A longitudinal standing wave $y=A\sin kx\cos \omega t$  is maintained in a homogenous rod with cross sectional area S, length $l$ and density  $\rho$. The potential energy stored in the rod at time  $t=\frac{\pi }{6\omega } is\alpha \left(\frac{\rho {\omega }^2{A}^{2}Sl}{32}\right)$. Assume the presence of odd as well as even harmonics are in the vibration. 
The value of $\alpha$ is _______

The equation of longitudinal standing wave is $y=A\sin kx\cos \omega t$

Now,  $dU=\frac{1}{2}p{V}^2{\left(\frac{\partial y}{\partial x}\right)}^{2}sdx$

$$\begin{gathered} dU=\frac{1}{2}\rho V^2{\left(kA\cos kx\cos \omega t\right)}^{2}sdx \\ dU=\frac{1}{2}\rho V^2{k}^2{A}^2{\cos }^2\omega t\cos KXsdx \\ \int dU=\frac{1}{2}\rho {\omega }^2{A}^{2}S{\cos }^2\omega t\underset{0}{\overset{\mathcal{l}}{\int }}{\cos }^{2}kxdx \\ U=\frac{1}{2}\rho {\omega }^2{A}^{2}S{\cos }^2\omega t\underset{0}{\overset{\mathcal{l}}{\int }}\left(\frac{1+\cos 2kx}{2}\right)dx \\ U=\frac{1}{4}\rho {\omega }^2{A}^{2}S\mathcal{l}{\cos }^2\omega t \\ At t=\frac{\pi }{6\omega } \\ U=\frac{1}{4}\rho {\omega }^2{A}^{2}S\mathcal{l}{\cos }^2\left(\frac{\pi }{6}\right) \\ U=\frac{3}{16}\rho {\omega }^2{A}^{2}S\mathcal{l} \\ Hence, \alpha =6 \end{gathered}$$