Frequency received by wall
\({f}_{1}={f}_{0}\left(\frac{{C}}{{C}-{V}}\right)\)
Again wall as a source
Frequency received by observer on car
\({f}_{2}={f}_{1}\left(\frac{{C}+{V}}{{C}}\right)\)
\(500=400\left(\frac{{C}+{v}}{{C}-{V}}\right)\)
\(\frac{5}{4}=\frac{{C}+{V}}{{C}-{V}}\)
\({C}=9 {V}\)
\({V}=\frac{{C}}{9}=\frac{330}{9}\,{m} / {s}\)
\({V}=\frac{330}{9} \times \frac{18}{5}=132\, {km} / {hr}\)
(જયાં $ {I_0} $ થ્રેશોલ્ડ તીવ્રતા)
$y_1=A \sin \left(k x-\omega t+\frac{\pi}{6}\right), \quad y_2=A \sin \left(k x-\omega t-\frac{\pi}{6}\right)$
પરિણામી તરંગનું સમીકરણ ક્યું છે.