A train sounding a whistle of frequency 510 Hz approaches a station at $72 \mathrm{~km} / \mathrm{hr}$. The…
- 544,480
- 480,544
- 612,544
- 544,612
Solution
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, $\begin{aligned} & \mathrm{n}_2=\mathrm{n}_0\left(\frac{\mathrm{v}}{\mathrm{v}+\mathrm{v}_{\mathrm{s}}}\right) \\ & \mathrm{n}_2=510 \times\left(\frac{320}{320+20}\right)=480 \mathrm{~Hz} \end{aligned}$
Asked in: MHT CET 2024 (04 May Shift 2)