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PHXI15:WAVES

354766 A car $P$ travelling at $20 m s^{- 1}$ sounds its horn at a frequency of $400 H z$ . Another car $Q$ is travelling behind the first car in the same direction with a velocity $40 m s^{- 1}$ . The frequency heard by the passenger of the car $Q$ is approximately [Take, velocity of sound $= 360 m s^{- 1}$ ]

1

471 H z

2

421 H z

3

514 H z

4

485 H z

Explanation:

Velocity of car $P (v_{s}) = 20 m / s$
Frequency of emitted sound $(v) = 400 H z$
Velocity of car $Q (v_{0}) = 360 m / s$
From Doppler's effect $v^{'} = v (\frac{v \pm v_{0}}{v \pm v_{s}})$
Since source and observer both are moving in the same direction,
$v^{'} = 400 (\frac{360 + 40}{360 + 20}) H z$ .
$v^{'} = 400 (\frac{400}{380}) = 421.05 H z$
$v^{'} = 421 H z$

JEE - 2023

PHXI15:WAVES

354767 A whistle is revolved with high speed in a horizontal circle of radius $R$ . To an observer at the centre of the circle the frequency of the whistle will appear to be

1 Decreasing

2 Increasing

3 Both

4 Constant

PHXI15:WAVES

354768 A source of sound having natural frequency $f_{0} = 1800 H z$ is moving uniformly along a line separated from a stationary listener by a distance $l = 250 m$ . The velocity of source is $n = 0.80$ times the velocity of sound. Find the frequency of sound received by the listener at the moment when the source gets closest to him.

1

2000 H z

2

5000 H z

3

7000 H z

4

9000 H z

Explanation:

As sound waves take finite time in reaching from one position to other position. The sound waves emitted by the source when it is closest to the listener (at $O$ ) does not reach instantaneously. The frequency heard at this instant should be emitted from some previous position of source. Let this sound (or disturbance) was created at position ' $S$ '.
supporting img
The direction of velocity vector of source makes an angle $θ$ with the line joining the disturbance position (source position) and observer position as shown in the figure. The time in which sound reaches the listener along $S L$ the source reaches point $O$ , closest to the listener. From figure,
$\cos θ = \frac{S O}{S L} = \frac{v_{s} t}{v} = \frac{v_{s}}{v} = n (1)$
Now using the formula for apparent frequency received by observer
$f_{app} = f_{original} (\frac{v - v_{0}}{v - v_{s}})$
As Doppler effect is applicable only along the line joining disturbance position (source position) and observer position,hence we should substitute the component of velocity of the source along this line ' $S L$ '.
Hence the frequency heard by the listener is $\begin{array}{l} f_{a p p} = f_{o} (\frac{v - 0}{v - v_{s} \cos θ}) (2) \\ \Rightarrow f_{a p p} = f_{o} (\frac{v}{v - (n v) n}) = \frac{f_{s}}{1 - n^{2}} (3) \end{array}$
On substituting numerical values, we get $f_{app} = 5000 H z$

PHXI15:WAVES

354769 A small source of sound moves on a circle (in anti clockwise) as shown in figure and an observer is sitting at $O$ . Let $f_{A}, f_{B}, f_{C}$ be the frequencies heard when the source is at $A$ , $B$ and $C$ respectively.
supporting img

1

f_{A} > f_{B} > f_{C}

2

f_{A} = f_{B} > f_{C}

3

f_{B} > f_{C} > f_{A}

4

f_{A} > f_{C} > f_{B}

Explanation:

Frequency heard by observer (stationary) will be maximum, if the source is moving towards the observer, which is the case when the source is at $B$ . Frequency heard by observer (stationary) will be mininum, if the source is moving away from the observer, which is the case when the source is at $A$ .
supporting img

For point $C$ , frequency heard is equal to the original frequency

PHXI15:WAVES

354770 A source of sound $S$ emitting waves of frequency $100 H z$ and an observer $O$ are located at some distance from each other. The source is moving with a speed of $19.4 m s^{- 1}$ at an angle of $60^{\circ}$ with the source observer line as shown in the figure. The observer is at rest. The apparent frequency observed by the observer (velocity of sound in air $330 m s^{- 1}$ ) is
supporting img

1

106 H z

2

103 H z

3

100 H z

4

97 H z

Explanation:

The frequency received by the observer is
$f^{'} = f (\frac{v}{v - v_{s} \cos 60^{\circ}}) = 100 (\frac{330}{330 - 19.4 \cos 60^{\circ}})$
$f^{'} = 100 (\frac{330}{330 - 9.7}) = 103 H z$

PHXI15:WAVES

354766 A car $P$ travelling at $20 m s^{- 1}$ sounds its horn at a frequency of $400 H z$ . Another car $Q$ is travelling behind the first car in the same direction with a velocity $40 m s^{- 1}$ . The frequency heard by the passenger of the car $Q$ is approximately [Take, velocity of sound $= 360 m s^{- 1}$ ]

1

471 H z

2

421 H z

3

514 H z

4

485 H z

Explanation:

Velocity of car $P (v_{s}) = 20 m / s$
Frequency of emitted sound $(v) = 400 H z$
Velocity of car $Q (v_{0}) = 360 m / s$
From Doppler's effect $v^{'} = v (\frac{v \pm v_{0}}{v \pm v_{s}})$
Since source and observer both are moving in the same direction,
$v^{'} = 400 (\frac{360 + 40}{360 + 20}) H z$ .
$v^{'} = 400 (\frac{400}{380}) = 421.05 H z$
$v^{'} = 421 H z$

JEE - 2023

PHXI15:WAVES

354767 A whistle is revolved with high speed in a horizontal circle of radius $R$ . To an observer at the centre of the circle the frequency of the whistle will appear to be

1 Decreasing

2 Increasing

3 Both

4 Constant

PHXI15:WAVES

354768 A source of sound having natural frequency $f_{0} = 1800 H z$ is moving uniformly along a line separated from a stationary listener by a distance $l = 250 m$ . The velocity of source is $n = 0.80$ times the velocity of sound. Find the frequency of sound received by the listener at the moment when the source gets closest to him.

1

2000 H z

2

5000 H z

3

7000 H z

4

9000 H z

Explanation:

As sound waves take finite time in reaching from one position to other position. The sound waves emitted by the source when it is closest to the listener (at $O$ ) does not reach instantaneously. The frequency heard at this instant should be emitted from some previous position of source. Let this sound (or disturbance) was created at position ' $S$ '.
supporting img
The direction of velocity vector of source makes an angle $θ$ with the line joining the disturbance position (source position) and observer position as shown in the figure. The time in which sound reaches the listener along $S L$ the source reaches point $O$ , closest to the listener. From figure,
$\cos θ = \frac{S O}{S L} = \frac{v_{s} t}{v} = \frac{v_{s}}{v} = n (1)$
Now using the formula for apparent frequency received by observer
$f_{app} = f_{original} (\frac{v - v_{0}}{v - v_{s}})$
As Doppler effect is applicable only along the line joining disturbance position (source position) and observer position,hence we should substitute the component of velocity of the source along this line ' $S L$ '.
Hence the frequency heard by the listener is $\begin{array}{l} f_{a p p} = f_{o} (\frac{v - 0}{v - v_{s} \cos θ}) (2) \\ \Rightarrow f_{a p p} = f_{o} (\frac{v}{v - (n v) n}) = \frac{f_{s}}{1 - n^{2}} (3) \end{array}$
On substituting numerical values, we get $f_{app} = 5000 H z$

PHXI15:WAVES

354769 A small source of sound moves on a circle (in anti clockwise) as shown in figure and an observer is sitting at $O$ . Let $f_{A}, f_{B}, f_{C}$ be the frequencies heard when the source is at $A$ , $B$ and $C$ respectively.
supporting img

1

f_{A} > f_{B} > f_{C}

2

f_{A} = f_{B} > f_{C}

3

f_{B} > f_{C} > f_{A}

4

f_{A} > f_{C} > f_{B}

Explanation:

Frequency heard by observer (stationary) will be maximum, if the source is moving towards the observer, which is the case when the source is at $B$ . Frequency heard by observer (stationary) will be mininum, if the source is moving away from the observer, which is the case when the source is at $A$ .
supporting img

For point $C$ , frequency heard is equal to the original frequency

PHXI15:WAVES

354770 A source of sound $S$ emitting waves of frequency $100 H z$ and an observer $O$ are located at some distance from each other. The source is moving with a speed of $19.4 m s^{- 1}$ at an angle of $60^{\circ}$ with the source observer line as shown in the figure. The observer is at rest. The apparent frequency observed by the observer (velocity of sound in air $330 m s^{- 1}$ ) is
supporting img

1

106 H z

2

103 H z

3

100 H z

4

97 H z

Explanation:

The frequency received by the observer is
$f^{'} = f (\frac{v}{v - v_{s} \cos 60^{\circ}}) = 100 (\frac{330}{330 - 19.4 \cos 60^{\circ}})$
$f^{'} = 100 (\frac{330}{330 - 9.7}) = 103 H z$

PHXI15:WAVES

354766 A car $P$ travelling at $20 m s^{- 1}$ sounds its horn at a frequency of $400 H z$ . Another car $Q$ is travelling behind the first car in the same direction with a velocity $40 m s^{- 1}$ . The frequency heard by the passenger of the car $Q$ is approximately [Take, velocity of sound $= 360 m s^{- 1}$ ]

1

471 H z

2

421 H z

3

514 H z

4

485 H z

Explanation:

Velocity of car $P (v_{s}) = 20 m / s$
Frequency of emitted sound $(v) = 400 H z$
Velocity of car $Q (v_{0}) = 360 m / s$
From Doppler's effect $v^{'} = v (\frac{v \pm v_{0}}{v \pm v_{s}})$
Since source and observer both are moving in the same direction,
$v^{'} = 400 (\frac{360 + 40}{360 + 20}) H z$ .
$v^{'} = 400 (\frac{400}{380}) = 421.05 H z$
$v^{'} = 421 H z$

JEE - 2023

PHXI15:WAVES

354767 A whistle is revolved with high speed in a horizontal circle of radius $R$ . To an observer at the centre of the circle the frequency of the whistle will appear to be

1 Decreasing

2 Increasing

3 Both

4 Constant

PHXI15:WAVES

354768 A source of sound having natural frequency $f_{0} = 1800 H z$ is moving uniformly along a line separated from a stationary listener by a distance $l = 250 m$ . The velocity of source is $n = 0.80$ times the velocity of sound. Find the frequency of sound received by the listener at the moment when the source gets closest to him.

1

2000 H z

2

5000 H z

3

7000 H z

4

9000 H z

Explanation:

As sound waves take finite time in reaching from one position to other position. The sound waves emitted by the source when it is closest to the listener (at $O$ ) does not reach instantaneously. The frequency heard at this instant should be emitted from some previous position of source. Let this sound (or disturbance) was created at position ' $S$ '.
supporting img
The direction of velocity vector of source makes an angle $θ$ with the line joining the disturbance position (source position) and observer position as shown in the figure. The time in which sound reaches the listener along $S L$ the source reaches point $O$ , closest to the listener. From figure,
$\cos θ = \frac{S O}{S L} = \frac{v_{s} t}{v} = \frac{v_{s}}{v} = n (1)$
Now using the formula for apparent frequency received by observer
$f_{app} = f_{original} (\frac{v - v_{0}}{v - v_{s}})$
As Doppler effect is applicable only along the line joining disturbance position (source position) and observer position,hence we should substitute the component of velocity of the source along this line ' $S L$ '.
Hence the frequency heard by the listener is $\begin{array}{l} f_{a p p} = f_{o} (\frac{v - 0}{v - v_{s} \cos θ}) (2) \\ \Rightarrow f_{a p p} = f_{o} (\frac{v}{v - (n v) n}) = \frac{f_{s}}{1 - n^{2}} (3) \end{array}$
On substituting numerical values, we get $f_{app} = 5000 H z$

PHXI15:WAVES

354769 A small source of sound moves on a circle (in anti clockwise) as shown in figure and an observer is sitting at $O$ . Let $f_{A}, f_{B}, f_{C}$ be the frequencies heard when the source is at $A$ , $B$ and $C$ respectively.
supporting img

1

f_{A} > f_{B} > f_{C}

2

f_{A} = f_{B} > f_{C}

3

f_{B} > f_{C} > f_{A}

4

f_{A} > f_{C} > f_{B}

Explanation:

Frequency heard by observer (stationary) will be maximum, if the source is moving towards the observer, which is the case when the source is at $B$ . Frequency heard by observer (stationary) will be mininum, if the source is moving away from the observer, which is the case when the source is at $A$ .
supporting img

For point $C$ , frequency heard is equal to the original frequency

PHXI15:WAVES

354770 A source of sound $S$ emitting waves of frequency $100 H z$ and an observer $O$ are located at some distance from each other. The source is moving with a speed of $19.4 m s^{- 1}$ at an angle of $60^{\circ}$ with the source observer line as shown in the figure. The observer is at rest. The apparent frequency observed by the observer (velocity of sound in air $330 m s^{- 1}$ ) is
supporting img

1

106 H z

2

103 H z

3

100 H z

4

97 H z

Explanation:

The frequency received by the observer is
$f^{'} = f (\frac{v}{v - v_{s} \cos 60^{\circ}}) = 100 (\frac{330}{330 - 19.4 \cos 60^{\circ}})$
$f^{'} = 100 (\frac{330}{330 - 9.7}) = 103 H z$

PHXI15:WAVES

354766 A car $P$ travelling at $20 m s^{- 1}$ sounds its horn at a frequency of $400 H z$ . Another car $Q$ is travelling behind the first car in the same direction with a velocity $40 m s^{- 1}$ . The frequency heard by the passenger of the car $Q$ is approximately [Take, velocity of sound $= 360 m s^{- 1}$ ]

1

471 H z

2

421 H z

3

514 H z

4

485 H z

Explanation:

Velocity of car $P (v_{s}) = 20 m / s$
Frequency of emitted sound $(v) = 400 H z$
Velocity of car $Q (v_{0}) = 360 m / s$
From Doppler's effect $v^{'} = v (\frac{v \pm v_{0}}{v \pm v_{s}})$
Since source and observer both are moving in the same direction,
$v^{'} = 400 (\frac{360 + 40}{360 + 20}) H z$ .
$v^{'} = 400 (\frac{400}{380}) = 421.05 H z$
$v^{'} = 421 H z$

JEE - 2023

PHXI15:WAVES

354767 A whistle is revolved with high speed in a horizontal circle of radius $R$ . To an observer at the centre of the circle the frequency of the whistle will appear to be

1 Decreasing

2 Increasing

3 Both

4 Constant

PHXI15:WAVES

354768 A source of sound having natural frequency $f_{0} = 1800 H z$ is moving uniformly along a line separated from a stationary listener by a distance $l = 250 m$ . The velocity of source is $n = 0.80$ times the velocity of sound. Find the frequency of sound received by the listener at the moment when the source gets closest to him.

1

2000 H z

2

5000 H z

3

7000 H z

4

9000 H z

Explanation:

As sound waves take finite time in reaching from one position to other position. The sound waves emitted by the source when it is closest to the listener (at $O$ ) does not reach instantaneously. The frequency heard at this instant should be emitted from some previous position of source. Let this sound (or disturbance) was created at position ' $S$ '.
supporting img
The direction of velocity vector of source makes an angle $θ$ with the line joining the disturbance position (source position) and observer position as shown in the figure. The time in which sound reaches the listener along $S L$ the source reaches point $O$ , closest to the listener. From figure,
$\cos θ = \frac{S O}{S L} = \frac{v_{s} t}{v} = \frac{v_{s}}{v} = n (1)$
Now using the formula for apparent frequency received by observer
$f_{app} = f_{original} (\frac{v - v_{0}}{v - v_{s}})$
As Doppler effect is applicable only along the line joining disturbance position (source position) and observer position,hence we should substitute the component of velocity of the source along this line ' $S L$ '.
Hence the frequency heard by the listener is $\begin{array}{l} f_{a p p} = f_{o} (\frac{v - 0}{v - v_{s} \cos θ}) (2) \\ \Rightarrow f_{a p p} = f_{o} (\frac{v}{v - (n v) n}) = \frac{f_{s}}{1 - n^{2}} (3) \end{array}$
On substituting numerical values, we get $f_{app} = 5000 H z$

PHXI15:WAVES

354769 A small source of sound moves on a circle (in anti clockwise) as shown in figure and an observer is sitting at $O$ . Let $f_{A}, f_{B}, f_{C}$ be the frequencies heard when the source is at $A$ , $B$ and $C$ respectively.
supporting img

1

f_{A} > f_{B} > f_{C}

2

f_{A} = f_{B} > f_{C}

3

f_{B} > f_{C} > f_{A}

4

f_{A} > f_{C} > f_{B}

Explanation:

Frequency heard by observer (stationary) will be maximum, if the source is moving towards the observer, which is the case when the source is at $B$ . Frequency heard by observer (stationary) will be mininum, if the source is moving away from the observer, which is the case when the source is at $A$ .
supporting img

For point $C$ , frequency heard is equal to the original frequency

PHXI15:WAVES

354770 A source of sound $S$ emitting waves of frequency $100 H z$ and an observer $O$ are located at some distance from each other. The source is moving with a speed of $19.4 m s^{- 1}$ at an angle of $60^{\circ}$ with the source observer line as shown in the figure. The observer is at rest. The apparent frequency observed by the observer (velocity of sound in air $330 m s^{- 1}$ ) is
supporting img

1

106 H z

2

103 H z

3

100 H z

4

97 H z

Explanation:

The frequency received by the observer is
$f^{'} = f (\frac{v}{v - v_{s} \cos 60^{\circ}}) = 100 (\frac{330}{330 - 19.4 \cos 60^{\circ}})$
$f^{'} = 100 (\frac{330}{330 - 9.7}) = 103 H z$

PHXI15:WAVES

354766 A car $P$ travelling at $20 m s^{- 1}$ sounds its horn at a frequency of $400 H z$ . Another car $Q$ is travelling behind the first car in the same direction with a velocity $40 m s^{- 1}$ . The frequency heard by the passenger of the car $Q$ is approximately [Take, velocity of sound $= 360 m s^{- 1}$ ]

1

471 H z

2

421 H z

3

514 H z

4

485 H z

Explanation:

Velocity of car $P (v_{s}) = 20 m / s$
Frequency of emitted sound $(v) = 400 H z$
Velocity of car $Q (v_{0}) = 360 m / s$
From Doppler's effect $v^{'} = v (\frac{v \pm v_{0}}{v \pm v_{s}})$
Since source and observer both are moving in the same direction,
$v^{'} = 400 (\frac{360 + 40}{360 + 20}) H z$ .
$v^{'} = 400 (\frac{400}{380}) = 421.05 H z$
$v^{'} = 421 H z$

JEE - 2023

PHXI15:WAVES

354767 A whistle is revolved with high speed in a horizontal circle of radius $R$ . To an observer at the centre of the circle the frequency of the whistle will appear to be

1 Decreasing

2 Increasing

3 Both

4 Constant

PHXI15:WAVES

354768 A source of sound having natural frequency $f_{0} = 1800 H z$ is moving uniformly along a line separated from a stationary listener by a distance $l = 250 m$ . The velocity of source is $n = 0.80$ times the velocity of sound. Find the frequency of sound received by the listener at the moment when the source gets closest to him.

1

2000 H z

2

5000 H z

3

7000 H z

4

9000 H z

Explanation:

As sound waves take finite time in reaching from one position to other position. The sound waves emitted by the source when it is closest to the listener (at $O$ ) does not reach instantaneously. The frequency heard at this instant should be emitted from some previous position of source. Let this sound (or disturbance) was created at position ' $S$ '.
supporting img
The direction of velocity vector of source makes an angle $θ$ with the line joining the disturbance position (source position) and observer position as shown in the figure. The time in which sound reaches the listener along $S L$ the source reaches point $O$ , closest to the listener. From figure,
$\cos θ = \frac{S O}{S L} = \frac{v_{s} t}{v} = \frac{v_{s}}{v} = n (1)$
Now using the formula for apparent frequency received by observer
$f_{app} = f_{original} (\frac{v - v_{0}}{v - v_{s}})$
As Doppler effect is applicable only along the line joining disturbance position (source position) and observer position,hence we should substitute the component of velocity of the source along this line ' $S L$ '.
Hence the frequency heard by the listener is $\begin{array}{l} f_{a p p} = f_{o} (\frac{v - 0}{v - v_{s} \cos θ}) (2) \\ \Rightarrow f_{a p p} = f_{o} (\frac{v}{v - (n v) n}) = \frac{f_{s}}{1 - n^{2}} (3) \end{array}$
On substituting numerical values, we get $f_{app} = 5000 H z$

PHXI15:WAVES

354769 A small source of sound moves on a circle (in anti clockwise) as shown in figure and an observer is sitting at $O$ . Let $f_{A}, f_{B}, f_{C}$ be the frequencies heard when the source is at $A$ , $B$ and $C$ respectively.
supporting img

1

f_{A} > f_{B} > f_{C}

2

f_{A} = f_{B} > f_{C}

3

f_{B} > f_{C} > f_{A}

4

f_{A} > f_{C} > f_{B}

Explanation:

Frequency heard by observer (stationary) will be maximum, if the source is moving towards the observer, which is the case when the source is at $B$ . Frequency heard by observer (stationary) will be mininum, if the source is moving away from the observer, which is the case when the source is at $A$ .
supporting img

For point $C$ , frequency heard is equal to the original frequency

PHXI15:WAVES

354770 A source of sound $S$ emitting waves of frequency $100 H z$ and an observer $O$ are located at some distance from each other. The source is moving with a speed of $19.4 m s^{- 1}$ at an angle of $60^{\circ}$ with the source observer line as shown in the figure. The observer is at rest. The apparent frequency observed by the observer (velocity of sound in air $330 m s^{- 1}$ ) is
supporting img

1

106 H z

2

103 H z

3

100 H z

4

97 H z

Explanation:

The frequency received by the observer is
$f^{'} = f (\frac{v}{v - v_{s} \cos 60^{\circ}}) = 100 (\frac{330}{330 - 19.4 \cos 60^{\circ}})$
$f^{'} = 100 (\frac{330}{330 - 9.7}) = 103 H z$

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