as \(L_{2}>L_{1},\) so \(n_{2}\)
For open pipe,
\(n=\frac{v}{2 L}\)
\(n_{1}-n_{2}=3\) beats \(/ \mathrm{s}\)
\(\frac{v}{2}\left(\frac{1}{L_{1}}-\frac{1}{L_{2}}\right)=3\)
\(\frac{v}{10^{-2}}\left(\frac{1}{50}-\frac{1}{50.5}\right)=6\)
\(v=\frac{6 \times 50 \times 50.5 \times 10^{-2}}{0.5}=303 \mathrm{m} / \mathrm{s}\)
$(a)$ $\left(x^2-v t\right)^2$
$(b)$ $\log \left[\frac{(x+v t)}{x_0}\right]$
$(c)$ $e^{\left\{-\frac{(x+v t)}{x_0}\right\}^2}$
$(d)$ $\frac{1}{x+v t}$
(હવામાં ધ્વનિની ઝડપ $340\, {ms}^{-1}$)