Answer:
Pressure amplitude, [tex]P=4.42\times 10^{-5}\ Pa[/tex]
Explanation:
It is given that,
The maximum displacement equal to the diameter of an oxygen molecule, [tex]d=3\times 10^{-10}\ m[/tex]
Speed of sound at 0 degrees is, v = 331.5 m/s
The pressure amplitude for the longitudinal wave is given by :
[tex]P=2\pi f\rho vd[/tex]
Where
[tex]\rho[/tex] is the density of air, [tex]\rho=1.2\ kg/m^3[/tex]
f is the frequency say 59 Hz
So, pressure amplitude is given by :
[tex]P=2\pi \times 59\times 1.2\times 331.5\times 3\times 10^{-10}[/tex]
[tex]P=4.42\times 10^{-5}\ Pa[/tex]
So, the pressure amplitude in a sound wave in the air must be [tex]P=4.42\times 10^{-5}\ Pa[/tex]. Hence, this is the required solution.
