$\therefore \lambda_{\mathrm{p}}=\frac{2 l}{\mathrm{p}}$ ........$(i)$
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$
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Assertion $A$ : When a rod lying freely is heated, no thermal stress is developed in it.
Reason $R :$ On heating the length of the rod increases.
In the light of the above statements, choose the correct answer from the options given below