A source of sound operates at 2.0 kHz and 20 W , radiating sound...

The intensity $I$ of a sound wave radiating uniformly in all directions from a point source is given by:

$I = \dfrac{P}{4 \pi r^2}$

The intensity is also related to the displacement amplitude $s_0$ of the sound wave by the formula:

$I = \dfrac{1}{2} \rho v \omega^2 s_0^2$

Since $\omega = 2 \pi f$, we can write:

$I = \dfrac{1}{2} \rho v (2 \pi f)^2 s_0^2 = 2 \pi^2 \rho v f^2 s_0^2$

Equating the two expressions for intensity:

$2 \pi^2 \rho v f^2 s_0^2 = \dfrac{P}{4 \pi r^2}$

$s_0^2 = \dfrac{P}{8 \pi^3 r^2 \rho v f^2}$

$s_0 = \dfrac{1}{2 \pi r f} \sqrt{\dfrac{P}{2 \pi \rho v}}$

Substituting the given values ($P = 20 \text{ W}$, $r = 6.0 \text{ m}$, $\rho = 1.2 \text{ kg m}^{-3}$, $v = 340 \text{ m s}^{-1}$, $f = 2000 \text{ Hz}$):

$s_0 = \dfrac{1}{2 \pi (6.0) (2000)} \sqrt{\dfrac{20}{2 \pi (1.2) (340)}}$

$s_0 = \dfrac{1}{24000 \pi} \sqrt{\dfrac{20}{816 \pi}} = \dfrac{1}{24000 \pi} \sqrt{\dfrac{5}{204 \pi}}$

Using $\pi \approx 3.1416$:

$s_0 \approx \dfrac{1}{75398} \sqrt{\dfrac{5}{640.89}} \approx \dfrac{0.0883}{75398} \approx 1.17 \times 10^{-6} \text{ m}$

Rounding to the nearest given option, we get $1.2 \times 10^{-6} \text{ m}$.