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2 years ago

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A woman is riding a bicycle at 18.0 m/s along a straight road that runs parallel to and right next to some railroad tracks. She

hears the whistle of a train that is behind. The frequency emitted by the train is 840 Hz, but the frequency the woman hears is 778 Hz. Take the speed of sound to be 340 m/s.
(a) What is the speed of the train, and is the train traveling away from or towardt he bicycle?

(b) What frequency is heard by a stationary observer located between the train and the bicycle?

1 answer:

mr_godi [17]2 years ago

8 0

Answer:

(a) Speed of train is 7.66 m/s and it is traveling away from the bicycle.

(b) Frequency heard by the stationary observer is 821.50 Hz.

Explanation:

Doppler effect is defines as the change in the frequency of the wave as the observer and/or source are moving away or towards each other.

(a) According to the problem,

Speed of the observer, v₀ = 18 m/s

Speed of sound in air, v = 340 m/s

Original frequency emitted by train, f = 840 Hz

Apparent frequency heard by the observer, f₀ = 778 Hz

Let v₁ be the speed of the train.

Since, apparent frequency is less than original frequency i.e. f₀ > f . Hence, train are travelling away from the bicycle.

Thus, the Doppler effect equation is :

$f_{0} = (\frac{v - v_{0} }{v+v_{1} } )f$

Substitute the suitable values in the above equation.

$778 = (\frac{340 -18 }{340+v_{1} } )840$

340 + v₁ = 347.66

v₁ = 7.66 m/s

(b) In this case, speed of observer, v₀ = 0 m/s

Apparent frequency, $f_{0} = (\frac{v }{v+v_{1} } )f$

Substitute the suitable values in the above equation.

$f_{0} = (\frac{340 }{340+7.66 } )840$

f₀ = 821.50 Hz

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