Two trains $A$ and $B$ are moving towards each other with speeds $72 \mathrm{kmh}^{-1}$ and $36…
- $500 \mathrm{~Hz}$
- $600 \mathrm{~Hz}$
- $700 \mathrm{~Hz}$
- $800 \mathrm{~Hz}$
Solution

Apparent frequency due to doppler's shift when both observer $O$ and source are approaching to each other, is given by apparent frequency, $f^{\prime}=f\left(\frac{v+v_0}{v-v_s}\right)$...(i) Here, $v_o=$ observers speed $=36 \mathrm{~km} / \mathrm{h}$ $=36 \times \frac{5}{18} \mathrm{~m} / \mathrm{s}=10 \mathrm{~m} / \mathrm{s}$ $v_s=$ source speed $=72 \mathrm{~km} / \mathrm{h}$ $=72 \times \frac{5}{18}=20 \mathrm{~m} / \mathrm{s}$ $\dot{v}=$ speed of sound $=340 \mathrm{~m} / \mathrm{s}$ Hence, by eq. (i) apparent frequency, $f^{\prime}=640\left(\frac{340+10}{340-20}\right)=\frac{640 \times 350}{320}$ $=700 \mathrm{~Hz}$
Asked in: AP EAMCET 2022 (07 Jul Shift 1)