Solve this : A source starts from rest at t = 0 and moves away from a stationary observer with a constant acceleration 5 m s - 2 . At t = 0 source is 160 m away from observer. If it emits sound of frequency 700 Hz the frequency heard by observer at t= 2.5s is _____ [velocity of sound = 340 m s - 1 ] 1) 600 Hz 2) 640 Hz 3) 680 Hz 4) 690 Hz Share with your friends Share 1 Pintu B. answered this Dear Student , When the source moves away from the observer then , Velocity of the source at t=2.5 s is ,vo=uo+at=0+5×2.5=12.5 m/schange in frequency,f'=fv-vsv=700×340-12.5340=674.26 m/sSo here the correct option is 3.Regards 2 View Full Answer
