Question

The speed of sound in air is 340 m/s, and the density of air is 1.2 kg/m3. If the displacement amplitude of a 330-Hz sound wave is 14 µm, what is its pressure-variation amplitude?

Answer 1

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Answer 2

The required value of pressure-variation amplitude of the given sound wave is 11.84 Pa.

Given data:

The speed of sound in air is, v = 340 m/s.

The density of air is, $\rho = 1.2 \;\rm kg/m^{3}$.

The frequency of sound wave is, f = 330 Hz.

The displacement amplitude of sound wave is, $A = 14 \;\rm \mu m= 14 \times 10^{-6} \;\rm m$.

The standard expression for the pressure variation amplitude for the sound wave propagating in air medium is,

$\Delta P= B \times A \times K$

Here,

B is the Bulk Modulus and its value is, $B = \rho \times v^{2}$.

K is the wave form constant and its value is, $K = \dfrac{2 \pi f}{v}$.

Solving as,

$\Delta P= (\rho \times v^{2}) \times A \times \dfrac{2 \pi f}{v}\\\\\Delta P= (\rho \times v) \times A \times (2 \pi f)\\\\\Delta P= (1.2 \times 340) \times (14 \times 10^{-6}) \times (2 \pi \times 330)\\\\\Delta P= 11.84 \;\rm Pa$

Thus, we can conclude that the required value of pressure-variation amplitude of the given sound wave is 11.84 Pa.

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