A whistling train approaches a junction. An observer standing at junction observes the frequency to

Question

A whistling train approaches a junction. An observer standing at junction observes the frequency to be 2.2 kHz and 1.8 kHz of the approaching and the receding train respectively. Find the speed of the train in m/s (speed of sound $=300 \mathrm{~m} / \mathrm{s}$ ).\

Professor PC Thomas Classes · Answer

According to Doppler's effect,\ $f^{\prime}=f \frac{v}{v-v_{T}}$ when train approaches,\ $f^{\prime \prime}=f \frac{v}{v+v_{T}}$ when train recedes. $$\begin{equation*} 	herefore \frac{f^{\prime}}{f^{\prime \prime}}=\frac{v+v_{T}}{v-v_{T}} 	ag{i} \end{equation*}$$ Given: $f^{\prime}=2.2 \mathrm{kHz}=2200 \mathrm{~Hz}$\ $f^{\prime \prime}=1.8 \mathrm{kHz}=1800 \mathrm{~Hz}$\ $v=$ Velocity of sound $=300 \mathrm{~m} / \mathrm{s}$\ $v_{T}=$ Velocity of train/source.\ $	herefore \frac{2200}{1800}=\frac{300+v_{T}}{300-v_{T}}$\ or $\quad 3300-11 v_{T}=2700+9 v_{T}$\ or $20 v_{T}=600, v_{T}=30 \mathrm{~ms}^{-1}$\