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  • A train has just completed a U-curve in a track which is a semicircle. The engine is at the forward end of the semi circular part of the track while the last carriage is

A train has just completed a U-curve in a track which is a semicircle. The engine is at the forward end of the semi circular part of the track while the last carriage is at the rear end of the semicircular track. The driver blows a whistle of frequency 150 Hz. Velocity of sound is 340 m/sec. Then the apparent frequency as observed by a passenger in the middle of a train when the speed of the train is 60 m/sec ?

Option: 1

150 \mathrm{~Hz}


Option: 2

200 \mathrm{~Hz}


Option: 3

250 \mathrm{~Hz}


Option: 4

300 \mathrm{~Hz}


Answers (1)

best_answer

The Doppler formula holds for non-collinear motion if vs and v0 are taken to be the resolved component along the line of sight, In this case, we have:

The component of velocity of source along the time joining the observer and source is:

\mathrm{V}_{\mathrm{s}}=-60 \cos 45^{\circ} \mathrm{m} / \mathrm{s}=\frac{-60}{\sqrt{2}} \mathrm{~m} / \mathrm{s}

The component of velocity of observer towards source is:

V_0=-60 \cos 45^{\circ} \mathrm{m} / \mathrm{s}=\frac{-60}{\sqrt{2}} \mathrm{~m} / \mathrm{s}

\mathrm{V}=340 \mathrm{~m} / \mathrm{s}

\mathrm{f}=150 \mathrm{~Hz}

By Doppler's effect formula, the apparent frequency f’ is given by:

\begin{aligned} f^{\prime} & =f\left(\frac{V-V_0}{V-V_s}\right) \\ f^{\prime} & =150\left(\frac{340-\left(\frac{-60}{\sqrt{2}}\right)}{340-\left(\frac{-60}{\sqrt{2}}\right)}\right) \end{aligned}

f^{\prime}=150 \mathrm{~Hz}

In this case, the apparent frequency observed by the passenger in the middle of the train is

150 \mathrm{~Hz}

Posted by

Devendra Khairwa

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