91Ó°ÊÓ

• Frequency,$\mathrm{f}=1665\text{Hz}$
• Pressure amplitude,$\mathrm{Δ\pm Ê}=1.13×{10}^{−3}\text{Pa}$

Apply the relation of pressure with amplitude, frequency, and speed of sound to find the displacement amplitude. Using the formula for the intensity of sound in terms of density, velocity, frequency, and amplitude, find the intensity.

The expression for the pressure amplitude is given by,

$\mathrm{P}=2\mathrm{Ï€´Ú}\mathrm{ÒÏ´¡\pm ¹}$

Where,$P$ is pressure,$f$ is frequency, $A$is area,$\mathrm{ÒÏ}$ is density and v is velocity.

Pressure amplitude is given as,

$\mathrm{Δ}P=2\mathrm{Ï€}f\mathrm{ÒÏ}Av$

$∴\mathrm{A}=\frac{\mathrm{Δ\pm Ê}}{2\mathrm{Ï€´Ú}\mathrm{ÒÏv}}$

Substitute $1.13×{10}^{−3}\text{Pa}$for $\mathrm{Δ}P$, $\text{1665Hz}$for $\mathrm{f}$, $1.225\text{ }\mathrm{kg}/{\mathrm{m}}^3$for $\mathrm{ÒÏ}$, $340\text{ }\mathrm{m}/\mathrm{s}$for $v$into the above equation,

$\begin{array}{c}\mathrm{A}=\frac{1.13×{10}^{−3}\text{pa}}{2\left(3.14\right)\left(\text{1665Hz}\right)\left(\frac{\text{1.225kg}}{{\text{m}}^{\text{3}}}\right)\left(\frac{\text{340m}}{\text{s}}\right)}\\ =0.259×{10}^{−9}\text{m}\\ ≈0.26\text{nm}\end{array}$

Hence, the displacement amplitude is $0.26\text{ }nm$.

The intensity of the sound wave is given as,

$\begin{array}{c}\mathrm{I}=\frac{1}{2}\mathrm{ÒÏv}{\left(2\mathrm{Ï€´Ú}\right)}^2{\mathrm{A}}^2\\ =0.5\left(\frac{\text{1.225kg}}{{\text{m}}^{\text{3}}}\right)\left(\frac{\text{340m}}{\text{s}}\right){(2×3.14×1665\text{Hz})}^2{(0.26×{10}^{−9}\text{m})}^2\\ =\frac{\left({\text{1.53×10}}^{\text{-9}}\right)\text{W}}{{\text{m}}^{\text{2}}}\\ ≈1.5{\text{nW/m}}^{\text{2}}\end{array}$

Hence, the intensity of the wave emitted by the ear is$1.5{\text{nW/m}}^{\text{2}}$ .