Question

If two sound waves, one in air and one in (fresh) water, are equal in intensity and angular frequency, what is the ratio of the pressure amplitude of the wave in water to that of the wave in air? Assume the water and the air are at $20 ° C$ .

(Given: Density of air $ρ_{air} = 1.20 kg / m^{3}$ Density of water $ρ_{water} = 1000 kg / m^{3}$ ,velocity of sound in air $v_{air} = 350 m / s$
and velocity of sound in water $v_{water} = 1512 m / s$ )

Accepted Answer

The intensity of sound wave is given by,

$I = \frac{1}{2} ω^{2} A^{2} ρ v$

where $ρ$ is the density of the medium,

$ρ_{water} = 1000 kg / m^{3}$ is the speed of sound,

$ω$ is the angular frequency,

And, $A$ is the displacement amplitude.

The displacement and pressure amplitudes are related by
$v_{air} = 350 m / s$

$∆ p_{m a x} = v ρ A ω$

The intensity of the wave in terms of the pressure amplitude,
$I = \frac{{(Δ p_{m})}^{2}}{2 ρ v}$

For waves of the same frequency, the ratio of the intensity for propagation in water to the intensity for propagation in the air is

$\frac{I_{w}}{I_{a}} = {(\frac{Δ p_{m w}}{Δ p_{m a}})}^{2} \frac{ρ_{a} v_{a}}{ρ_{w} v_{w}},$ where the subscript $a$ denotes air and subscript $w$ denotes water.
Since $I_{a} = I_{w}, \frac{Δ p_{m w}}{Δ p_{m a}} = \sqrt{\frac{ρ_{w} v_{w}}{ρ_{a} v_{a}}}$

$= \sqrt{\frac{(1000 kg / m^{3}) (1512 m / s)}{(1.20 kg / m^{3}) (350 m / s)}} = 60$

Hence, the ratio of pressure amplitude of wave in water to that of wave in air is $60 ˸ 1$

B M Sharma Solutions for Chapter: Sound Waves and Doppler Effect, Exercise 1: ILLUSTRATIONS

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Sound Waves and Doppler Effect, Exercise 1: ILLUSTRATIONS

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