A train moves towards a stationary observer with a speed of 34 m/s. The train sounds a whistle and its frequency registered by the observer is f₁. If the train’s speed is reduced to 17 m/s, the frequency registered is f₂. If the speed of sound is 340 m/s, then the ratio $\frac{f_{1}}{f_{2}}$ is

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Solution

The correct option is D
$f_{1} = (\frac{340}{340 - 34}) f = \frac{10}{9} f f_{2} = (\frac{340}{340 - 17}) f = \frac{20}{19} f ∴ \frac{f_{1}}{f_{2}} = \frac{\frac{10}{9}}{\frac{20}{19}} = \frac{19}{18}$

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A train moves towards a stationary observer with speed 34 m/s. The train sounds a whistle and its frequency registered by the observer is $f_{1}$ . If the train's speed is reduced to 17 m/s, the frequency registered is $f_{2}$ . If the speed of sound is 340 m/s then the ratio $\frac{f_{1}}{f_{2}}$ is :

A train moves towards a stationary observer with speed 34 m/s. The train sounds a whistle and its frequency registered is f1. If the train’s speed is reduced to 17 m/s, the frequency registered is $f_{2}$ . If the speed of sound is 340 m/s then the ratio $\frac{f_{1}}{f_{2}}$ is

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A train moves towards a stationary observer with a speed of 34 m/s. The train sounds a whistle and its frequency registered by the observer is f1. If the train’s speed is reduced to 17 m/s, the frequency registered is f2. If the speed of sound is 340 m/s, then the ratio f1f2 is

The correct option is D f1=(340340−34)f=109ff2=(340340−17)f=2019f∴f1f2=1092019=1918

The correct option is D
$f_{1} = (\frac{340}{340 - 34}) f = \frac{10}{9} f f_{2} = (\frac{340}{340 - 17}) f = \frac{20}{19} f ∴ \frac{f_{1}}{f_{2}} = \frac{\frac{10}{9}}{\frac{20}{19}} = \frac{19}{18}$