Question

A point $A$ is located at a distance $r = 1.5$ $m$ from a point isotropic source of sound of frequency $f = 600$ $Hz$ . The sonic power of the source is $P = 0.80$ $W$ . Neglecting the damping of the waves and assuming the velocity of sound in air to be equal to $v = 340$ $m s^{- 1}$ and density of air to be $1.293 kg m^{- 3}$ , find at point $A$ :

(a) the pressure oscillation amplitude $(Δ p)_{m}$ and its ratio to the air pressure;

(b) the oscillation amplitude of particles of the medium; compare it with the wavelength of sound.

Accepted Answer

Let the power of the point source be $P = 0.80 W$ and since it is a point source, it will create spherical wavefront. Therefore, the wave intensity at a distance $r = 1.5 m$ from the point source is given by,

$I = \frac{P}{4 π r^{2}}$

$m s^{- 1}$ We have intensity, $I = \frac{(Δ p)_{m}^{2}}{2 ρ v} = \frac{P}{4 π r^{2}}$

or $(Δ p)_{m} = \sqrt{\frac{ρ v P}{2 π r^{2}}}$

$= \sqrt{\frac{1.293 kg m^{- 3} \times 340 m s^{- 1} \times 0.80 W}{2 π \times 1.5 \times 1.5 m^{2}}} = \sqrt{\frac{1.293 \times 340 \times 0.8}{2 π \times 1.5 \times 1.5}} {(\frac{kg kg m^{2} s^{- 3} m s^{- 1}}{m^{5}})}^{\frac{1}{2}}$

$= 4.9877 (kg m^{- 1} s^{- 2}) \approx 5$ $Pa$ .

Ratio with air pressure, $p = 10^{5} Pa$ ,

$\frac{(Δ p)_{m}}{p} = 5 \times 10^{- 5}$

$(b)$ We have bulk modulus, $B = - \frac{Δ p}{\frac{\partial ξ}{\partial x}} = ρ v^{2}$

$\Rightarrow Δ p = - ρ v^{2} \frac{\partial ξ}{\partial x}$

Relation between pressure amplitude and displacement amplitude is,

$(Δ p)_{m} = ρ v^{2} (k ξ_{m}) = ρ v . (\frac{ω}{k}) (k ξ_{m}) = ρ v (2 π f) ξ_{m}$

$ξ_{m} = a = \frac{(Δ p)_{m}}{2 π ρ v f} = \frac{5}{2 π \times 1.293 \times 340 \times 600} m = 3 μm$

Comparison of oscillation amplitude with wavelength, $\frac{ξ_{m}}{λ} = \frac{3 \times 10^{- 6}}{\frac{340}{600}} = \frac{1800}{340} \times 10^{- 6} = 5 \times 10^{- 6}$

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