A person observes two moving trains, 'A' reaching the station and 'B' leaving the station with equal speed of $30 \mathrm{~m} / \mathrm{s}$. If both trains emit sounds with frequency $300 \mathrm{~Hz}$, (Speed of sound: $330 \mathrm{~m} / \mathrm{s}$) approximate difference of frequencies heard by the person will be:
Solution
By doppler effect : $f^{\prime}=f_{0}\left[\frac{v-v_{0}}{v-v_{s}}\right]$
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$\Rightarrow \quad f_{A}^{\prime}=300\left[\frac{330}{330-30}\right] \mathrm{Hz}$
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$=330 \mathrm{~Hz}$
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And $f_{B}^{\prime}=300\left[\frac{330}{330+30}\right] \mathrm{Hz}$
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$=\frac{5}{6} \times 330 \mathrm{~Hz}=275 \mathrm{~Hz}$<br/><br/>
$\therefore \Delta \mathrm{f}=330-275=55 \mathrm{~Hz}$.
About this question
Subject: Physics · Chapter: Waves · Topic: Wave Motion and Types
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