Question

Due to a point isotropic sound source, the sound level at a point is observed as $40dB$. The density of air is $\rho$ = $\frac{15}{11}kg/m^3$ and velocity of sound in air is $330m/s$. (The threshold intensity is ${10}^{-12}\frac{W}{m^2}$)

The pressure amplitude at the observation point is $x\left({10}^{-3}\right)N/m^2$. Then $x$ = _____

• A. 9
• B. 6
• C. 2
• D. 3

Answer 1

The correct option is D

3

In the propagation of sound waves, let pressure amplitude be $\Delta p_0$ and displacement amplitude be A.

$\Delta p_0$ = BAK where symbols have their usual meanings

From $SL$ = $10log\frac{I}{I_0}$

$40$ = $10log\frac{I}{{10}^{-12}}$ $\Rightarrow I$ = ${10}^{-8}W/m^2$

From $I$ = $\frac{\Delta p_0^2}{2\rho v}$

$\Delta p_0$ = $\sqrt{I\times 2\rho v}$ = $\sqrt{{10}^{-8}\times 2\times \frac{15}{11}\times 330N/m^2}$