Two men are walking along a horizontal straight line in the same direction. The man in front walks at a speed 1.0ms−1 and the man behind walks at a speed 2.0ms−1. A third man is standing at a height 12m 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 330ms−1. At the instant, when the moving men are 10m apart, the stationary man is equidistant from them. The frequency of beats in Hz, heard by the stationary man at this instant, is ____.
Answer: 5
Solution
The geometry gives cosθ=5/13 where θ is the angle from vertical. The observed frequencies are fA=1430330−2cosθ330 and fB=1430330−1cosθ330. The beat frequency is fA−fB=1430×330[330−10/131−330+5/131]≈5.00 Hz.
JEE Advanced · 2018 · 20 May 2018 · Paper 1 · Numerical answer
