91Ó°ÊÓ

Formula is$\mathrm{P}(\mathrm{x},\mathrm{t})=\mathrm{BkA}\sin (\mathrm{kx}-\mathrm{Ï–³Ù})$ where,P The pressure, B Bulk modulus of the medium, A Displacement amplitude, k the wave number, c The position at which we want the pressure, t time,$\mathrm{Ï–}=2\mathrm{Ï€´Ú}$angular frequency of the wave

$\mathrm{P}(\mathrm{x},\mathrm{t})=\mathrm{BkA}\sin (\mathrm{kx}-\mathrm{Ï–³Ù})$is maximum when sin has maximum value,

the maximum value for sin function =1${\mathrm{P}}_{\max }=\mathrm{BKA}$

P is the maximum pressure, B is Bulk modulus of the medium, A is the displacement amplitude,$\mathrm{k}=\frac{2\mathrm{π}}{\mathrm{λ}}$ the wave number

$$\begin{gathered} {\mathrm{B}}_{\mathrm{air}}=1.42×{10}^5\mathrm{Pa} \\ \mathrm{v}=344\mathrm{m}/\mathrm{s} \\ \mathrm{k}=\frac{2\mathrm{Ï€}}{\mathrm{λ}} \\ \mathrm{v}=\frac{\mathrm{λ}}{\mathrm{T}}=\mathrm{λ´Ú} \\ \mathrm{λ}=\frac{\mathrm{v}}{\mathrm{f}} \\ \mathrm{k}=\frac{2\mathrm{Ï€}}{\mathrm{λ}}=\frac{2\mathrm{Ï€}}{\mathrm{v}}=\frac{2\mathrm{Ï€´Ú}}{\mathrm{v}}=\frac{2×3.14×320\mathrm{Hz}}{344\mathrm{m}/\mathrm{s}}=5.845\mathrm{rad}/\mathrm{m} \\ {\mathrm{P}}_{\max }=\mathrm{BkA}=\mathrm{B}×\frac{2\mathrm{Ï€´Ú}}{\mathrm{v}}\mathrm{A} \\ =\left(1.42×{10}^5\right)\left(\frac{2×3.14×320}{344}\right)\left(5×{10}^{-3}×{10}^{-3}\right)=4.15\mathrm{Pa} \end{gathered}$$

Therefore, the pressure amplitude is 4.15 Pa

$\mathrm{I}=\frac{1}{2}{\mathrm{\mu þÏ–°ì´¡}}^2=\frac{1}{2}\sqrt{\mathrm{ÒÏ\mu þ}}×{\mathrm{Ï–}}^2{\mathrm{A}}^2$

B is Bulk modulus of the medium, A is Displacement amplitude, k The wave number,The density of medium,$\mathrm{Ï–}=2\mathrm{Ï€´Ú}$angular frequency of the wave

$$\begin{gathered} \mathrm{I}=\frac{1}{2}{\mathrm{\mu þÏ–°ì´¡}}^2=\frac{1}{2}\mathrm{B}×2{\mathrm{Ï€´ÚkA}}^2 \\ =\frac{1}{2}\left(1.42×{10}^5\right)\left(2×3.14×320\right)\left(5.845\right)\left(5×{10}^{-3}×{10}^{-3}\right) \\ =0.02026\mathrm{W}/{\mathrm{m}}^2 \end{gathered}$$

Therefore, the intensity is$0.02026\mathrm{W}/{\mathrm{m}}^2$

The decibel scale$\mathrm{β}=\left(10\mathrm{dB}\right)\log \frac{\mathrm{I}}{{\mathrm{I}}_0}$were,$\mathrm{β}$ sound intensity level, I intensity of sound wave${\mathrm{l}}_0={10}^{-12}$ is the reference intensity.

Substitute the value in equation$\mathrm{β}=\left(10\mathrm{dB}\right)\log \frac{\mathrm{l}}{{\mathrm{l}}_0}$

$\left(10\mathrm{dB}\right)\log \frac{0.02086}{{10}^{-12}}=103.2\mathrm{dB}$

Therefore, the intensity in decibels is 103.2dB