1. Physics
  2. Oscillations and Waves
  3. Superposition of Waves

Oscillations and Waves · JEE & NEET Physics

Superposition of Waves: notes and previous year questions

Waves add: interference, standing waves, harmonics of strings and organ pipes, and beats.

Superposition of Waves in short

  • When waves meet, displacements add; each wave then carries on unchanged.
  • Same-frequency waves interfere: A² = A₁² + A₂² + 2A₁A₂ cos φ.
  • Constructive where the path difference is nλ; destructive where it is an odd number of half wavelengths.
  • Two equal waves travelling opposite ways make a standing wave y = 2A sin kx cos ωt.

1The principle of superposition

Two pulses on a rope run into each other. Where they overlap, the displacement is simply the sum of the two: two upright pulses make one twice as tall; an upright and an upside-down pulse cancel. Then they pass through each other and go on unchanged.

y=y1+y2+y3+…y = y_1 + y_2 + y_3 + \ldotsAdd with signs. It holds for small amplitudes, where the medium responds in proportion.

Example: a 3 cm crest meeting a 1 cm trough gives +3+(−1)=+2+3 + (-1) = +2 cm at that moment.

2Interference

Two waves of the same frequency, y1=A1sin⁡ωty_1 = A_1\sin\omega t and y2=A2sin⁡(ωt+ϕ)y_2 = A_2\sin(\omega t + \phi), add to a wave of the same frequency. Draw each as an arrow as long as its amplitude, the second turned by ϕ\phi and starting at the tip of the first; the sum is the arrow from start to end, and the cosine rule gives:

A2=A12+A22+2A1A2cos⁡ϕA^2 = A_1^2 + A_2^2 + 2A_1A_2\cos\phi
tan⁡α=A2sin⁡ϕA1+A2cos⁡ϕ\tan\alpha = \frac{A_2\sin\phi}{A_1 + A_2\cos\phi}The phase α of the resultant.
φResultant amplitudeType
0,2π,…0, 2\pi, \ldotsA1+A2A_1 + A_2 (equal: 2A02A_0)constructive
π/2\pi/2A12+A22\sqrt{A_1^2 + A_2^2} (equal: A02A_0\sqrt{2})in between
2π/32\pi/3equal amplitudes: A0A_0in between
π,3π,…\pi, 3\pi, \ldots∣A1−A2∣|A_1 - A_2| (equal: 0)destructive

Example: 5sin⁡ωt+5sin⁡(ωt+π/3)5\sin\omega t + 5\sin(\omega t + \pi/3): A2=25+25+50cos⁡60∘=75A^2 = 25 + 25 + 50\cos 60^\circ = 75, so A=53≈8.66A = 5\sqrt{3} \approx 8.66.

Two speakers playing in step: the phase difference comes from the path difference, Δϕ=(2π/λ)Δx\Delta\phi = (2\pi/\lambda)\Delta x. Walking across the room you hear loud and quiet places in turn.

loud: Δx=nλ,quiet: Δx=(2n+1)λ2\text{loud: } \Delta x = n\lambda, \quad \text{quiet: } \Delta x = (2n + 1)\frac{\lambda}{2}
  • λ=0.8\lambda = 0.8 m and Δx=1.2\Delta x = 1.2 m =1.5λ= 1.5\lambda: quiet.
  • Intensity goes as A2A^2.
  • Noise-cancelling headphones make a wave exactly opposite to the noise: A sin ωt + A sin(ωt + π) = 0.

3Standing waves

Two identical waves travelling in opposite directions, y1=Asin⁡(kx−ωt)y_1 = A\sin(kx - \omega t) and y2=Asin⁡(kx+ωt)y_2 = A\sin(kx + \omega t), add (using sin⁡C+sin⁡D=2sin⁡C+D2cos⁡C−D2\sin C + \sin D = 2\sin\tfrac{C+D}{2}\cos\tfrac{C-D}{2}) to a wave that does not travel:

y=2Asin⁡kx cos⁡ωty = 2A\sin kx\,\cos\omega t
  • Every point moves as cos⁡ωt\cos\omega t with its own amplitude 2Asin⁡kx2A\sin kx.
  • Nodes never move: sin⁡kx=0\sin kx = 0, at x=0,λ/2,λ,…x = 0, \lambda/2, \lambda, \ldots
  • Antinodes move most: sin⁡kx=±1\sin kx = \pm1, at x=λ/4,3λ/4,…x = \lambda/4, 3\lambda/4, \ldots
  • No energy flows along; it sloshes between kinetic and potential inside each loop.
BetweenDistance
Node and next nodeλ/2\lambda/2
Antinode and next antinodeλ/2\lambda/2
Node and nearest antinodeλ/4\lambda/4

4Vibrating strings

A string fixed at both ends must have nodes at both ends, so it holds a whole number of loops, each λ/2\lambda/2 long: L=nλ/2L = n\lambda/2.

λn=2Ln,fn=nv2L,n=1,2,3,…\lambda_n = \frac{2L}{n}, \quad f_n = \frac{nv}{2L}, \quad n = 1, 2, 3, \ldots
f1=12LTμf_1 = \frac{1}{2L}\sqrt{\frac{T}{\mu}}The fundamental (first harmonic); the others are 2f₁, 3f₁, …
  • L=1L = 1 m, v=200v = 200 m/s: f1=100f_1 = 100 Hz, f2=200f_2 = 200 Hz, and 3 loops need 300 Hz.
  • 5 loops on 2 m at 500 Hz: λ=0.8\lambda = 0.8 m, v=400v = 400 m/s.
  • One end free (an antinode): only odd harmonics, f=(2n−1)v/4Lf = (2n - 1)v/4L.

5Organ pipes

In air the closed end of a pipe is a displacement node (the air cannot move) and an open end an antinode.

fn=(2n−1)v4Lf_n = (2n - 1)\frac{v}{4L}Closed at one end: a quarter wave fits first; only odd harmonics f₁, 3f₁, 5f₁ …
fn=nv2Lf_n = \frac{nv}{2L}Open at both ends: all harmonics, like a string.
SystemEndsHarmonicsFundamental
String, both ends fixednode – nodeallv/2L
String, one end freenode – antinodeodd onlyv/4L
Closed pipenode – antinodeodd onlyv/4L
Open pipeantinode – antinodeallv/2L

6Beats

Two notes of slightly different frequency drift in and out of step, so their sum swells and fades: we hear beats.

y=2Acos⁡ ⁣(2πf1−f22t)sin⁡ ⁣(2πf1+f22t)y = 2A\cos\!\Big(2\pi\tfrac{f_1 - f_2}{2}t\Big)\sin\!\Big(2\pi\tfrac{f_1 + f_2}{2}t\Big)A note at the average frequency whose amplitude follows the slow cosine.
fbeat=∣f1−f2∣f_{beat} = |f_1 - f_2|The loudness peaks twice in each cycle of the slow cosine (at + and at −).
  • Tuning: adjust until the beats slow down and disappear; then the frequencies are equal.
  • Beats are heard clearly only when |f₁ − f₂| is small (below about 10 Hz); farther apart you hear two notes.
  • 6 beats with a 256 Hz fork: the other is 250 or 262 Hz.
  • Fork A 256 Hz, 4 beats with B; wax on B (lowering it) makes 6 beats: B moved away from 256, so it was 252 Hz.
  • Two 200 Hz strings, one tightened by 2%: f∝Tf \propto \sqrt{T} rises about 1% to 202 Hz, giving 2 beats per second.

Summary

Key ideas

  • When waves meet, displacements add; each wave then carries on unchanged.
  • Same-frequency waves interfere: A² = A₁² + A₂² + 2A₁A₂ cos φ.
  • Constructive where the path difference is nλ; destructive where it is an odd number of half wavelengths.
  • Two equal waves travelling opposite ways make a standing wave y = 2A sin kx cos ωt.
  • Nodes never move and antinodes move most; node to node is λ/2, node to antinode λ/4.
  • A standing wave carries no net energy.
  • A string fixed at both ends plays all harmonics fₙ = nv/2L.
  • A closed pipe plays only odd harmonics of v/4L; an open pipe plays all harmonics of v/2L.
  • The mix of harmonics gives each instrument its timbre.
  • Two close frequencies give beats at |f₁ − f₂| per second, used for tuning.

Every equation

Superposition
y=y1+y2y = y_1 + y_2
Resultant amplitude
A2=A12+A22+2A1A2cos⁡ϕA^2 = A_1^2 + A_2^2 + 2A_1A_2\cos\phi
Resultant phase
tan⁡α=A2sin⁡ϕA1+A2cos⁡ϕ\tan\alpha = \tfrac{A_2\sin\phi}{A_1 + A_2\cos\phi}
Constructive
Δx=nλ\Delta x = n\lambda
Destructive
Δx=(2n+1)λ/2\Delta x = (2n + 1)\lambda/2
Path and phase
Δϕ=(2π/λ) Δx\Delta\phi = (2\pi/\lambda)\,\Delta x
Standing wave
y=2Asin⁡kxcos⁡ωty = 2A\sin kx\cos\omega t
Nodes
x=nλ/2x = n\lambda/2
Antinodes
x=(2n+1)λ/4x = (2n + 1)\lambda/4
String modes
λn=2L/n\lambda_n = 2L/n
String harmonics
fn=nv/2Lf_n = nv/2L
Fundamental
f1=12LT/μf_1 = \tfrac{1}{2L}\sqrt{T/\mu}
Closed pipe
fn=(2n−1)v/4Lf_n = (2n - 1)v/4L
Open pipe
fn=nv/2Lf_n = nv/2L
Beats
fbeat=∣f1−f2∣f_{beat} = |f_1 - f_2|

Previous year questions with solutions

Real JEE and NEET questions on superposition of waves. Try each one before you open the solution.

Q1JEE Main 2026One correct option

Match List I with List II

List-I | List-II

A. sin⁡2ωt{\sin}^{2}\omega t | I. Periodic with time period T=πωT=\frac{\pi }{\omega } but not simple harmonic motion (SHM)

B. sin⁡3(2ωt){\sin}^{3}(2\omega t) | II. Periodic with time period T=2πωT=\frac{2\pi }{\omega } but Not SHM

C. sin⁡(ωt)+cos⁡(πωt)\sin (\omega t)+\cos (\pi \omega t) | III. Periodic with time period T=πωT=\frac{\pi }{\omega } and SHM

D. cos⁡ωt+cos⁡2ωt\cos \omega t+\cos 2\omega t | IV. Non-periodic

Choose the correct answer from the options given below:

  1. AA-III, B-I, C-IV, D-II
  2. BA-II, B-I, C-III, D-IV
  3. CA-III, B-II, C-IV, D-I
  4. DA-II, B-I, C-IV, D-III
Show answer and solution

Answer: Option A

Rewrite each and count the frequencies.

A: sin⁡2ωt=12−12cos⁡2ωt\sin^{2}\omega t = \dfrac{1}{2} - \dfrac{1}{2}\cos 2\omega t — one frequency, 2ω2\omega, so SHM, with period 2π2ω=πω\dfrac{2\pi}{2\omega} = \dfrac{\pi}{\omega}: III.

B: sin⁡3(2ωt)=34sin⁡2ωt−14sin⁡6ωt\sin^{3}(2\omega t) = \dfrac{3}{4}\sin 2\omega t - \dfrac{1}{4}\sin 6\omega t — two frequencies, so not SHM; both terms repeat after 2π2ω=πω\dfrac{2\pi}{2\omega} = \dfrac{\pi}{\omega}: I.

C: sin⁡ωt+cos⁡πωt\sin\omega t + \cos\pi\omega t — angular frequencies ω\omega and πω\pi\omega, whose ratio π\pi is not a ratio of whole numbers, so it never repeats: IV.

D: cos⁡ωt+cos⁡2ωt\cos\omega t + \cos 2\omega t — ratio 1:21 : 2, period 2πω\dfrac{2\pi}{\omega}, but two frequencies, so not SHM: II.

So A-III, B-I, C-IV, D-II: option A.

The trap is option C, which swaps I and II: it gives sin⁡3(2ωt)\sin^{3}(2\omega t) the period 2πω\dfrac{2\pi}{\omega} of sin⁡ωt\sin\omega t, but its argument is 2ωt2\omega t, so it repeats twice as often. Options B and D both call sin⁡2ωt\sin^{2}\omega t "not SHM", but written as 12−12cos⁡2ωt\dfrac{1}{2} - \dfrac{1}{2}\cos 2\omega t it is one clean cosine about a shifted centre. B also calls sin⁡ωt+cos⁡πωt\sin\omega t + \cos\pi\omega t SHM, and D calls cos⁡ωt+cos⁡2ωt\cos\omega t + \cos 2\omega t SHM, though each has two different frequencies.

Q2JEE Advanced 2023Numerical answer

A string of length 1m1 m and mass 2×10−5kg2\times {10}^{-5} \mathrm{kg} is under tension TT. When the string vibrates, two successive harmonics are found to occur at frequencies 750Hz750 \mathrm{Hz} and 1000Hz1000 \mathrm{Hz}. The value of tension TT is ________ Newton.

Show answer and solution

Answer: 5

Successive harmonics differ by the fundamental: f1=1000−750=250 Hzf_{1} = 1000 - 750 = 250\ \mathrm{Hz} (750750 and 1000 Hz1000\ \mathrm{Hz} are the third and fourth harmonics).

The wave speed is v=2Lf1=2×1×250=500 m/sv = 2Lf_{1} = 2 \times 1 \times 250 = 500\ \mathrm{m/s}, and μ=2×10−51=2×10−5 kg/m\mu = \dfrac{2 \times 10^{-5}}{1} = 2 \times 10^{-5}\ \mathrm{kg/m}. From v=Tμv = \sqrt{\dfrac{T}{\mu}},

T=μv2=2×10−5×2.5×105=5 NT = \mu v^{2} = 2 \times 10^{-5} \times 2.5 \times 10^{5} = 5\ \mathrm{N}

The trap is to take 750 Hz750\ \mathrm{Hz} as the fundamental: that gives v=1500 m/sv = 1500\ \mathrm{m/s} and a tension of 45 N45\ \mathrm{N}. Another is T=μv=0.01 NT = \mu v = 0.01\ \mathrm{N}, forgetting to square the speed.

Q3NEET 2015One correct option

When two displacements represented by y1=asin⁡(ωt)y_{1} = a\sin(\omega t) and y2=bcos⁡(ωt)y_{2} = b\cos(\omega t) are superimposed the motion is

  1. Asimple harmonic with amplitude a2+b2\sqrt{a^{2}+b^{2}}
  2. Bsimple harmonic with amplitude (a+b)2\frac{(a+b)}{2}
  3. Cnot a simple harmonic
  4. Dsimple harmonic with amplitude ab\frac{a}{b}
Show answer and solution

Answer: Option A

Write the cosine as a sine: y2=bcos⁡ωt=bsin⁡(ωt+π2)y_{2} = b\cos\omega t = b\sin\left(\omega t + \dfrac{\pi}{2}\right). So the two are SHMs of the same ω\omega, a phase π2\dfrac{\pi}{2} apart, and

y1+y2=asin⁡ωt+bcos⁡ωt=a2+b2 sin⁡(ωt+δ)y_{1} + y_{2} = a\sin\omega t + b\cos\omega t = \sqrt{a^{2} + b^{2}}\,\sin(\omega t + \delta), with tan⁡δ=ba\tan\delta = \dfrac{b}{a}

That is SHM, of amplitude a2+b2\sqrt{a^{2} + b^{2}}: option A.

The trap is C: a sine plus a cosine looks like two different motions, but with the same ω\omega they merge into one SHM. B, a+b2\dfrac{a + b}{2}, averages the amplitudes, and D, ab\dfrac{a}{b}, is not even a length. Neither comes from adding the displacements.

Practice questions, easy to hard

Three questions from the superposition of waves practice ladder: one easy, one medium, one hard.

Q4Numerical answer

Now two waves of the same frequency travelling the same way through the same place:

y1=A1sin⁡(kx−ωt)y_{1} = A_{1}\sin(kx - \omega t) and y2=A2sin⁡(kx−ωt+ϕ)y_{2} = A_{2}\sin(kx - \omega t + \phi)

At any one point, each is an SHM of the same ω\omega, with a phase difference ϕ\phi between them, and you already know how two such SHMs add: the result is one SHM. So the sum is one wave, with the same kk and ω\omega, of amplitude

A=A12+A22+2A1A2cos⁡ϕA = \sqrt{A_{1}^{2} + A_{2}^{2} + 2A_{1}A_{2}\cos\phi}

and with a phase δ\delta, measured from the first wave, given by tan⁡δ=A2sin⁡ϕA1+A2cos⁡ϕ\tan\delta = \dfrac{A_{2}\sin\phi}{A_{1} + A_{2}\cos\phi} (the phasor sum).

Two waves of amplitudes 3 mm3\ \mathrm{mm} and 4 mm4\ \mathrm{mm} and the same frequency travel together along a string, with a phase difference of π2\dfrac{\pi}{2}. What is the amplitude of the resultant wave, in mm?

Show answer and solution

Answer: 5 mm

cos⁡π2=0\cos\dfrac{\pi}{2} = 0, so A=32+42=25=5 mmA = \sqrt{3^{2} + 4^{2}} = \sqrt{25} = 5\ \mathrm{mm}.

The trap is 7 mm7\ \mathrm{mm}, adding the amplitudes as if the waves were in step; that happens only when ϕ=0\phi = 0. Another is 1 mm1\ \mathrm{mm}, the difference, which happens only when ϕ=π\phi = \pi.

Q5One or more correct options

Two different wires, joined end to end and clamped at their far ends under the same tension, can vibrate with a node at the joint. Then each piece is a string fixed at both ends, with a whole number of loops of its own, and both pieces vibrate at the same frequency. So the frequency must be a harmonic of BOTH pieces, and the lowest possible one is the smallest frequency that is a whole-number multiple of both fundamentals.

The joint can instead be an antinode. Then each piece is like a string with one free end: a node at its clamp and an antinode at the joint, so it has only its odd harmonics, (2n−1)v4L\dfrac{(2n - 1)v}{4L}. Its lowest note, v4L\dfrac{v}{4L}, is half its fundamental with both ends clamped. Such a mode exists only if some frequency is on BOTH pieces' lists of odd harmonics.

On its own, clamped at both ends, wire 1 would have a fundamental of 100 Hz100\ \mathrm{Hz}, and wire 2 one of 250 Hz250\ \mathrm{Hz}. They are joined as described and vibrate with a node at the joint. Which statements are correct?

  1. AThe lowest possible frequency is 500 Hz500\ \mathrm{Hz}
  2. BAt that frequency, wire 1 has 55 loops and wire 2 has 22 loops
  3. CWith an antinode at the joint instead, the lowest possible frequency would be 125 Hz125\ \mathrm{Hz}
  4. DThe lowest possible frequency is 250 Hz250\ \mathrm{Hz}
Show answer and solution

Answer: Options A, B

Wire 1 allows 100,200,300,400,500,… Hz100, 200, 300, 400, 500, \dots\ \mathrm{Hz}; wire 2 allows 250,500,750,… Hz250, 500, 750, \dots\ \mathrm{Hz}. The first frequency on both lists is 500 Hz500\ \mathrm{Hz} (statement A), where wire 1 is in its fifth harmonic, 55 loops, and wire 2 in its second, 22 loops (statement B). Counting the ends and the shared node at the joint, the whole string then has 6+3−1=86 + 3 - 1 = 8 nodes.

C is false. With an antinode at the joint, wire 1 allows only 50,150,250,350,… Hz50, 150, 250, 350, \dots\ \mathrm{Hz} (odd multiples of 5050) and wire 2 only 125,375,625,… Hz125, 375, 625, \dots\ \mathrm{Hz} (odd multiples of 125125). 125 Hz125\ \mathrm{Hz} is wire 2's lowest note but is not on wire 1's list. In fact no frequency is on both: wire 1's are all EVEN multiples of 25 Hz25\ \mathrm{Hz} and wire 2's all ODD multiples, so these wires have no mode at all with an antinode at the joint. D is the trap of taking the higher fundamental: at 250 Hz250\ \mathrm{Hz} wire 1 would need 2.52.5 loops, which would put an antinode, not a node, at the joint.

Q6Numerical answer

Damping drains the energy of an oscillation as E=E0e−bt/mE = E_{0}e^{-bt/m}, where E0=12kA02E_{0} = \dfrac{1}{2}kA_{0}^{2} is the energy it started with. What is gone has been turned into heat in the medium.

A 0.5 kg0.5\ \mathrm{kg} block on a spring with k=50 N/mk = 50\ \mathrm{N/m} is pulled 10 cm10\ \mathrm{cm} from equilibrium and released in oil, with damping constant b=0.1 kg/sb = 0.1\ \mathrm{kg/s}. How much energy has the oil taken in the first 5 s5\ \mathrm{s}? Take e−1=0.368e^{-1} = 0.368 and give the answer in millijoules, to the nearest millijoule.

Show answer and solution

Answer: 158 mJ

E0=12×50×0.12=0.25 JE_{0} = \dfrac{1}{2} \times 50 \times 0.1^{2} = 0.25\ \mathrm{J}. After 5 s5\ \mathrm{s}, btm=0.1×50.5=1\dfrac{bt}{m} = \dfrac{0.1 \times 5}{0.5} = 1, so E=0.25×0.368=0.092 JE = 0.25 \times 0.368 = 0.092\ \mathrm{J}. The oil has taken 0.25−0.092=0.158 J=158 mJ0.25 - 0.092 = 0.158\ \mathrm{J} = 158\ \mathrm{mJ}.

The trap is using the amplitude's exponent, bt2m=0.5\dfrac{bt}{2m} = 0.5, for the energy: that gives e−0.5≈0.61e^{-0.5} \approx 0.61 left and only about 98 mJ98\ \mathrm{mJ} taken. Another is answering 92 mJ92\ \mathrm{mJ}, the energy still left rather than the energy taken.