Question

A sound wave in air at 20°C has a frequency of 320 Hz and a displacement amplitude of $\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{mm}$For this sound wave calculate the (a) pressure amplitude (in Pa); (b) intensity $\mathbf{W}\mathbf{/}{\mathbf{m}}^{\mathbf{2}}$(c) sound intensity level (in decibels).

Short answer

A) $4.15\mathrm{Pa}$ b) $0.02086\mathrm{W}/{\mathrm{m}}^2$ c) 103.2dB

Step-by-step solution

STEP 1 Instantons pressure fluctuation for sinusoidal sound wave:

Formula is$\mathrm{P}(\mathrm{x},\mathrm{t})=\mathrm{BkA}\sin (\mathrm{kx}-\mathrm{\varpi t})$ 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,$\mathrm{\varpi }=2\mathrm{\pi f}$angular frequency of the wave

$\mathrm{P}(\mathrm{x},\mathrm{t})=\mathrm{BkA}\sin (\mathrm{kx}-\mathrm{\varpi t})$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,$\mathrm{k}=\frac{2\mathrm{\pi }}{\mathrm{\lambda }}$ the wave number

Calculate the pressure amplitude

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

Therefore, the pressure amplitude is 4.15 Pa

Calculate the Intensity of a sinusoidal sound wave in a fluid

$\mathrm{I}=\frac{1}{2}{\mathrm{B\varpi kA}}^2=\frac{1}{2}\sqrt{\mathrm{\rho B}}\times {\mathrm{\varpi }}^2{\mathrm{A}}^2$

B is Bulk modulus of the medium, A is Displacement amplitude, k The wave number,The density of medium,$\mathrm{\varpi }=2\mathrm{\pi f}$angular frequency of the wave

$$\begin{gathered} \mathrm{I}=\frac{1}{2}{\mathrm{B\varpi kA}}^2=\frac{1}{2}\mathrm{B}\times 2{\mathrm{\pi fkA}}^2 \\ =\frac{1}{2}\left(1.42\times {10}^5\right)\left(2\times 3.14\times 320\right)\left(5.845\right)\left(5\times {10}^{-3}\times {10}^{-3}\right) \\ =0.02026\mathrm{W}/{\mathrm{m}}^2 \end{gathered}$$

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

Calculate the sound intensity level in decibels

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

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

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

Therefore, the intensity in decibels is 103.2dB