Corresponding to displacement equation, $y=A\sin (kx+\omega t)$ of a longitudinal wave make its pressure and density wave also. Bulk modulus of the medium is $B$ and density is $\rho$.

$\text{ Given, }y(x,t)=A\sin (kx+\omega t)$

$\frac{\partial y}{\partial x}=kA\cos (kx+\omega t)$

Now, pressure equation is given by

$\Delta p=-B\left(\frac{\partial y}{\partial x}\right)$

Substituting the value of $\frac{\partial y}{\partial x}$ from Eq. (i), we have

$\Delta p=-BAk\cos (kx+\omega t)$

Similarly, density equation is given by

$\left(\Delta \rho \right)=\frac{\rho }{B}\left(\Delta p\right)=\frac{\rho }{B}\left(-B\frac{\partial y}{\partial x}\right)=-\rho \left(\frac{\partial y}{\partial x}\right)$

Substituting the values of $\frac{\partial y}{\partial x}$ from Eq. (i), we have,

$\Delta \rho =-\rho Ak\cos (kx+\omega t)$