Question

A source of sound operates at$20kHz,20W$ emitting sound uniformly in all directions. The speed of sound in air is $340m{s}^{-1}$ and the density of air is $1.2kg{m}^{-3}$

(a) What is the intensity at a distance of $6m$ from the source?

(b) What will be the pressure amplitude at this point?

(c) What will be the displacement amplitude at this point?

Answer 1

Step 1: Given data

Distance from the source, $r=6m$

Power, $P=20W$

Frequency of source, $f=20KHz$

Density of air, $\rho =1.2kg{m}^{-3}$

Velocity of sound in air, $v=340m{s}^{-1}$

Let area from source $=A$

Where,

Area, $A=4{\mathrm{\pi r}}^2$

Step 2: (a) Finding the intensity

Let intensity $=I$

Intensity at a distance of $6m$ from the source is,

$$\begin{gathered} I=\frac{P}{A} \\ I=\frac{P}{4\pi r^2} \\ I=\frac{20}{4\pi \times 6^2} \\ I=44mW/m^2 \end{gathered}$$

Hence, Intensity is $44mW/m^2$

Step 3: (b) Find pressure amplitude

Let amplitude of pressure be $p_0$.

The intensity of sound is related to its pressure amplitude as

$$\begin{gathered} I=p_0^2/2\rho v \\ {p}_0=\surd \left(2I\rho v\right) \\ {p}_0=\surd (2x44x{10}^{-3}x1.2x340) \\ {p}_0=6Pa \end{gathered}$$

Hence, pressure amplitude is $p_0=6Pa$

Step 4: (c) Find displacement amplitude

Let displacement amplitude be $S_0$.

The intensity of sound is related to its displacement amplitude as

$$\begin{gathered} I=2{\pi }^2{S}_0^2{v}^2\rho v \\ {S}_0^2=\frac{I}{2{\pi }^2{v}^2\rho v} \\ {S}_0=\sqrt{\frac{0.044}{2{\pi }^2{\times \left(340\right)}^2\times 1.2\times 340}} \\ {S}_0=1.2x{10}^{-6}m \end{gathered}$$

Hence, displacement amplitude is $S_0=1.2x{10}^{-6}m$.