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

Step-by-step solution

v s = r ω v s = 1.2 × 2 × 3.14 × 400 60 = 50 m s - 1 v min ⁡   = v v + v s ν = 340 340 + 50 × 500 = 436   Hz   v max ⁡   = v v - v s v = 340 340 - 50 × 500 = 586   Hz