English: So, this left hand side would become simply $\frac{\partial^{2}P}{\partial t^{2}}$ and that is equal to $\frac{B}{\rho}$ and here again it is the same thing So, I am going to have $\frac{\partial^{2}P}{\partial x^{2}}$ and it is very important to note that this P is related to displacement through this equation and this negative sign is very important.
Telugu: కాబట్టి, ఈ ఎడమ వైపు కేవలం $\frac{\partial^{2}P}{\partial t^{2}}$ అవుతుంది మరియు అది B/ ρ   కి సమానం మరియు ఇక్కడ మళ్లీ అదే విషయం P = $- B\frac{\partial\eta}{\partial x}$.