A car sounding a horn of frequency 1000 Hz passes a stationary observer. The ratio of frequencies of the…

A car sounding a horn of frequency 1000 Hz passes a stationary observer. The ratio of frequencies of the horn noted by the observer before and after passing of the car is $11: 9$. The speed of car is (Speed of sound $v=340 \mathrm{~ms}^{-1}$ )
  1. $34 \mathrm{~ms}^{-1}$
  2. $17 \mathrm{~ms}^{-1}$
  3. $170 \mathrm{~ms}^{-1}$
  4. $340 \mathrm{~ms}^{-1}$

Solution

$\mathrm{f}=1000 \mathrm{~Hz}$
When the car approaching the observer, $f_1=\left(\frac{v}{v-v_s}\right) f$
When the car moving away from the observer, $\begin{aligned} & f_2=\left(\frac{v}{v+v_s}\right) f \\ & \therefore \frac{f_1}{f_2}=\frac{v+v_s}{v-v_s} \Rightarrow \frac{11}{9}=\frac{340+v_S}{340-v_s} \\ & \therefore \quad \text { Velocity of car, } v_s=34 \mathrm{~ms}^{-1} \end{aligned}$

Asked in: AP EAMCET 2024 (21 May Shift 1)

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