1. Physics
  2. Oscillations and Waves
  3. Doppler Effect

Oscillations and Waves · JEE & NEET Physics

Doppler Effect: notes and previous year questions

Moving sources and listeners, the general formula and its signs, echoes and radar, light, angles and shock waves.

Doppler Effect in short

  • The received frequency changes when the source or the listener moves; the source's own frequency does not.
  • A moving source squeezes the waves ahead of it and stretches them behind.
  • A moving listener meets unchanged waves faster or slower.
  • General formula: f′ = f(v ± v_o)/(v ∓ v_s), with signs chosen so that approaching raises the pitch.

1The changing pitch

A moving ambulance sends out each crest from where it is at that moment. Ahead of it the crests are bunched: a shorter wavelength, more crests per second, a higher pitch. Behind it they are spread out: a lower pitch. The siren itself never changes.

The Doppler effect is the change in the frequency you receive when the source, the listener, or both move. It happens with all waves: sound, water waves and light.

2A moving source

The source sends a crest every period T=1/fT = 1/f. In that time a crest moves vTvT (the speed of sound in the air), but the source moves vsTv_s T after it. So the gap ahead is only (v−vs)T(v - v_s)T. The listener still receives the crests at speed vv, so f′=v/λ′f' = v/\lambda':

λ′=v−vsf,f′=f vv−vs\lambda' = \frac{v - v_s}{f}, \quad f' = f\,\frac{v}{v - v_s}Source approaching a still listener: higher.
f′′=f vv+vsf'' = f\,\frac{v}{v + v_s}Source moving away: the waves behind are stretched to (v + v_s)/f; lower.

3A moving observer

With the source still, the crests keep their normal spacing λ=v/f\lambda = v/f. A listener moving toward the source at vov_o meets them at v+vov + v_o:

f′=f v+vovf' = f\,\frac{v + v_o}{v}Toward the source; moving away, use v − v_o.
  • 500 Hz, walking toward it at 17 m/s: 500×357/340=525500 \times 357/340 = 525 Hz. For 550 Hz you need vo=34v_o = 34 m/s.
  • Moving source and moving listener are not the same: at 34 m/s toward a 500 Hz sound, a moving listener hears 550 Hz but a moving source gives 500×340/306≈555.6500 \times 340/306 \approx 555.6 Hz. A moving source changes the wavelength in the air itself; a moving listener only meets the same waves faster.

4The general formula

f′=f v±vov∓vsf' = f\,\frac{v \pm v_o}{v \mp v_s}All speeds measured relative to the air.
  • Listener moving toward the source: +vo+v_o on top; away: −vo-v_o.
  • Source moving toward the listener: −vs-v_s below; away: +vs+v_s.
  • Check: coming together must raise the pitch, moving apart must lower it.

5Echoes and radar

A bat flying at vbv_b toward a wall: first the wall is a still listener receiving f1=f v/(v−vb)f_1 = f\,v/(v - v_b); then the wall is a still source and the bat a listener moving toward it: f2=f1(v+vb)/vf_2 = f_1(v + v_b)/v.

f′′=f v+vrv−vrf'' = f\,\frac{v + v_r}{v - v_r}Echo from a reflector approaching at v_r (or a source approaching a still reflector).
Δff≈2vrv\frac{\Delta f}{f} \approx \frac{2v_r}{v}For small speeds; the 2 is for the trip there and back.
  • A bat at 10 m/s sending 40 kHz hears 40×350/330≈42.440 \times 350/330 \approx 42.4 kHz.
  • Radar and ultrasound use Δf=2fv/c\Delta f = 2fv/c: 10 GHz radar and a 2 kHz shift give v=Δf c/2f=30v = \Delta f\,c/2f = 30 m/s.
  • Uses: police radar, weather radar (rain and wind), ultrasound scans of blood flow and baby heartbeats, bat echolocation.

6Doppler effect for light

Light needs no medium, so only the relative speed v of source and observer matters (v positive when approaching), and the sound formula does not apply.

Δff=vc\frac{\Delta f}{f} = \frac{v}{c}For v much less than c.
f′=fc+vc−vf' = f\sqrt{\frac{c + v}{c - v}}Any speed (special relativity).
z=λ′−λλ≈vcz = \frac{\lambda' - \lambda}{\lambda} \approx \frac{v}{c}Redshift: receding, λ′ > λ. Blueshift: approaching, λ′ < λ.
  • A 600 nm line seen at 660 nm: z=0.1z = 0.1, v≈3×107v \approx 3 \times 10^7 m/s, moving away.
  • A 656 nm line seen at 655.3 nm: blueshift, approaching at about c×0.7/656≈320c \times 0.7/656 \approx 320 km/s.
  • Distant galaxies are all redshifted, the farther the more (Hubble's law, v = H₀d): the universe is expanding.
  • Astronomers also use it for star speeds, planets that make their stars wobble, and rotating galaxies.

7Angles and shock waves

f′=f v+vocos⁡αv−vscos⁡βf' = f\,\frac{v + v_o\cos\alpha}{v - v_s\cos\beta}α, β: angles between each velocity and the line joining source and observer.

Only motion along the line of sight changes the pitch. A passing ambulance sounds highest far away while coming, exactly its true note at the closest point (cos⁡β=0\cos\beta = 0), then lower and lower: the pitch slides down smoothly.

A source faster than sound (vs>vv_s > v) outruns its crests and the formula breaks down. The crests pile up on a cone, a shock wave, heard as a sonic boom. In the time the source moves vstv_s t a crest spreads vtvt:

M=vsv,sin⁡θ=vvs=1MM = \frac{v_s}{v}, \quad \sin\theta = \frac{v}{v_s} = \frac{1}{M}Mach number M; θ is the half-angle of the cone. M = 2 gives 30°.

Summary

Key ideas

  • The received frequency changes when the source or the listener moves; the source's own frequency does not.
  • A moving source squeezes the waves ahead of it and stretches them behind.
  • A moving listener meets unchanged waves faster or slower.
  • General formula: f′ = f(v ± v_o)/(v ∓ v_s), with signs chosen so that approaching raises the pitch.
  • The same speed gives different shifts for a moving source and a moving listener.
  • No relative motion (both moving the same way at the same speed) means no shift.
  • Echoes from moving reflectors are shifted twice: Δf/f ≈ 2v/v_sound, or 2v/c for radar.
  • For light only the relative speed matters: Δf/f = v/c; receding gives a redshift, approaching a blueshift.
  • At an angle, only the velocity component along the line of sight counts.
  • Faster than sound, crests form a shock cone with sin θ = 1/M.

Every equation

Source approaching
f′=f v/(v−vs)f' = f\,v/(v - v_s)
Source receding
f′=f v/(v+vs)f' = f\,v/(v + v_s)
Wavelength ahead
λ′=(v−vs)/f\lambda' = (v - v_s)/f
Listener approaching
f′=f (v+vo)/vf' = f\,(v + v_o)/v
Listener receding
f′=f (v−vo)/vf' = f\,(v - v_o)/v
General
f′=f (v±vo)/(v∓vs)f' = f\,(v \pm v_o)/(v \mp v_s)
Moving reflector
f′′=f (v+vr)/(v−vr)f'' = f\,(v + v_r)/(v - v_r)
Small reflector speed
Δf/f≈2vr/v\Delta f/f \approx 2v_r/v
Radar
Δf=2fv/c\Delta f = 2fv/c
Light, slow
Δf/f=v/c\Delta f/f = v/c
Light, any speed
f′=f(c+v)/(c−v)f' = f\sqrt{(c + v)/(c - v)}
Redshift
z=Δλ/λ≈v/cz = \Delta\lambda/\lambda \approx v/c
Hubble's law
v=H0dv = H_0 d
At an angle
f′=f v+vocos⁡αv−vscos⁡βf' = f\,\tfrac{v + v_o\cos\alpha}{v - v_s\cos\beta}
Mach number
M=vs/vM = v_s/v
Shock cone
sin⁡θ=1/M\sin\theta = 1/M

Previous year questions with solutions

Real JEE and NEET questions on doppler effect. Try each one before you open the solution.

Q1JEE Main 2026Numerical answer

Two tuning forks A and B are sounded together giving rise to 8 beats in 2 s. When fork A is loaded with wax, the beat frequency is reduced to 4 beats in 2 s. If the original frequency of tuning fork B is 380 Hz then original frequency of tuning fork A is _________ Hz.

Show answer and solution

Answer: 384

88 beats in 2 s2\ \mathrm{s} is 44 beats per second, so A is 380+4=384 Hz380 + 4 = 384\ \mathrm{Hz} or 380−4=376 Hz380 - 4 = 376\ \mathrm{Hz}.

Wax lowers A's frequency, and the beats fall to 44 in 2 s2\ \mathrm{s}, that is 22 per second. A must have moved towards 380380: from 384 Hz384\ \mathrm{Hz} down to 382 Hz382\ \mathrm{Hz}. From 376 Hz376\ \mathrm{Hz} it would have moved further away and the beats would have grown.

So A was 384 Hz384\ \mathrm{Hz}.

The trap is 376 Hz376\ \mathrm{Hz}: it assumes the wax fork must be the lower one. Another slip is reading 88 beats in 2 s2\ \mathrm{s} as 88 per second, which gives 388388 or 372 Hz372\ \mathrm{Hz}.

Q2JEE Advanced 2021One or more correct options

A source, approaching with speed uu towards the open end of a stationary pipe of length LL, is emitting a sound of frequency fsf_{s}. The farther end of the pipe is closed. The speed of sound in air is vv and f0f_{0} is the fundamental frequency of the pipe. For which of the following combination(s) of uu and fsf_{s}, will the sound reaching the pipe lead to a resonance?

  1. Au=0.8vu = 0.8v and fs=f0f_{s} = f_{0}
  2. Bu=0.8vu = 0.8v and fs=2f0f_{s} = 2f_{0}
  3. Cu=0.8vu = 0.8v and fs=0.5f0f_{s} = 0.5f_{0}
  4. Du=0.5vu = 0.5v and fs=1.5f0f_{s} = 1.5f_{0}
Show answer and solution

Answer: Options A, D

A pipe closed at one end resonates only at odd multiples of its fundamental: f0f_{0}, 3f03f_{0}, 5f05f_{0}, ... . The pipe is still and the source approaches it, so the pipe receives

f′=fs vv−uf' = f_{s}\,\dfrac{v}{v - u}

A: vv−0.8v=5\dfrac{v}{v - 0.8v} = 5, so f′=5f0f' = 5f_{0}, an odd multiple: resonance.

B: f′=5×2f0=10f0f' = 5 \times 2f_{0} = 10f_{0}, an even multiple: no resonance.

C: f′=5×0.5f0=2.5f0f' = 5 \times 0.5f_{0} = 2.5f_{0}: not a multiple at all.

D: vv−0.5v=2\dfrac{v}{v - 0.5v} = 2, so f′=2×1.5f0=3f0f' = 2 \times 1.5f_{0} = 3f_{0}: resonance.

So A and D.

The trap is B, which would work for an open pipe (all harmonics) but not for a closed one. Another is to test fsf_{s} itself, forgetting the Doppler shift: then only A's f0f_{0} would seem to fit.

Q3NEET 2012One correct option

Two sources of sound placed close to each other, are emitting progressive waves given by

y1=4sin⁡600πty_{1} = 4\sin 600\pi t and y2=5sin⁡608πty_{2} = 5\sin 608\pi t

An observer located near these two sources of sound will hear

  1. A4 beats per second with intensity ratio 25 : 16 between waxing and waning.
  2. B8 beats per second with intensity ratio 25 : 16 between waxing and waning.
  3. C8 beats per second with intensity ratio 81 : 1 between waxing and waning.
  4. D4 beats per second with intensity ratio 81 : 1 between waxing and waning.
Show answer and solution

Answer: Option D

Beats: f1=600π2π=300 Hzf_{1} = \dfrac{600\pi}{2\pi} = 300\ \mathrm{Hz} and f2=608π2π=304 Hzf_{2} = \dfrac{608\pi}{2\pi} = 304\ \mathrm{Hz}, so fb=4f_{b} = 4 beats per second.

Intensity: the amplitudes are 44 and 55. At the loud moment (waxing) the amplitude is 5+4=95 + 4 = 9; at the soft moment (waning) it is 5−4=15 - 4 = 1. Intensity goes as amplitude squared, so

ImaxImin=9212=81:1\dfrac{I_{max}}{I_{min}} = \dfrac{9^{2}}{1^{2}} = 81 : 1

That is D.

The trap is A, 25:1625 : 16, which squares the two amplitudes separately: that compares the two sources, not the loud and soft moments. B and C take 608π−600π608\pi - 600\pi and call it 88 beats per second, forgetting to divide by 2π2\pi.

Practice questions, easy to hard

Three questions from the doppler effect practice ladder: one easy, one medium, one hard.

Q4Numerical answer

Often the two sounds are given as equations. A wave y=asin⁡(ωt)y = a\sin(\omega t) has angular frequency ω\omega in rad/s, and its frequency is f=ω2πf = \dfrac{\omega}{2\pi} in Hz. So read each ω\omega, divide by 2π2\pi, and subtract.

Two sources close together give, at a listener, y1=asin⁡(2000πt)y_{1} = a\sin(2000\pi t) and y2=asin⁡(2008πt)y_{2} = a\sin(2008\pi t), with tt in seconds. How many beats does the listener hear in 5 s5\ \mathrm{s}?

Show answer and solution

Answer: 20

f1=2000π2π=1000 Hzf_{1} = \dfrac{2000\pi}{2\pi} = 1000\ \mathrm{Hz} and f2=2008π2π=1004 Hzf_{2} = \dfrac{2008\pi}{2\pi} = 1004\ \mathrm{Hz}. So fb=4f_{b} = 4 beats per second, and in 5 s5\ \mathrm{s} there are 4×5=204 \times 5 = 20 beats.

The trap is subtracting the angular frequencies, 2008π−2000π=8π2008\pi - 2000\pi = 8\pi, and calling that 88 (or 8π≈258\pi \approx 25) beats per second: the beat frequency is a difference of ff, not of ω\omega. Another slip is to stop at 44, which is the number in one second, not in five.

Q5One correct option

A train sounding a whistle of frequency ff passes a still listener at speed vsv_{s}. Coming towards him he hears f vv−vsf\,\dfrac{v}{v - v_{s}}; going away, f vv+vsf\,\dfrac{v}{v + v_{s}}. The pitch drops suddenly as the train passes. Dividing the two, ff and the top vv cancel:

fapproachfrecede=v+vsv−vs\dfrac{f_{approach}}{f_{recede}} = \dfrac{v + v_{s}}{v - v_{s}}

A listener hears a train's whistle at 360 Hz360\ \mathrm{Hz} as it approaches and at 300 Hz300\ \mathrm{Hz} after it has passed. The speed of sound is 330 m/s330\ \mathrm{m/s}. How fast is the train moving?

  1. A66 m/s66\ \mathrm{m/s}
  2. B30 m/s30\ \mathrm{m/s}
  3. C55 m/s55\ \mathrm{m/s}
  4. D15 m/s15\ \mathrm{m/s}
Show answer and solution

Answer: Option B

360300=1.2=330+vs330−vs\dfrac{360}{300} = 1.2 = \dfrac{330 + v_{s}}{330 - v_{s}}, so 1.2(330−vs)=330+vs1.2(330 - v_{s}) = 330 + v_{s}, which gives 396−330=2.2vs396 - 330 = 2.2v_{s} and vs=30 m/sv_{s} = 30\ \mathrm{m/s}.

(Check: the whistle is 300×360330≈327 Hz300 \times \dfrac{360}{330} \approx 327\ \mathrm{Hz}; 327×330300=360327 \times \dfrac{330}{300} = 360 and 327×330360=300327 \times \dfrac{330}{360} = 300.)

The trap is A, 66 m/s66\ \mathrm{m/s}, which is 360−300300×330\dfrac{360 - 300}{300} \times 330, as if the drop in pitch compared with the lower note measured the speed directly. C, 55 m/s55\ \mathrm{m/s}, does the same with the higher note. D, 15 m/s15\ \mathrm{m/s}, halves the right answer.

Q6One correct option

A pipe resonates when the sound reaching it has one of the pipe's own frequencies. If the source is moving, the pipe (a still observer) receives the Doppler-shifted frequency, not the source's own.

An open pipe 1.0 m1.0\ \mathrm{m} long stands still; the speed of sound is 340 m/s340\ \mathrm{m/s}. A source of 300 Hz300\ \mathrm{Hz} moves straight towards its open end at speed uu. For which uu does the sound reaching the pipe make it resonate?

  1. A20 m/s20\ \mathrm{m/s}
  2. B34 m/s34\ \mathrm{m/s}
  3. C40 m/s40\ \mathrm{m/s}
  4. D60 m/s60\ \mathrm{m/s}
Show answer and solution

Answer: Option C

The open pipe resonates at nv2L=170n Hz\dfrac{nv}{2L} = 170n\ \mathrm{Hz}: 170170, 340340, 510510, ... . The pipe receives 300×340340−u300 \times \dfrac{340}{340 - u}. Setting this equal to 340 Hz340\ \mathrm{Hz} gives 340−u=300340 - u = 300, so u=40 m/su = 40\ \mathrm{m/s}.

The others miss every harmonic: 20 m/s20\ \mathrm{m/s} gives about 319 Hz319\ \mathrm{Hz}, 34 m/s34\ \mathrm{m/s} gives about 333 Hz333\ \mathrm{Hz} and 60 m/s60\ \mathrm{m/s} gives about 364 Hz364\ \mathrm{Hz}. None of these is a multiple of 170170. The trap is to aim for the fundamental, 170 Hz170\ \mathrm{Hz}: that is below 300 Hz300\ \mathrm{Hz}, and an approaching source can only raise the frequency, so the first harmonic it can reach is 340 Hz340\ \mathrm{Hz}.