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JEE Main 202226 JulyMorning Shift
Exam
JEE Main
Year
2022
Date
26 July
Shift
Morning Shift
Subject
Physics
Chapter
Waves and Sound

JEE Main 2022 · 26 July · Morning Shift · Physics · Waves and Sound

When a car is approaching the observer, the frequency of horn is...

When a car is approaching the observer, the frequency of horn is 100Hz100Hz. After passing the observer, it is 50Hz50Hz. If the observer moves with the car, the frequency will be x3Hz\dfrac{x}{3}Hz where x=x =_____ .

Correct answer

0

Step-by-step solution

Apparent frequency when source of moving towards observer, f1=100=f0(cc−vs)...(1)</p><p>Here,{f}_{1} = 100 = {f}_{0}\left(\dfrac{c}{c - {v}_{s}}\right) . . . \left(1\right)</p><p>Here,{f}_{0} =frequencyofobserver,frequency of observer,c =speedofsoundandspeed of sound and{v}_{s} =speedofsource.</p><p>Apparentfrequencywhensourceofmovingawayfromobserver,</p><p>speed of source.</p><p>Apparent frequency when source of moving away from observer, </p><p>{f}_{2} = 50 = {f}_{0}(cc+vs)\left(\dfrac{c}{c + {v}_{s}}\right) . . . \left(2\right)</p><p>Fromabovetwoequations,wehave</p><p></p><p>From above two equations, we have </p><p>\dfrac{{f}_{1}}{{f}_{2}} = 2 = \dfrac{c + {v}_{s}}{c - {v}_{s}}</p><p></p><p>⇒ 2c - 2{v}_{s} = c + {v}_{s}</p><p></p><p>⇒ 3{v}_{s} = c</p><p>Puttingthevalueinequation(1)</p><p>Putting the value in equation \left(1\right), we have

100=f0c2c3=32f0100 = {f}_{0}\dfrac{c}{\dfrac{2c}{3}} = \dfrac{3}{2}{f}_{0}

⇒f0=2003Hz⇒ {f}_{0} = \dfrac{200}{3}Hz

Hence, x=200x = 200.

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