91Ó°ÊÓ

Formula is ${\mathrm{p}}_{\max }=\sqrt{2\mathrm{ÒϳÕ\pm ô}}$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 $\mathrm{A}=\frac{\left({\mathrm{p}}_{\max }\right)\mathrm{v}}{\mathrm{B}\left(2\mathrm{Ï€´Ú}\right)}$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.

The pressure amplitude of the sound is given as${\mathrm{p}}_{\max }=\sqrt{2\mathrm{ÒÏ\pm ¹\pm ô}}$ and the intensity of the level of sound is$\mathrm{β}=\left(10\mathrm{dB}\right)\log \left(\frac{\mathrm{l}}{{\mathrm{l}}_0}\right)$ where,$\mathrm{β}$ 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 $\mathrm{β}$, $1×{10}^{-12}\mathrm{W}/{\mathrm{m}}^2$for${\mathrm{l}}_0$ in the above equation to find I.

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

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

Substitute the values in${\mathrm{p}}_{\max }=\sqrt{2\mathrm{ÒÏ\pm ¹\pm ô}}$

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

Therefore, the pressure amplitude of the trumpet is 0.0114 Pa

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$1.14×{10}^{-2}\mathrm{Pa}$

Substitute the values in the equation$\mathrm{A}=\frac{\left({\mathrm{p}}_{\max }\right)\mathrm{v}}{\mathrm{B}\left(2\mathrm{Ï€´Ú}\right)}$

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

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

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 $1.14×{10}^{-2}\mathrm{Pa}$and the intensity of the sound at 5.0 m is $1.585×{10}^{-7}\mathrm{W}/{\mathrm{m}}^2$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$\mathrm{β}=10\log \left(\frac{{\mathrm{l}}_1}{{\mathrm{l}}_0}\right)$

Substitute the values in the above equation

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

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

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

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

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