Doppler Effect

Q: What is the Doppler effect in sound?

The Doppler effect is the change in frequency heard when a source of sound and a listener move relative to the medium and to each other. When they approach, more waves reach the listener per second and the pitch rises; when they separate, the pitch falls. A passing car's horn is the everyday example.

Q: What is the general formula for the Doppler effect?

For motion along the line joining them, f prime = f (v plus or minus vo) / (v minus or plus vs), where v is the speed of sound, vo the observer's speed and vs the source's speed. Use the upper signs when each moves towards the other and the lower signs when it moves away.

Q: Why is the Doppler effect different for a moving source and a moving observer?

A moving source changes the wavelength of the sound in the air, squeezing it ahead and stretching it behind. A moving observer leaves the wavelength unchanged but meets the waves faster or slower. Because sound travels relative to the air, the two cases give different frequencies for the same speed.

Q: How do you find the frequency of an echo from a moving car?

Treat the wall first as a stationary observer receiving f v / (v minus u), then as a stationary source of that frequency heard by the approaching driver. The result is f (v + u) / (v minus u), about twice the one-way shift. The image-source method gives the same answer.

Q: Does wind cause a Doppler effect?

If both source and observer are at rest, wind does not change the frequency heard; it only changes the speed and wavelength of the sound. When they move, wind blowing from source to observer is included by replacing the speed of sound v with v plus the wind speed.

Q: What happens when a source moves faster than sound?

A supersonic source outruns its own waves. The wavefronts pile up on a cone whose half angle alpha satisfies sine alpha equals v over the source speed, forming a shock wave heard as a sonic boom. The ordinary Doppler formula does not apply in this case.

Q: How is the Doppler effect tested in JEE Main?

JEE Main asks for frequencies with moving sources and observers, finding a train's speed from approach and recession frequencies, echoes from walls, beats between direct and reflected sound, and velocity components at an angle. Getting the signs right from the towards or away rule is the key step.

Q: What Doppler effect questions come in NEET?

NEET questions are mostly direct: a source or observer approaching or receding, both moving, and the echo from a wall. Remember that observer speed goes in the numerator and source speed in the denominator, and that approach always raises the frequency.

PhysicsWavesFor NEET aspirants

The Doppler effect is the change in the frequency heard when a source of sound and a listener move relative to the medium: a car horn sounds higher as it approaches and lower as it moves away. For motion along the line joining them, $f^{'} = f \frac{v \pm v _{o}}{v \mp v _{s}}$ , with the upper signs for approach. This page derives the Doppler effect for a moving observer, a moving source and both, then covers reflected sound, accelerated motion, motion at an angle and the supersonic limit. A scoring topic in JEE Main, JEE Advanced and NEET.

On this page1Idea2Moving observer3Moving source4General formula5Reflected sound6Accelerated motion7At an angle8Supersonic

Key Formulas - Quick Reference

★ Must learnGeneral: $f^{'} = f \frac{v \pm v _{o}}{v \mp v _{s}}$ ; upper signs when moving towards the other
Signed form ( $S \to O$ positive): $f^{'} = f \frac{v - v _{o}}{v - v _{s}}$
Moving observer: $f^{'} = f \frac{v \pm v _{o}}{v}$ ( $λ$ unchanged)
★ Must learnMoving source: $λ^{'} = \frac{v \mp v _{s}}{f}$ , $f^{'} = f \frac{v}{v \mp v _{s}}$
★ Must learnReflection from a fixed wall, car at $v_{c}$ : $f^{'} = f \frac{v + v _{c}}{v - v _{c}}$
Wind $w$ along $S \to O$ : replace $v$ by $v + w$
At an angle: use components along the line joining, $f^{'} = f \frac{v + v _{o} cos θ _{o}}{v - v _{s} cos θ _{s}}$
Accelerated motion: $v_{s}$ at the instant of emission, $v_{o}$ at the instant of reception
Supersonic ( $v_{s} > v$ ): shock cone, $sin α = \frac{v}{v _{s}}$ ; the formula does not apply

1. What Is the Doppler Effect?

Stand by a road while a car at rest sounds its horn: you hear its true frequency. When the car approaches with the horn sounding, the pitch is higher, and it drops as the car passes and moves away. This was first explained by the Austrian physicist Christian Doppler (1842).

★ Must learnDoppler effect: when a source of sound and a listener are in motion relative to each other (and to the medium), the frequency heard by the listener differs from the frequency of the source. Approach raises the frequency; separation lowers it.

Uses: speed guns and weather radar, ultrasound scans of blood flow (echocardiography), SONAR tracking of submarines, and, with light, measuring how fast stars and galaxies move towards or away from us.

Figure 1: A moving source. Each front is centred where the siren was when it was emitted, so the fronts crowd ahead and spread behind. Drawn for

v_{s} = 0.4 v

λ^{'} = 0.6 λ

ahead and

1.4 λ

behind. A real ambulance (

v_{s} = 20 m s^{- 1}

v = 340 m s^{- 1}

) shifts a

700 Hz

siren to about

744 Hz

ahead and

661 Hz

behind.

Figure 2: Each wavefront is centred where the source was when it emitted it (grey dots). A moving source squeezes the waves ahead (

λ^{'} = (v - v_{s}) / f

, higher pitch at

O_{1}

) and stretches them behind (lower pitch at

O_{2}

). The speed of sound itself is unchanged.

1.1 Source and observer both at rest

A source of frequency $f$ at rest sends waves of wavelength $λ_{0} = \frac{v}{f}$ through still air at speed $v$ . A stationary observer receives them at speed $v$ , so the frequency heard is $\frac{v}{λ _{0}} = f$ . With no relative motion there is no Doppler effect. (Sound is longitudinal, but the figures draw it as crests to make the spacing visible.)

2. Stationary Source, Moving Observer

Figure 3: A moving observer does not change

λ

; it changes how fast the fronts are met. Approaching:

f^{'} = f \frac{v + v _{o}}{v}

; receding:

f^{'} = f \frac{v - v _{o}}{v}

The source is at rest, so the wavelength in air is unchanged: $λ = \frac{v}{f}$ .
An observer moving towards the source at $v_{o}$ meets the waves at relative speed $v + v_{o}$ .
Frequency heard $=$ fronts met per second:
$f^{'} = \frac{v + v _{o}}{λ} = f \frac{v + v _{o}}{v}$
Moving away from the source: $f^{'} = f \frac{v - v _{o}}{v}$ .

3. Moving Source, Stationary Observer

Now the source moves towards the observer at $v_{s}$ while emitting waves (Figure 2).

In one period $\frac{1}{f}$ the source emits one wave, whose front travels $\frac{v}{f}$ . In that time the source itself moves $\frac{v _{s}}{f}$ in the same direction.
So the waves ahead are squeezed. Apparent wavelength:
$λ^{'} = \frac{v}{f} - \frac{v _{s}}{f} = \frac{v - v _{s}}{f}$
These waves still travel at $v$ to the observer at rest:
$f^{'} = \frac{v}{λ ^{'}} = f \frac{v}{v - v _{s}}$
Source moving away: the waves behind it are stretched, $λ^{''} = \frac{v + v _{s}}{f}$ , and $f^{'} = f \frac{v}{v + v _{s}}$ .

Moving observer

Wavelength in air unchanged. Only the speed at which waves are met changes: $f^{'} = f \frac{v \pm v _{o}}{v}$ . The change is linear in $v_{o}$ .

Moving source

Wavelength in air itself changes. Waves arrive at the usual speed $v$ : $f^{'} = f \frac{v}{v \mp v _{s}}$ . The change is not linear; it blows up as $v_{s} \to v$ .

So a source approaching at speed $u$ and an observer approaching at the same speed $u$ do not give the same frequency: motion relative to the medium matters, not just relative motion.

4. Source and Observer Both Moving: the General Formula

Combine both effects: the source sets the wavelength, the observer sets the speed at which it is met. For a source and an observer moving along the line joining them:

★ Must learn

f^{'} = f \frac{v \pm v _{o}}{v \mp v _{s}}

Upper signs (

+ v_{o}

in the numerator,

- v_{s}

in the denominator) when each moves towards the other; lower signs when moving away.

Figure 4: Take

S \to O

as positive for both velocities:

f^{'} = f \frac{v - v _{o}}{v - v _{s}}

. Check: approaching makes the fraction larger, receding smaller.

Situation	Frequency heard
Source towards stationary observer	$f \frac{v}{v - v _{s}}$ (higher)
Source away from stationary observer	$f \frac{v}{v + v _{s}}$ (lower)
Observer towards stationary source	$f \frac{v + v _{o}}{v}$ (higher)
Observer away from stationary source	$f \frac{v - v _{o}}{v}$ (lower)
Both approaching	$f \frac{v + v _{o}}{v - v _{s}}$
Both moving the same way, source behind	$f \frac{v - v _{o}}{v - v _{s}}$

Exam Trick

Signs by common sense. Write $f^{'} = f \frac{v □ v _{o}}{v □ v _{s}}$ and fill each box so that motion towards the other raises $f^{'}$ : a plus in the numerator, a minus in the denominator. Motion away does the opposite. No need to memorise a sign table. Wind of speed $w$ blowing from source towards observer simply changes $v$ to $v + w$ in both places; if both are at rest, wind causes no change.

Key idea

Source motion changes the wavelength in air; observer motion changes how fast the fronts are met. Both are measured relative to the medium.

Figure 5: Doppler problems in five checks: reflector (two steps), components along the line, wind, signs by common sense, and a final sanity check.

5. Doppler Effect in Reflected Sound

A car moving at $v_{c}$ towards a stationary wall sounds a horn of frequency $f$ . The driver hears the echo at a higher pitch. Solve in two steps.

Figure 6: Two steps (wall as observer, then as source), or one step with the image car behind the wall:

f^{'} = f \frac{v + v _{c}}{v - v _{c}}

Wall as a stationary observer, car as a moving source: $f_{1} = f \frac{v}{v - v _{c}}$ .
Wall as a stationary source of $f_{1}$ , car (driver) as an observer approaching at $v_{c}$ :
$f^{'} = f_{1} \frac{v + v _{c}}{v} = f \frac{v + v _{c}}{v - v _{c}}$

Image method. The echo behaves as if it came from an image of the car behind the wall, approaching at the same speed $v_{c}$ ; the moving-source, moving-observer formula then gives the same result directly. For a stationary reflector this is exact. For a moving reflector, use the two steps: the reflector is first an observer, then a source, with its own velocity each time.

Figure 7: A moving reflector: two steps. The moth first receives as an observer,

f^{'} = f \frac{v + u _{m}}{v - u _{b}}

, then re-emits as a source; the bat hears

f^{''} = f^{'} \frac{v + u _{b}}{v - u _{m}} \approx 43.7 kHz

for

f = 40 kHz

u_{b} = 10

u_{m} = 5 m s^{- 1}

v = 340 m s^{- 1}

(Solved Example 13).

Key idea

Echo from a stationary wall, approaching at

u

f^{'} = f \frac{v + u}{v - u}

. For

u ≪ v

this is about

f (1 + \frac{2 u}{v})

: twice the one-way shift.

6. Accelerated Motion and Motion at an Angle

6.1 Accelerated source or observer

The general formula still holds if we use the right instants: $v_{s}$ is the velocity of the source at the moment it emitted the sound, and $v_{o}$ is the velocity of the observer at the moment it receives it. Alternatively, work out the compressed or stretched wavelength from the source's motion and the speed of sound relative to the observer.

6.2 Source and observer not on the same line

If the velocities are not along the line joining source and observer, only their components along that line matter. Two cars on perpendicular roads, for example:

Figure 8: Only velocity components along the line joining source and observer matter:

f^{'} = f \frac{v + v _{2} cos θ _{2}}{v - v _{1} cos θ _{1}}

(both components pointing towards the other). Perpendicular motion gives no shift.

f^{'} = f \frac{v + v _{2} cos θ _{2}}{v - v _{1} cos θ _{1}}

where $θ_{1}$ and $θ_{2}$ are the angles between each velocity and the line joining them (at the instant of emission). A velocity perpendicular to the line produces no Doppler shift: a source moving in a circle around a listener at the centre is heard at its true frequency.

Figure 9: Computed for

v_{s} = 30 m s^{- 1}

v = 340 m s^{- 1}

. The pitch falls from

f \frac{v}{v - v _{s}}

f \frac{v}{v + v _{s}}

as the source goes by; the closer it passes, the more sudden the drop (the familiar 'neeee-yowww' of a passing car).

Quick Recall: tap to check

Is the pitch of a passing car's horn exactly

f

at the moment it is nearest to you?

Not exactly. The sound you hear then left the car a moment earlier, while it was still approaching, so it is slightly above

f

. Sound emitted at closest approach (velocity perpendicular to the line joining you) arrives at exactly

f

A source approaches at

v /2

. What is

f^{'}

f \frac{v}{v /2} = 2 f

An observer approaches a source at

v /2

. What is

f^{'}

f \frac{v + v /2}{v} = 1.5 f

Exam Trick

Small speeds: add them up. When $v_{s}, v_{o} ≪ v$ , $\frac{Δ f}{f} \approx \frac{v _{o} + v _{s}}{v}$ for approach (the same size, negative, for separation). An echo from a wall approached at $u$ doubles it: $\frac{Δ f}{f} \approx \frac{2 u}{v}$ . A $500 Hz$ horn at $u = 17 m s^{- 1}$ : $Δ f \approx 500 \times \frac{34}{340} = 50 Hz$ (exact: $52.6 Hz$ ).

7. Limits: Supersonic Sources

Figure 10: For

v_{s} > v

the source outruns its waves; the fronts pile up on a cone with

sin α = v / v_{s}

(Mach number

v_{s} / v

). The Doppler formula no longer applies; a sonic boom is heard as the cone sweeps past.

JEE Advanced

As $v_{s} \to v$ the waves in front of the source pile up and $f^{'} = f v / (v - v_{s})$ grows without limit. If $v_{s} > v$ (a supersonic aircraft), the source outruns its own waves, which are enclosed by a cone of half angle $α$ with $sin α = \frac{v}{v _{s}}$ . The ratio $v_{s} / v$ is the Mach number. The concentrated pressure along the cone is a shock wave, heard on the ground as a sonic boom. The Doppler formula is valid only for $v_{s} < v$ and $v_{o} < v$ .

Quick Recall: tap to check

A car approaches a wall at speed

u

and sounds its horn. Is the echo higher or lower for the driver?

Higher:

f \frac{v + u}{v - u}

Does wind change the frequency heard if source and observer are both at rest?

No:

v

changes equally in the numerator and the denominator.

What is the half angle of the shock cone of a jet flying at Mach 2?

sin α = \frac{1}{2}

, so

α = 3 0^{\circ}

Figure 11: Revision map: observer and source motion, the general formula, reflection, angles and the supersonic limit.

8. Solved Examples

Solved Example 1

A train approaches a platform at

30 m s^{- 1}

, sounding a

500 Hz

whistle. What frequency does a person on the platform hear before and after the train passes? (

v = 330 m s^{- 1}

)

Solution:

Approaching: $f^{'} = 500 \times \frac{330}{330 - 30} = 550 Hz$ . Receding: $f^{'} = 500 \times \frac{330}{330 + 30} \approx 458 Hz$ .

Answer: $550 Hz$ then about $458 Hz$ .

Solved Example 2

A person moves at

30 m s^{- 1}

towards, and then away from, a stationary

500 Hz

siren. Find the frequencies heard (

v = 330 m s^{- 1}

) and compare with Solved Example 1.

Solution:

Towards: $500 \times \frac{330 + 30}{330} \approx 545 Hz$ . Away: $500 \times \frac{330 - 30}{330} \approx 455 Hz$ .

Answer: $545 Hz$ and $455 Hz$ , different from $550$ and $458 Hz$ : moving the observer is not the same as moving the source.

Solved Example 3

Car A sounds a

400 Hz

horn while moving at

20 m s^{- 1}

towards car B, which moves at

10 m s^{- 1}

towards A. What frequency does the driver of B hear? (

v = 340 m s^{- 1}

)

Solution:

Both approach: $f^{'} = 400 \times \frac{340 + 10}{340 - 20} = 400 \times \frac{350}{320} = 437.5 Hz$ .

Answer: $437.5 Hz$ .

Solved Example 4

The whistle of a train is heard as

560 Hz

while it approaches a platform and

480 Hz

after it passes. Find the speed of the train and the true frequency (

v = 340 m s^{- 1}

Solution:

$\frac{f _{1}}{f _{2}} = \frac{v + v _{s}}{v - v _{s}} = \frac{560}{480} = \frac{7}{6}$ , so $6 (v + v_{s}) = 7 (v - v_{s})$ and $v_{s} = \frac{v}{13} \approx 26.2 m s^{- 1}$ .

$f = 560 \times \frac{v - v _{s}}{v} = 560 \times \frac{12}{13} \approx 517 Hz$ .

Answer: $\approx 26 m s^{- 1}$ ( $94 km h^{- 1}$ ); $f \approx 517 Hz$ .

Solved Example 5

A bat flies at

10 m s^{- 1}

towards a wall, emitting

40 kHz

ultrasound. What frequency does it hear in the echo? (

v = 340 m s^{- 1}

)

Solution:

$f^{'} = f \frac{v + u}{v - u} = 40 \times \frac{350}{330} \approx 42.4 kHz$ .

Answer: about $42.4 kHz$ .

Solved Example 6

A person runs at

2 m s^{- 1}

towards a tall wall holding a vibrating

512 Hz

fork. How many beats per second does he hear between the direct sound and the echo? (

v = 340 m s^{- 1}

)

Solution:

Direct sound: no relative motion between fork and ear, so $512 Hz$ . Echo: $512 \times \frac{342}{338} \approx 518.1 Hz$ .

Answer: about $6$ beats per second.

Solved Example 7

500 Hz

fork is carried at

2 m s^{- 1}

away from a stationary listener, towards a wall. How many beats does the listener hear per second? (

v = 340 m s^{- 1}

)

Solution:

Direct (source receding): $500 \times \frac{340}{342} \approx 497.1 Hz$ . The wall receives $500 \times \frac{340}{338} \approx 503.0 Hz$ and reflects it unchanged (stationary wall and listener).

Answer: $503.0 - 497.1 \approx 6$ beats per second.

Solved Example 8

A source approaching a stationary observer at

v /2

is heard at frequency

f_{1}

; an observer approaching a stationary source at

v /2

hears

f_{2}

. The ratio

f_{1} : f_{2}

is
(A)

1 : 1

(B)

4 : 3

(C)

3 : 4

(D)

2 : 1

Solution:

$f_{1} = f \frac{v}{v - v /2} = 2 f$ ; $f_{2} = f \frac{v + v /2}{v} = 1.5 f$ . $f_{1} : f_{2} = 4 : 3$ .

Answer: (B).

Solved Example 9

A car moves at

30 m s^{- 1}

along a straight road, sounding a

500 Hz

horn. At the moment of emission, the line from the car to a stationary listener makes

6 0^{\circ}

with the car's velocity. Find the frequency heard (

v = 340 m s^{- 1}

Solution:

Component of the source velocity towards the listener: $30 cos 6 0^{\circ} = 15 m s^{- 1}$ .

$f^{'} = 500 \times \frac{340}{340 - 15} \approx 523 Hz$ .

Answer: about $523 Hz$ .

Solved Example 10

A source moves in a circle at constant speed; a listener sits at the centre. The frequency heard is
(A) higher
(B) lower
(C) the same as emitted
(D) alternately higher and lower

Solution:

The velocity is always perpendicular to the radius, the line joining source and listener, so its component along that line is zero.

Answer: (C).

Solved Example 11

A car accelerates towards a stationary listener. It gives a short

400 Hz

honk when its speed is

20 m s^{- 1}

; when the honk reaches the listener the car is at

30 m s^{- 1}

. What frequency is heard? (

v = 340 m s^{- 1}

)

Solution:

Use the source speed at emission: $f^{'} = 400 \times \frac{340}{340 - 20} = 425 Hz$ .

Answer: $425 Hz$ (the later speed does not matter for this sound).

Solved Example 12

A steady wind blows at

10 m s^{- 1}

from a stationary

600 Hz

siren towards a stationary listener. What frequency is heard?

Solution:

With the wind, $v \to v + w$ in both numerator and denominator: $f^{'} = 600 \times \frac{v + w}{v + w} = 600 Hz$ .

Answer: $600 Hz$ . Wind changes the wave speed and wavelength, not the frequency, when neither source nor listener moves.

Solved Example 13

A bat flying at

10 m s^{- 1}

emits

40 kHz

ultrasound towards a moth that flies towards it at

5 m s^{- 1}

. What frequency does the bat hear in the echo? (

v = 340 m s^{- 1}

)

Solution:

Step 1 (moth as observer): $f^{'} = 40 \times \frac{340 + 5}{340 - 10} = 40 \times \frac{345}{330} \approx 41.82 kHz$ .

Step 2 (moth as source, bat as observer): $f^{''} = f^{'} \times \frac{340 + 10}{340 - 5} = 41.82 \times \frac{350}{335} \approx 43.69 kHz$ .

Answer: about $43.7 kHz$ (Figure 7). The image method does not apply to a moving reflector: use the two steps.

Practice Questions

A $600 Hz$ source approaches a stationary listener at $40 m s^{- 1}$ ( $v = 340 m s^{- 1}$ ). Find $f^{'}$ .Answer: $680 Hz$ .
A listener approaches a stationary $600 Hz$ source at $40 m s^{- 1}$ . Find $f^{'}$ .Answer: $\approx 671 Hz$ .
A police car at $20 m s^{- 1}$ chases a car at the same speed ahead of it, siren $1000 Hz$ . What does the driver ahead hear?Answer: $1000 Hz$ (both move the same way at the same speed).
A car at $20 m s^{- 1}$ approaches a cliff sounding $1000 Hz$ . Frequency of the echo heard by the driver? ( $v = 340 m s^{- 1}$ )Answer: $1000 \times 360/320 = 1125 Hz$ .
A source of $1200 Hz$ recedes at $40 m s^{- 1}$ from a stationary listener. Find $f^{'}$ and the wavelength heard ( $v = 340 m s^{- 1}$ ).Answer: $\approx 1074 Hz$ ; $λ^{'} = 380/1200 \approx 0.32 m$ .
A jet flies at Mach 2. Find the half angle of its shock cone.Answer: $sin α = 1/2$ : $α = 3 0^{\circ}$ .

Common Mistakes to Avoid

Watch out

Using relative speed only. A moving source and a moving observer give different results; motion relative to the air matters.
Putting $v_{o}$ in the denominator or $v_{s}$ in the numerator. Observer terms go on top, source terms at the bottom.
Getting signs backwards. Towards each other must raise the frequency: $+ v_{o}$ on top, $- v_{s}$ below.
Thinking the wavelength changes for a moving observer. Only a moving source changes $λ$ in the medium.
Using the full velocity when it is at an angle. Take the component along the line joining source and observer.
Applying the one-way formula to an echo. Reflection needs two steps (or the image source): $f (v + u) / (v - u)$ .
Assuming wind changes the frequency when source and observer are at rest. It does not.
Using the formula for supersonic sources. It holds only for $v_{s} < v$ .

Frequently Asked Questions

What is the Doppler effect in sound?

The Doppler effect is the change in frequency heard when a source of sound and a listener move relative to the medium and to each other. When they approach, more waves reach the listener per second and the pitch rises; when they separate, the pitch falls. A passing car's horn is the everyday example.

What is the general formula for the Doppler effect?

For motion along the line joining them, $f^{'} = f (v \pm v_{o}) / (v \mp v_{s})$ , where $v$ is the speed of sound, $v_{o}$ the observer's speed and $v_{s}$ the source's speed. Use the upper signs when each moves towards the other and the lower signs when it moves away.

Why is the Doppler effect different for a moving source and a moving observer?

A moving source changes the wavelength of the sound in the air, squeezing it ahead and stretching it behind. A moving observer leaves the wavelength unchanged but meets the waves faster or slower. Because sound travels relative to the air, the two cases give different frequencies for the same speed.

How do you find the frequency of an echo from a moving car?

Treat the wall first as a stationary observer receiving $f v / (v - u)$ , then as a stationary source of that frequency heard by the approaching driver. The result is $f (v + u) / (v - u)$ , about twice the one-way shift. The image-source method gives the same answer.

Does wind cause a Doppler effect?

If both source and observer are at rest, wind does not change the frequency heard; it only changes the speed and wavelength of the sound. When they move, wind blowing from source to observer is included by replacing the speed of sound v with v plus the wind speed.

What happens when a source moves faster than sound?

A supersonic source outruns its own waves. The wavefronts pile up on a cone whose half angle alpha satisfies sine alpha equals v over the source speed, forming a shock wave heard as a sonic boom. The ordinary Doppler formula does not apply in this case.

How is the Doppler effect tested in JEE Main?

JEE Main asks for frequencies with moving sources and observers, finding a train's speed from approach and recession frequencies, echoes from walls, beats between direct and reflected sound, and velocity components at an angle. Getting the signs right from the towards or away rule is the key step.

What Doppler effect questions come in NEET?

NEET questions are mostly direct: a source or observer approaching or receding, both moving, and the echo from a wall. Remember that observer speed goes in the numerator and source speed in the denominator, and that approach always raises the frequency.

Previous year questions on Doppler Effect

3 questions from past papers, each with a step-by-step solution.

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