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 )

Question

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 ) (A) 386Hz-436Hz (B) 426Hz-474Hz (C) 426Hz-586Hz (D) 436Hz-586Hz

Tardigrade Admin · Accepted Answer

vs=rω vs=1.2× 2× 3.14× ((400/60))=50ms- 1 vmin =(v/v + vs)ν =(340/340 + 50)× 500=436 Hz vmax =(v/v - vs)v =(340/340 - 50)× 500 =586Hz

Q. A whistle of frequency $500 Hz$ , tied to the end of a string of length $1.2 m$ , revolves at $400 re v / min$ . A listener standing some distance away in the plane of rotation of the whistle, hears frequency in the range of (speed of sound = $340 m s^{- 1}$ )

$386 Hz - 436 Hz$

$426 Hz - 474 Hz$

$426 Hz - 586 Hz$

$436 Hz - 586 Hz$

$v_{s} = r ω$
$v_{s} = 1.2 \times 2 \times 3.14 \times (\frac{400}{60}) = 50 m s^{- 1}$
$v_{min} = \frac{v}{v + v _{s}} ν$
$= \frac{340}{340 + 50} \times 500 = 436 Hz$
$v_{ma x} = \frac{v}{v - v _{s}} v$
$= \frac{340}{340 - 50} \times 500$
$= 586 Hz$

Q. A whistle of frequency 500Hz , tied to the end of a string of length 1.2m , revolves at 400rev/min . A listener standing some distance away in the plane of rotation of the whistle, hears frequency in the range of (speed of sound = 340ms−1 )

386Hz−436Hz

426Hz−474Hz

426Hz−586Hz

436Hz−586Hz