    Question

# Two trains A and B simultaneously start moving along parallel tracks from a station along same direction. A starts with constant acceleration 2 m/s2 from rest while B with the same acceleration but with initial velocity of 40 m/s. Twenty seconds after the start, passenger of A hears whistle of B. If frequency of whistle is 1194 Hz and velocity of sound in air is 322 m/s, then the frequency observed by the passenger is

A
1186 Hz
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B
1086 Hz
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C
986 Hz
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D
886 Hz
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Solution

## The correct option is B 1086 Hz fapp=f0(v±v0v±vs) PR=0+12.2.202=400 m PQ=t2 PS=40+122t2 QS=QR+RS ⟹PS−PQ=PR−PQ+xsound(20−t) So, (40t+t2)−t2=400−t2+322(20−t) 40t=400−t2+6440−322t t2+362t−6840=0t=18 fapp=f0(v±v0v±vs)v0→20 s,vs→t=18 s. v0=0+2×20=40m/svs=u+at=40+2×18=76m/s fapp=1194(322+40322+76) ⟹fapp=1086 Hz  Suggest Corrections  0      Explore more