Question

If y= A sin (wt-kx) , then the value of d2y/dt2 by d2y/dx2

Answer 1

$y=A\sin (\omega t-kx)\\ \Rightarrow \dfrac{\partial y}{\partial t}=A\omega\sin (\omega t-kx)\\ \Rightarrow \dfrac{{\partial }^2y}{\partial t^2}=A{\omega }^2\sin (\omega t-kx)\\ y=A\sin (\omega t-kx)\\ \Rightarrow \dfrac{\partial y}{\partial x}=-Ak\sin (\omega t-kx)\\ \Rightarrow \dfrac{{\partial }^2 y}{\partial x^2}=Ak^2\sin (\omega t-kx)\\ \Rightarrow \dfrac{\dfrac{{\partial }^2 y}{\partial t^2}}{\dfrac{{\partial }^2 y}{\partial x^2}}=\dfrac{{\omega }^2}{k^2}=c^2$