}^{-12}\mathrm{W}/{\mathrm{m}}^2$ for ${\mathrm{l}}_0$ in the above equation to find I. 

$$\begin{gathered} \left(52.0\mathrm{dB}\right)=\left(10\mathrm{dB}\right)\log \left(\frac{\mathrm{I}}{1\times {10}^{-12}\mathrm{W}/{\mathrm{m}}^2}\right) \\ \left(\frac{\mathrm{I}}{1\times {10}^{-12}\mathrm{W}/{\mathrm{m}}^2}\right)={10}^{52} \\ \mathrm{l}=1.585\times {10}^{-7}\mathrm{W}/{\mathrm{m}}^2 \end{gathered}$$

 

Thus, the intensity of the sound at that point is $1.585\times {10}^{-7}\mathrm{W}/{\mathrm{m}}^2$

Substitute the values in ${\mathrm{p}}_{\max }=\sqrt{2\mathrm{\rho vl}}$

$$\begin{gathered} {\mathrm{p}}_{\max }=\sqrt{2\left(1.20\mathrm{kg}/{\mathrm{m}}^3\right)}\left(344\mathrm{m}/\mathrm{s}\right)\left(1.585\times {10}^{-7}\mathrm{W}/{\mathrm{m}}^2\right) \\ =1.14\times {10}^{-2}\mathrm{Pa} \\ =0.0114\mathrm{Pa} \end{gathered}$$

Therefore, the pressure amplitude of the trumpet is 0.0114 Pa

STEP 3 Calculate the displacement amplitude of the sound

The sound from the trumpet radiates uniformly in all directions, the sound intensity level of the trumpet at a distance of 5.0 m is 52.0 dB, the frequency of the sound is 587 Hz, the speed of the sound in air is 344m/s and the pressure amplitude of the sound is $1.14\times {10}^{-2}\mathrm{Pa}$

Substitute the values in the equation $\mathrm{A}=\frac{\left({\mathrm{p}}_{\max }\right)\mathrm{v}}{\mathrm{B}\left(2\mathrm{\pi f}\right)}$

$$\begin{gathered} \mathrm{A}=\frac{\left(1.14\times {10}^{-2}\mathrm{Pa}\right)\left(344\mathrm{m}/\mathrm{s}\right)}{\left(1.42\times {10}^2\mathrm{Pa}\right)\left(2\mathrm{\pi }587\mathrm{Hz}\right)} \\ =0.0749\times {10}^{-7}\mathrm{m} \\ =7.49\times {10}^{-7}\mathrm{m} \end{gathered}$$

Therefore, the displacement amplitude of the sound is $7.49\times {10}^{-7}\mathrm{m}$

STEP 4 Calculate the distance at which the sound intensity level is equal to 30.0 dB

The sound from the trumpet radiates uniformly in all directions, the sound intensity level of the trumpet at a distance of 5.0 m is 52.0 dB, the frequency of the sound is 587 Hz, the speed of the sound in air is 344 m/s, the pressure amplitude of the sound is $1.14\times {10}^{-2}\mathrm{Pa}$ and the intensity of the sound at 5.0 m is  $1.585\times {10}^{-7}\mathrm{W}/{\mathrm{m}}^2$ The distance at which the intensity level of the sound is equal to 30.0 dB by applying 

inverse square rule is $\frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}=\sqrt{\frac{{\mathrm{l}}_2}{{\mathrm{l}}_1}}$ and the intensity level of sound is $\mathrm{\beta }=10\log \left(\frac{{\mathrm{l}}_1}{{\mathrm{l}}_0}\right)$

Substitute the values in the above equation

$$\begin{gathered} \left(30.0\mathrm{dB}\right)=\left(10\mathrm{dB}\right)\log \left(\frac{\mathrm{l}}{1\times {10}^{-12}\mathrm{W}/{\mathrm{m}}^2}\right) \\ \left(\frac{\mathrm{l}}{1\times {10}^{-12}\mathrm{W}/{\mathrm{m}}^2}\right)={10}^{30} \\ \mathrm{l}=1\times {10}^{-9}\mathrm{W}/{\mathrm{m}}^2 \end{gathered}$$

Therefore, the intensity at distance ${\mathrm{r}}_1$ is $1\times {10}^{-9}\mathrm{W}/{\mathrm{m}}^2$

Substitute the values in the equation $\frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}=\sqrt{\frac{{\mathrm{l}}_2}{{\mathrm{l}}_1}}$  to find ${\mathrm{r}}_1$

 

$$\begin{gathered} \frac{{\mathrm{r}}_1}{5.0\mathrm{m}}=\sqrt{\frac{1.585\times {10}^{-7}\mathrm{W}/{\mathrm{m}}^2}{1\times {10}^{-9}\mathrm{W}/{\mathrm{m}}^2}} \\ =62.9\mathrm{m} \\ =63.0\mathrm{m} \end{gathered}$$

Therefore, the distance at which the sound intensity level is equal to 30.0 dB is 63.0 m.