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JEE Main 20236 AprilMorning Shift
Exam
JEE Main
Year
2023
Date
6 April
Shift
Morning Shift
Subject
Physics
Chapter
Waves and Sound

JEE Main 2023 · 6 April · Morning Shift · Physics · Waves and Sound

A person driving car at a constant speed of 15m s ^ - 1 is...

A person driving car at a constant speed of 15ms−115m{s}^{- 1} is approaching a vertical wall. The person notices a change of 40Hz40Hz in the frequency of his car’s horn upon reflection from the wall. The frequency of horn is_______Hz .

(Given: Speed of sound: 30ms−130m{s}^{- 1})

Correct answer

0

Step-by-step solution

The formula to calculate the frequency (f′)\left(f '\right) as heard by an observer with respect to the frequency (f)\left(f\right) produced from a source is given by

f′=v±vov∓vsf....(1)f ' = \dfrac{v \pm {v}_{o}}{v ∓ {v}_{s}}f . . . . ( 1 )

where, vv is the speed of sound in air, vo{v}_{o} is the speed of the observer and vs{v}_{s} is the speed of the source.

In the first situation, the source is the moving car and the observer is the wall. Hence, the frequency (fw)\left({f}_{w}\right) as heard on the wall can be written as

{fw=v−0v−vcf=vv−vcf....(2)\begin{cases}{f}_{w} & = & \dfrac{v - 0}{v - {v}_{c}}f \\ & = & \dfrac{v}{v - {v}_{c}}f . . . . ( 2 )\end{cases}

where, vc{v}_{c} is the speed of the car.

In the second situation, when the sound reflects back from the wall, the source of sound is the wall and the observer is the car.

Hence, the frequency (fc)\left({f}_{c}\right) of sound as heard by the car after the reflection is given by

{fc=v+vcv−0fw=v+vcvfw....(3)\begin{cases}{f}_{c} & = & \dfrac{v + {v}_{c}}{v - 0}{f}_{w} \\ & = & \dfrac{v + {v}_{c}}{v}{f}_{w} . . . . ( 3 )\end{cases}

From equation (2) and (3), it follows that

{fc=v+vcv×vv−vcf=v+vcv−vcf....(4)\begin{cases}{f}_{c} & = & \dfrac{v + {v}_{c}}{v} \times \dfrac{v}{v - {v}_{c}}f \\ & = & \dfrac{v + {v}_{c}}{v - {v}_{c}}f . . . . ( 4 )\end{cases}

It is given in the problem that

fc−f=40.....(5){f}_{c} - f = 40 . . . . . ( 5 )

Substitute the expression from equation (4) into equation (5) and solve to calculate the frequency of the horn.

{v+vcv−vcf−f=40⇒2vcv−vcf=40\begin{cases}\dfrac{v + {v}_{c}}{v - {v}_{c}}f - f & = & 40 \\ & ⇒ & \dfrac{2{v}_{c}}{v - {v}_{c}}f = 40\end{cases}

{⇒\begin{cases} & ⇒ & \end{cases}\begin{cases} & & f\end{cases}{=\begin{cases} & = & \end{cases}\begin{cases} & & \end{cases}{40\begin{cases} & & 40\end{cases}\begin{cases} & & \dfrac{v - {v}_{c}}{2{v}_{c}}\end{cases}{\begin{cases} & & \end{cases}\begin{cases} & = & \end{cases}{\begin{cases} & & \end{cases}\begin{cases} & & 40\end{cases}{×\begin{cases} & & \times\end{cases}\begin{cases} & & \dfrac{330 - 15}{2 \times 15}\end{cases}{\begin{cases} & & \end{cases}\begin{cases} & & m\end{cases}{\begin{cases} & & \end{cases}\begin{cases} & & s\end{cases}{−1\begin{cases} & & {}^{- 1}\end{cases}\begin{cases} & & \end{cases}{=\begin{cases} & = & \end{cases}\begin{cases} & & \end{cases}{420\begin{cases} & & 420\end{cases}\begin{cases} & & \end{cases}{m\begin{cases} & & m\end{cases}\begin{cases} & & \end{cases}{s\begin{cases} & & s\end{cases}\begin{cases} & & {}^{- 1}\end{cases}$$

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