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NEET
Exam
NEET
Subject
Physics
Chapter
Waves and Sound

NEET · Physics · Waves and Sound

Two loudspeakers M and N are located 20m apart and emit sound at...

Two loudspeakers M and N are located 20m apart and emit sound at frequencies 118 Hz and 121 Hz, respectively. A car is initially at a point P, 1800 m away from the midpoint Q of the line MN and moves towards Q constantly at 60 km/hr along the perpendicular bisector of MN. It crosses Q and eventually reaches a point R, 1800 m away from Q. Let υυ\left(t\right)representthebeatfrequencymeasuredbyapersonsittinginthecarattimet.Letrepresent the beat frequency measured by a person sitting in the car at time t. Let{ν}_{p}, {ν}_{Q}and{ν}_{R}bethebeatfrequenciesmeasuredatlocationsP,QandR,respectively.Thespeedofsoundinairisbe the beat frequencies measured at locations P, Q and R, respectively. The speed of sound in air is330m{s}^{- 1} . Which of the following statement(s) is false regarding the sound heard by the person ?Two loudspeakers M and N are located 20m apart and emit sound at frequencies 118 Hz and 121 Hz, respectively. A car isTwo loudspeakers M and N are located 20m apart and emit sound at frequencies 118 Hz and 121 Hz, respectively. A car isTwo loudspeakers M and N are located 20m apart and emit sound at frequencies 118 Hz and 121 Hz, respectively. A car is

Options

  1. AThe plot below represents schematically the variation of beat frequency with time
    What is the role of sodium in the Wurtz-Fittig reaction?
  2. BThe plot below represents schematically the variations of beat frequency with time
    What is the role of sodium in the Wurtz-Fittig reaction?
  3. CThe rate of change in beat frequency is maximum when the car passes through Q
  4. DNone of the above

Correct answer

D. None of the above

Step-by-step solution

Two loudspeakers M and N are located 20m apart and emit sound at frequencies 118 Hz and 121 Hz, respectively. A car is
fM=C+VcosθCf1{f}_{M} = \dfrac{C + Vcos \theta}{C}{f}_{1}
fN=C+VcosθCf2{f}_{N} = \dfrac{C + Vcos \theta}{C}{f}_{2}
\Deltaf=fN−fM\Deltaf = {f}_{N} - {f}_{M}
=C+VcosθC(f2−f1)= \dfrac{C + Vcos \theta}{C}\left({f}_{2} - {f}_{1}\right)
d(\Deltaf)dt=−VC(f2−f1)sinθdθdt\dfrac{d\left(\Deltaf\right)}{dt} = - \dfrac{V}{C}\left({f}_{2} - {f}_{1}\right)sin \theta\dfrac{d\theta}{dt}
∴∴ and d(\Deltaf)dt\dfrac{d\left(\Deltaf\right)}{dt} is maximum when θ=90o\theta = {90}^{o}
\left[∴ \left(iii\right)iscorrect\right]<br/><br/>{v}_{P} = \left(1 + \dfrac{V}{C}cos \theta\right)\Deltaf<br/><br/>{v}_{Q} = \Deltaf<br/><br/>{v}_{R} = \left(1 - \dfrac{V}{C}cos \theta\right)\Deltaf<br/><br/>∴ {v}_{P} + {v}_{r} = 2{v}_{Q}
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