A whistle of frequency 500 Hz tied to the end of a string of length 1 . 2 m revolves at 400 rev min - 1 . A listener standing some distance away in the plane of rotation of whistle hears frequencies in the range. (Speed of sound = 340 m s - 1 )

Step-by-step solution

v s = r ω = 1 . 2 × 2 π f ∵ ω = 2 π f = 1 . 2 × 2 × 3 . 14 × 400 60 = 50   m   s - 1 v min = v v + v s · v = 340 340 + 50 × 500 = 436   Hz v max = v v - v s · v = 340 340 - 50 × 500 = 586   Hz