Question
A train has just completed a U-curve in a track which is a semi-circle. 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 semi-circular track. The driver blows a whistle of frequency 200 Hz. Velocity of sound is 340 m/s. Then the apparent frequency as observed by a passenger in the middle of the train, when the speed of the train is 30 m/s, is
Answer: Option C
:
C
The Doppler formula holds for non-collinear motion if vs and v0 are taken to be the resolved component along the line of slight, In this case, we have
v0=−vtsin45∘=−30√2m/s
vs=−vtsin45∘=30√2m/s
We have, v = 340 m/s, n = 200 Hz. The apparent frequency n’ is given by
n′=n[v−v0v−vs]=200[340+(30√2)340+(30√2)]=200Hz
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:
C
The Doppler formula holds for non-collinear motion if vs and v0 are taken to be the resolved component along the line of slight, In this case, we have
v0=−vtsin45∘=−30√2m/s
vs=−vtsin45∘=30√2m/s
We have, v = 340 m/s, n = 200 Hz. The apparent frequency n’ is given by
n′=n[v−v0v−vs]=200[340+(30√2)340+(30√2)]=200Hz
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