Problem 6.7#
Starting with equation 6.54
\[\begin{equation*}
\sin^2 (\omega\, \Delta t) = C^2 \, \sin^2 (k\,\Delta x)
\end{equation*}\]
where
\[\begin{equation*}
C^2 = \frac{\Delta t^2 \,\mu}{\rho \, \Delta x^2}
\end{equation*}\]
which can be seen by setting the determinant of the matrix in equation 6.53 to equal 0. Taking the derivative of this with respect to \(k\), and remembering that \(\omega\) is a function of \(k\), we use the chain rule to get
\[\begin{equation*}
2\, \Delta t\, \cos(\omega\, \Delta t)\, \sin (\omega\, \Delta t) \frac{\partial \omega}{\partial k} = 2\, C^2 \, \Delta x \, \cos (k\,\Delta x) \, \sin (k\,\Delta x)
\end{equation*}\]
Using the trigonometric identity that
\[\begin{equation*}
\sin(2\theta) = 2\sin(\theta)\cos(\theta)
\end{equation*}\]
this may be simplified to
\[\begin{equation*}
\Delta t\, \sin (2 \omega\, \Delta t) \frac{\partial \omega}{\partial k} = C^2 \, \Delta x \, \sin (2 k\,\Delta x)
\end{equation*}\]
followed by division
\[\begin{equation*}
\frac{\partial \omega}{\partial k} = \frac{C^2 \, \Delta x}{\Delta t} \frac{\sin (2 k\,\Delta x) }{ \sin (2 \omega\, \Delta t) }
\end{equation*}\]
and finally substituting the definition of \(C^2\)
\[\begin{equation*}
\frac{\partial \omega}{\partial k} = \frac{\Delta t^2 \,\mu}{\rho \, \Delta x^2} \frac{\Delta x}{\Delta t} \frac{\sin (2 k\,\Delta x) }{ \sin (2 \omega\, \Delta t) } = \frac{c^2\, \Delta t }{\Delta x} \frac{\sin (2 k\,\Delta x) }{ \sin (2 \omega\, \Delta t) }
\end{equation*}\]