Q139 / 160 · TS EAMCET 2024 (09 May Shift 2) · Physics

A boy standing on a platform observes the frequency of a train horn as it passes by. The change in the frequency noticed as the train approaches and recedes him with a velocity of 108 kmph is (speed of sound in air $=330 \mathrm{~ms}^{-1}$ )

  • Correct answer
    A
    $18.33 \%$
  • B
    $16.67 \%$
  • C
    $21.27 \%$
  • D
    $15.23 \%$

Solution

Using Doppler's effect formula: 1) Approaching Train: $\begin{aligned} & f_{\text {approach }}=f_0 \times \frac{v}{v-v_0} \\ & F_{\text {recede }}=f_0 \times \frac{v}{v+v_0} \\ \therefore \Delta f & =f_{\text {app }}-f_{r e c e d e} \\ & =f_0\left(\frac{v}{v-v_0}-\frac{v}{v+v_0}\right) \\ \Rightarrow \Delta f & =f_0\left(\frac{\left(v+v_0\right)-\left(v-v_0\right)}{\left(v+v_0\right)\left(v-v_0\right)}\right) \\ \Rightarrow \Delta f & =f_0\left(\frac{2 v_0}{v^2-v_s{ }^2}\right) \cdot v \\ v & =330 \mathrm{~m} / \mathrm{s} \\ v_s & =108 \mathrm{~km} / \mathrm{hr}=\frac{108}{1} \times \frac{5}{18}=30 \mathrm{~m} / \mathrm{s} \\ \Rightarrow \Delta f & =f_0 .330 . \frac{2 \times 30}{330^2-30^2} \\ \Delta f & =f_0 \times 330 \times \frac{60}{108000} \end{aligned}$ $\begin{aligned} \Rightarrow \Delta f & =f_0 \cdot \frac{19800}{108000}=f_0 \cdot \frac{11}{60} \\ \% \text { Change } & =\frac{\Delta f}{f_0} \times 100 \\ & =\frac{11}{60} \times 100=18.33 \% \end{aligned}$ $\therefore$ ANS: $(A) \%$ Change $=18.33 \%$

Hence, the correct answer is option: 'A'.

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