Question

The faintest sound, the human ear can detect at a frequency of $1kHz$ (for which ear is most sensitive) corresponds to an intensity of about ${10}^{-12}W/m^2$. Assuming the density of air $\cong 1.5kg/m^3$ and velocity of sound in air $\cong 300m/s$, the pressure amplitude and displacement amplitude of the sound will be respectively ___ $N/m^2$ and ____$m$.

• A. $2\times {10}^{-5}Pa,\frac{2}{3\pi }\times {10}^{-10}m$
• B. $5\times {10}^{-5}Pa,\frac{1}{\pi }\times {10}^{-10}m$
• C. $3\times {10}^{-5}Pa,\frac{1}{3\pi }\times {10}^{-10}m$
• D. $5\times {10}^{-5}Pa,\frac{4}{3\pi }\times {10}^{-10}m$

Answer 1

The correct option is A $2\times {10}^{-5}Pa,\frac{2}{3\pi }\times {10}^{-10}m$

Step 1: Find pressure amplitude of the sound.

Formula used:$I=\frac{1}{2}\frac{P_0^2}{\rho v}$

Given, intensity of the sound, $I={10}^{-12}W/m^2$

Density of the air, $\rho =1.5kg/m^3$

Velocity of sound in air, $v=300m/s$

As we know, intensity of the sound

$P_0=(2I\rho v{)}^{\frac{1}{2}}$

$P_0=(2\times {10}^{-12}\times 1.5\times 300{)}^{\frac{1}{2}}$

$P_0=3\times {10}^{-5}Pa$

Step 2: Find displacement amplitude of the sound.

Formula used: $P_0=BAK$

Given, frequency detect by the human ear, $f=1KHz=1\times {10}^{3}Hz$

As we know, pressure wave is given by,

$P_0=BAK=\left(\rho v^2\right)A\frac{2\pi }{\left(\frac{v}{f}\right)}{P}_0$

$=\left(2\pi \rho vf\right)A3\times {10}^{-5}$

$=\left(2\pi \right)\left(1.5\right)\left(300\right)\left({10}^3\right)AA$

$=\left(\frac{1}{3\pi }\right)\times {10}^{-10}m$

Final answer: (a)