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PhysicsNumerical+3Question 15 of 54

JEE Advanced 2019 · Paper 1, Physics question 15

A train S1, moving with a uniform velocity of 108 km/h, approaches another train S2 standing on a platform. An observer O moves with a uniform velocity of 36 km/h towards S2, as shown in figure. Both the trains are blowing whistles of same frequency 120 Hz. When O is 600 m away from S2 and distance between S1 and S2 is 800 m, the number of beats heard by O is ______.

[Speed of the sound = 330 m/s]

S2 at a corner; S1 is 800 m to the right of S2, moving towards S2 at 108 km/h; observer O is 600 m below S2, moving towards S2 at 36 km/h (the two paths are perpendicular).
Answer and worked solution

Official answer: 7

Speeds. S1 moves at 30 m/s and O at 10 m/s.

Whistle of S2 (stationary, observer approaching head-on):
f2=120⋅330+10330=123.6 Hz.f_2=120\cdot\frac{330+10}{330}=123.6\text{ Hz}.

Whistle of S1. S1 is 1000 m from O. The line S1→O has direction cosines 0.8 along S1's motion and 0.6 along O's.

  • S1's velocity component towards O is 30×0.8=2430\times0.8=24 m/s.
  • O's velocity component towards S1 is 10×0.6=610\times0.6=6 m/s.

f1=120⋅330+6330−24=131.8 Hz.f_1=120\cdot\frac{330+6}{330-24}=131.8\text{ Hz}.

Beats.
f1−f2≈8.1 per second.f_1-f_2\approx8.1\text{ per second}.
Answer: 8.13 (key: 7.00 to 10.00)

Topics:Doppler effectBeats
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