Q143 / 150 · MHT CET 2024 (04 May Shift 2) · Physics
A train sounding a whistle of frequency 510 Hz approaches a station at $72 \mathrm{~km} / \mathrm{hr}$. The frequency of the note heard by an observer on the platform as the train (1) approaches the station and then (2) recedes the station are respectively (in hertz) (velocity of sound in air $=320 \mathrm{~m} / \mathrm{s}$ )
- Correct answerA544,480
- B480,544
- C612,544
- D544,612
Solution
Given,
Frequency of source $\left(n_0\right)=510 \mathrm{~Hz}$
Velocity of source $=72 \mathrm{~km} / \mathrm{hr}$
$=72 \times \frac{5}{18}=20 \mathrm{~m} / \mathrm{s}$
Velocity of sound in air $=320 \mathrm{~m} / \mathrm{s}$ Doppler formula for apparent frequency, when source is approaching a stationary listener, $n_1=n_0\left(\frac{v}{v-v_s}\right)$ $\therefore \quad \mathrm{n}_1=510 \times\left(\frac{320}{320-20}\right)=544 \mathrm{~Hz}$ Doppler formula for apparent frequency, when source is moving away from a stationary listener, $\begin{aligned} & \mathrm{n}_2=\mathrm{n}_0\left(\frac{\mathrm{v}}{\mathrm{v}+\mathrm{v}_{\mathrm{s}}}\right) \\ & \mathrm{n}_2=510 \times\left(\frac{320}{320+20}\right)=480 \mathrm{~Hz} \end{aligned}$
Velocity of sound in air $=320 \mathrm{~m} / \mathrm{s}$ Doppler formula for apparent frequency, when source is approaching a stationary listener, $n_1=n_0\left(\frac{v}{v-v_s}\right)$ $\therefore \quad \mathrm{n}_1=510 \times\left(\frac{320}{320-20}\right)=544 \mathrm{~Hz}$ Doppler formula for apparent frequency, when source is moving away from a stationary listener, $\begin{aligned} & \mathrm{n}_2=\mathrm{n}_0\left(\frac{\mathrm{v}}{\mathrm{v}+\mathrm{v}_{\mathrm{s}}}\right) \\ & \mathrm{n}_2=510 \times\left(\frac{320}{320+20}\right)=480 \mathrm{~Hz} \end{aligned}$
Hence, the correct answer is option: 'A'.