Two trains $A$ and $B$ are moving towards each other with speeds $72 \mathrm{kmh}^{-1}$ and $36…

Two trains $A$ and $B$ are moving towards each other with speeds $72 \mathrm{kmh}^{-1}$ and $36 \mathrm{kmh}^{-1}$ respectively. The train $A$ whistles at $640 \mathrm{~Hz}$ frequency. Before the trains meet, frequency of sound heard by a passenger in Train $B$ is ( Speed of sound in air $=340 \mathrm{~ms}^{-1}$ ).
  1. $500 \mathrm{~Hz}$
  2. $600 \mathrm{~Hz}$
  3. $700 \mathrm{~Hz}$
  4. $800 \mathrm{~Hz}$

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)

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