JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Waves & Sound

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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

]

Answer: 8.13

Solution

Convert velocities:

S1: 108 km/h = 30 m/s (approaching)
O: 36 km/h = 10 m/s (towards S2)

Frequency from S2 (stationary source, moving observer):

f_{2} = \frac{v + v _{O}}{v} f_{0} = \frac{330 + 10}{330} \times 120 = \frac{340}{330} \times 120 = 123.63 Hz

Frequency from S1 (moving source, moving observer):
S1 approaches at 30 m/s, O moves towards S2 (away from S1 at relative velocity).
Actually, O moves at 10 m/s towards S2, which is 6 m/s component towards S1.

f_{1} = \frac{v + v _{O}}{v - v _{S}} f_{0} = \frac{330 + 6}{330 - 24} \times 120 = \frac{336}{306} \times 120 \approx 131.76 Hz

Beat frequency:

f_{b e a t} = ∣ f_{1} - f_{2} ∣ = ∣131.76 - 123.63∣ = 8.13 Hz

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Numerical answer

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JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Waves & Sound

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