Answer
(a) The pressure amplitude is $0.051~Pa$
(b) The displacement amplitude is $1.5\times 10^{-7}~m$
Work Step by Step
(a) We can find the intensity of the sound:
$\beta = 10~log\frac{I}{I_0}$
$65.0 = 10~log\frac{I}{I_0}$
$6.50 = log\frac{I}{I_0}$
$10^{6.50} = \frac{I}{I_0}$
$I = (10^{6.50})~I_0$
$I = (10^{6.50})~(1.0\times 10^{-12}~W/m^2)$
$I = 3.16\times 10^{-6}~W/m^2$
We can use $343~m/s$ as the speed of sound in air.
We can use $\rho = 1.2~kg/m^3$ as the density of air.
We can find the pressure amplitude:
$P_0 = \sqrt{2I\rho v}$
$P_0 = \sqrt{(2)(3.16\times 10^{-6}~W/m^2)(1.2~kg/m^3)(343~m/s)}$
$P_0 = 0.051~Pa$
The pressure amplitude is $0.051~Pa$
(b) We can find the displacement amplitude:
$s_0 = \sqrt{\frac{I}{2\pi^2 \rho f^2 v}}$
$s_0 = \sqrt{\frac{3.16\times 10^{-6}~W/m^2}{(2\pi^2)(1.2~kg/m^3)(131~Hz)^2(343~m/s)}}$
$s_0 = 1.5\times 10^{-7}~m$
The displacement amplitude is $1.5\times 10^{-7}~m$
