InterviewSolution
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InterviewSolution
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A violin player riding on a slow train plays a 440 Hz note. Another violin player standing near the track plays the same note. When the two are close by and the train approaches the person on the ground, he hears 4 beats per second. The speed of sound in air = 340 m/s. (a) Calculate the speed of the train. (b) What beat frequency is heard by the player in the train? |
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Answer» (a) The apparent frequency heard by the standing person will be more than 440 Hz due to Doppler's effect. Since he hears 4 beats per second, the apparent frequency of the violin in the train, ν' = 440+4 Hz =444 Hz. Here ν =440 Hz V = 340 m/s u =? ν' = Vν/(V-u) →444 =340*440/(340-u) →340-u = 340*440/444 =336.94 →u = 340 - 336.94 m/s = 3.06 m/s =3.06*3600/1000 km/h ≈11 km/h (b) For the person in the train, the source on the ground is stationary and observer approaches. The apparent frequency heard by the person in the train ν'' =ν(V+u)/V = 440*(340+3.06)/340 Hz =440*1.009 Hz =443.96 Hz The beat frequency heard when the train approaches =443.96 - 440 = 3.96 beats/s. i.e. a little less than 4 beats/s |
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