Question
Question: Two men are walking along a horizontal straight line in the same direction. The man in front walks a...
Two men are walking along a horizontal straight line in the same direction. The man in front walks at a speed of 1.0ms−1 and the man behind walks at a speed of 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 men walking are blowing identical whistles which emit a sound of frequency 1430Hz. 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 ___________.
Solution
Beat frequency is the difference of frequencies of two waves of slightly different frequencies. The two frequencies need to be in the same medium.
Complete step by step answer:
According to Doppler’s Effect, apparent frequency (ν′) can be calculated by,
υ′=(v−vSv−vO)υ
Where, v= speed of sound in medium
vO= speed of observer
vS= speed of source
υ= real frequency
The component of velocity which the observers comprehend are,
⇒vA=vAcosθ
⇒vB=vBcosθ
The apparent frequency of sound from A as heard by the observer,
⇒υA′=(v−vAv)υ
⇒υ′A=(330−1cosθ330)1430 ⇒υ′A=1−330cosθ11430 ⇒υ′A=(1−330cosθ)1430
The apparent frequency of sound from B as heard from observer,
⇒υ′B=(v−vBv)υ ⇒υ′B=(330−2cosθ330)1430 ⇒υ′B=1−3302cosθ11430 ⇒υ′B=(1+3302cosθ)1430
So, the beat frequency (Δυ) can be calculated by,