A car is traveling with a velocity magnitude of p_1 eastward, parallel to a road

Question

A car is traveling with a velocity magnitude of p_1 eastward, parallel to a road next to a set of railroad tracks. Two trains are approaching each other, and they are both traveling v_1 relative to the ground. Both trains blow their whistles, one after another, as they approach the cars initial location. The whistles have a natural frequency integral, and let the speed of sound in air to be v. Calculate the frequencies, integral^-1_E for eastbound the train and integral^-1_W for the westbound train, as heard by the driver of the car. Express your answer in terms of the following variables: v_1, v_2, integral, and/or v. The trains blow their whistles again after they have passed by each other and the car Calculate the new frequencies, integral^-1_E for the eastbound train and integral^-1_W for the westbound train, as heard by the driver of the car, Express your answer in terms of the following variables: v_1, v_2 integral, and/or v. Calculate the numerical answers for parts (a) and (b) using the following values: v_1 = 10 m/s, v_2 = 50 m/s, f = 7500 Hz, and v = 350 m/s.

Explanation / Answer

a) fE' = f*(v - v1)/(v - v2)

fW' = f*(v + v2)/(v - v2)

b) fE' = f*(v - v1 )/(v + v2)

fW' = f*(v + v2)/(v + v2)

c)

a)

fE' = f*(v - v1)/(v - v2)

= 7500*(350 - 10)/(350 - 50)

= 8500 Hz

fW' = f*(v + v2)/(v - v2)

= 7500*(350 + 10)/(350 - 50)

= 9000 Hz

b) fE' = f*(v - v1 )/(v + v2)

= 7500*(350 - 10)/(350 + 50)

= 6375 Hz

fW' = f*(v + v2)/(v - v2)

= 7500*(350 + 10)/(350 + 50)

= 6750 Hz

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