Wave amplitude

Two point sources emit surface waves that interfere with each other. The amplitudes of the waves emitted by the two sources are given by,

$z_1 = A_1\cos (k|\vec{r}-\vec{r}_1|-\omega t)/\sqrt{|\vec{r}-\vec{r}_1|}$ cm

$z_2 = A_2\cos (k|\vec{r}-\vec{r}_2|-\omega t)/\sqrt{|\vec{r}-\vec{r}_2|}$ cm

Where $|\vec{r}-\vec{r}_1|$ is the distance from source 1 and $|\vec{r}-\vec{r}_2|$ is the distance from source 2. The wavenumber $k$ is related to the wavelength $\lambda$, $k=\frac{2\pi}{\lambda}$, and the angular frequency $\omega$ is related to the period $T$, $\omega=\frac{2\pi}{T}$.

Point source 1 emits waves from position cm where $A_1$=  cm^3/2. The quantity $A_1$ has the unusual units of cm^3/2 because $z$ is measured in centimeters and $z \propto A/\sqrt{r}$.

Point source 2 emits waves from position cm where $A_2$=  cm^3/2.

The speed of the waves is $c=$  cm/s, and the angular frequency of the oscillations is $\omega=$  rad/s.

The interference pattern caused by these waves produces vertical simple harmonic oscillations at every point. What is the amplitude of these oscillations at position  cm?