Physics · Sound Waves

JEE Advanced 2018 — Paper 1 — Question 8

Two men are walking along a horizontal straight line in the same direction. The man in front walks at a speed 1.0 ms−11.0 \mathrm{~ms}^{-1} and the man behind walks at a speed 2.0 ms−12.0 \mathrm{~ms}^{-1}. A third man is standing at a height 12 m above the same horizontal line such that all three men are in a vertical plane. The two walking men are blowing identical whistles which emit a sound of frequency 1430 Hz . The speed of sound in air is 330 ms−1330 \mathrm{~ms}^{-1}. At the instant, when the moving men are 10 m apart, the stationary man is equidistant from them. The frequency of beats in Hz , heard by the stationary man at this instant, is \qquad -.

Answer: 5

Numerical answer — enter this value.

Step-by-step solution

Δf=f0(330330−2cos⁡θ−330330+1cos⁡θ)=f0(330330−1013−330330+513)≃f0(1+1013×330−1+513×330)=1513×330×1430=5 Hz\begin{aligned} & \Delta f=f_{0}\left(\frac{330}{330-2 \cos \theta}-\frac{330}{330+1 \cos \theta}\right) \\& \quad=f_{0}\left(\frac{330}{330-\frac{10}{13}}-\frac{330}{330+\frac{5}{13}}\right) \\& \simeq f_{0}\left(1+\frac{10}{13 \times 330}-1+\frac{5}{13 \times 330}\right) \\& =\frac{15}{13 \times 330} \times 1430=5 \mathrm{~Hz} \end{aligned}
Solution figure

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Exam
JEE Advanced 2018
Paper
Paper 1
Subject
Physics
Chapter
Sound Waves
Topic
Doppler Effect of Sound