# Wave equation and pressure waves

Describe the wave equation and the pressure wave given that $$T=263 \text{ K}$$, $$\rho=1.34 \text{ kg/m^3}$$, $$f=200 \text{ Hz}$$. It is known that $$500 \text{ m}$$ away, the sound intensity level $$\beta=43 \text{ dB}$$. The source is a point source and the waves can be seen as spherical. The medium is air.

My attempt:

The waves are spherical. This means that the wave equation can be written as $$s(r,t)=\frac{A_s}{r}cos(kr-\omega t)$$. Using the given data and that the medium is air, $$v=\sqrt{\frac{\gamma R T}{M}}=106 *10^3 \text{ m/s}$$, $$k=\frac{2\pi f}{v}=0.011 \text{1/m}$$, $$\omega=2\pi f=1256 \text{1/s}$$. Using the sound intensity formula $$\beta=10 db*log(\frac{I}{I_0})$$ we get that $$I=1.996*10^{-8} \text{W/m^2}$$ and using $$\beta=20 db*log(\frac{p}{p_0})$$ we get $$p=0.0028 \text{ Pa}$$. My question is how to proceed from here.

My initial thought was to use the relationships $$I=\frac{p_{max}^2}{2\rho v}$$, $$p_max=BkA$$, $$v=\sqrt\frac{B}{\rho}$$ where this $$A$$ would equal $$\frac{A_s}{r}$$ at $$r=500$$. But I don't know whether those formulas are valid for spherical waves or not.

My other thought was to use the pressure wave equation, which I think can be written as $$p(r,t)=\frac{A_p}{r}cos(kr-\omega t)$$. Then use that $$p=0.0028 \text{ Pa} = \frac{A_p}{r}$$ at $$r=500$$. But I don't know whether that is the correct form of the pressure wave equation or whether $$p=0.0028 \text{ Pa} = \frac{A_p}{r}$$ is correct.

Any help on how to move forward?