$\therefore \lambda_{\mathrm{p}}=\frac{2 l}{\mathrm{p}}$        ........$(i)$

Given that, $y=2 \sin \left(\frac{4 \pi x}{15}\right) \cos (96 \pi t)$

$=2\left[\sin \left(\frac{4 \pi \mathrm{x}}{15}+96 \pi \mathrm{t}\right)+\sin \left(\frac{4 \pi \mathrm{x}}{15}-96 \pi \mathrm{t}\right)\right]$

Comparing it with standard equation, we get

$\mathrm{v}=\frac{\omega}{2 \pi}=\frac{96 \pi}{2 \pi}=48 \mathrm{\,Hz}$

 and $ \quad \mathrm{K}=\frac{4 \pi}{15}=\frac{2 \pi}{\lambda_{\mathrm{p}}} \quad$ or 

$\lambda_{\mathrm{p}}=\frac{15}{2}$

from eq. $(i)$ we get,

$\frac{15}{2}=\frac{2 \times 60}{\mathrm{p}} $ or $ \mathrm{p}=16$