\({n_a} = n\,\left( {\frac{v}{{v - {v_S}}}} \right)\)
\(⇒\) \(219 = n\,\left( {\frac{{340}}{{340 - {v_S}}}} \right)\) …\((i)\)
when train is receding (goes away), frequency heard by the observer is
\({n_r} = n\,\left( {\frac{v}{{v + {v_s}}}} \right)\)
\(⇒\) \(184 = n\left( {\frac{{340}}{{340 + {v_s}}}} \right)\) …\((ii)\)
On solving equation \((i)\) and \((ii)\) we get \(n = 200Hz\)
and \({v_S} = 29.5m/s.\)
(બંને ઉદગમની આવૃતિ $F_1= F_2=500\, Hz$ અને હવામાં ધ્વનિનો વેગ $=330\, m / s$ છે.)