Superposition of Waves

Q: What are the conditions for constructive and destructive interference?

Constructive interference needs a phase difference of 2n pi, or a path difference of a whole number of wavelengths, giving amplitude A1 + A2. Destructive interference needs a phase difference of (2n+1) pi, or a path difference of an odd number of half wavelengths, giving amplitude A1 minus A2.

Q: How do you find the ratio of maximum to minimum intensity?

Use the square of (root I1 plus root I2) divided by (root I1 minus root I2), or with the amplitude ratio r, the square of (r + 1) over (r minus 1). For intensities in the ratio 4 to 1, the amplitude ratio is 2 and the answer is 9 to 1.

PhysicsWavesFor NEET aspirants

The superposition of waves says that when two or more waves meet, the resultant displacement is the sum of the individual displacements, $y = y_{1} + y_{2}$ . Superposition of waves explains interference (resultant amplitude $A^{2} = A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} cos ϕ$ ), the reflection of pulses at fixed and free ends, the splitting of a wave at the junction of two strings, and later stationary waves and beats. This page covers the principle, interference conditions, path difference, Quincke's tube and reflection and transmission. A high-weight topic in JEE Main and JEE Advanced.

On this page1Principle2Interference3Resultant amplitude4Path difference5Quincke's tube6Reflection at ends7Two strings

Key Formulas - Quick Reference

★ Must learnSuperposition: $y = y_{1} + y_{2} + \dots$ ; collinear: $y = y_{1} + y_{2} + \dots$
★ Must learnResultant amplitude: $A^{2} = A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} cos ϕ$ ; intensity $I = I_{1} + I_{2} + 2 I_{1} I_{2} cos ϕ$
Phase and path difference: $ϕ = \frac{2 π}{λ} Δ x$
★ Must learnConstructive: $ϕ = 2 nπ$ , $Δ x = nλ$ , $A = A_{1} + A_{2}$ . Destructive: $ϕ = (2 n + 1) π$ , $Δ x = (2 n + 1) \frac{λ}{2}$ , $A = ∣ A_{1} - A_{2} ∣$
$\frac{I _{m a x}}{I _{m i n}} = (\frac{I _{1} + I _{2}}{I _{1} - I _{2}})^{2} = (\frac{A _{1} + A _{2}}{A _{1} - A _{2}})^{2}$
Equal intensities: $I = 4 I_{0} cos^{2} \frac{ϕ}{2}$ (from $0$ to $4 I_{0}$ )
Two sources, far point: $Δ x = d sin θ \approx \frac{d x}{D}$
Quincke's tube: successive maxima when tube moves $x = \frac{λ}{2}$ , so $v = 2 f x$
★ Must learnTwo strings: $A_{t} = \frac{2 v _{2}}{v _{1} + v _{2}} A_{i}$ , $A_{r} = \frac{v _{2} - v _{1}}{v _{1} + v _{2}} A_{i}$ (fixed end: $A_{r} = - A_{i}$ ; free end: $A_{r} = + A_{i}$ )

1. The Principle of Superposition

★ Must learnPrinciple of superposition: when two or more waves superpose at a particle of the medium, the resultant displacement of that particle is the vector sum of the displacements that each wave would produce there on its own:

y = y_{1} + y_{2} + y_{3} + \dots + y_{N}

If all the displacements are along the same line, this becomes the algebraic sum

y = y_{1} + y_{2} + \dots + y_{N}

Each wave travels on as if the other were not there; they add only where they overlap. The principle holds for small displacements (linear media), which covers strings, sound and light in all exam problems.

Figure 1: Principle of superposition. While the pulses overlap, the string shape (solid) is the algebraic sum of the two pulses (dashed):

y = y_{1} + y_{2}

. Afterwards each pulse moves on unchanged, as if the other had never been there.

When two pulses on the same side of a string overlap, the string bulges more; when they are on opposite sides, they partly or fully cancel. After crossing, each pulse continues with its original shape and speed. Superposition leads to several important phenomena:

Interference: two coherent waves travelling in the same direction (this page).
Stationary waves: two identical waves travelling in opposite directions (Stationary Waves concept).
Beats: two waves of slightly different frequencies (Beats concept).
Lissajous figures: two perpendicular oscillations (not needed here in detail).

2. Interference of Waves

Let two sinusoidal waves of the same wavelength travel in the same direction along a string. The resultant depends on how far one waveform is shifted from the other. If they are exactly in phase, every particle's displacement doubles: constructive interference. If they are exactly in opposite phase, the displacements cancel and the string stays straight: destructive interference.

Figure 2: Two coherent waves of amplitude

A

travelling the same way. In phase (

ϕ = 0

): resultant amplitude

2 A

(constructive). Opposite phase (

ϕ = π

): resultant zero (destructive).

Coherent waves and sources. Waves of the same frequency (and form) with a constant phase relation are coherent; only they give a steady interference pattern. Two independent sources are practically never coherent, so coherent sources are made by splitting the wave from one source (as in Quincke's tube) or by driving two sources from one oscillator. Interference is the superposition of two coherent waves travelling in the same direction.

3. Analytical Treatment: Resultant Amplitude and Intensity

Let the waves from sources $S_{1}$ and $S_{2}$ reach a point after travelling $x_{1}$ and $x_{2}$ :

y_{1} = A_{1} sin (ω t - k x_{1}), y_{2} = A_{2} sin (ω t - k x_{2})

Phase difference at the point: $ϕ = k (x_{2} - x_{1}) = \frac{2 π}{λ} Δ x$ , where $Δ x$ is the path difference.
Two SHMs of the same $ω$ add like vectors (phasors) at angle $ϕ$ :
$A^{2} = A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} cos ϕ$
Since intensity $I \propto A^{2}$ :
$I = I_{1} + I_{2} + 2 I_{1} I_{2} cos ϕ$

Figure 3: Add amplitudes like vectors at angle

ϕ

A^{2} = A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} cos ϕ

. Drawn for

A_{1} = 3

A_{2} = 2

ϕ = 6 0^{\circ}

A = 19 \approx 4.36

	Constructive interference	Destructive interference
Condition on phase	$cos ϕ = 1$ : $ϕ = 2 nπ$	$cos ϕ = - 1$ : $ϕ = (2 n + 1) π$
Condition on path	$Δ x = nλ$	$Δ x = (2 n + 1) \frac{λ}{2}$
Amplitude	$A_{1} + A_{2}$	$∣ A_{1} - A_{2} ∣$
Intensity	$I_{m a x} = (I_{1} + I_{2})^{2}$	$I_{m i n} = (I_{1} - I_{2})^{2}$
Equal intensities $I_{0}$	$4 I_{0}$	$0$

Here $n = 0, 1, 2, \dots$ In general, for two waves of equal intensity $I_{0}$ : $I = 2 I_{0} (1 + cos ϕ) = 4 I_{0} cos^{2} \frac{ϕ}{2}$ .

Figure 4:

I = I_{1} + I_{2} + 2 I_{1} I_{2} cos ϕ

. Maxima at

ϕ = 2 nπ

, minima at

(2 n + 1) π

. Energy is only redistributed: the average stays

I_{1} + I_{2}

. For

I_{1} = 4 I_{2}

I_{m a x} / I_{m i n} = 9

(Solved Example 1).

Exam Trick

Work with the amplitude ratio. If $\frac{A _{1}}{A _{2}} = r$ (so $\frac{I _{1}}{I _{2}} = r^{2}$ ), then $\frac{I _{m a x}}{I _{m i n}} = (\frac{r + 1}{r - 1})^{2}$ . Intensity ratio $4 : 1$ means $r = 2$ and $I_{m a x} : I_{m i n} = 9 : 1$ . Going backwards: $I_{m a x} / I_{m i n} = 25$ means $\frac{r + 1}{r - 1} = 5$ , so $r = 1.5$ .

Key idea

Interference only redistributes energy: bright (loud) where

cos ϕ > 0

, dark (quiet) where

cos ϕ < 0

, and the average stays

I_{1} + I_{2}

4. Interference of Sound from Two Sources

Two loudspeakers driven by the same oscillator are coherent sources. At any point, the path difference $Δ x = S_{2} P - S_{1} P$ decides whether the sound is loud or soft. For a point far away ( $D ≫ d$ ) the two paths are almost parallel and the path difference is the short side $S_{2} Q$ of a thin right triangle.

Figure 5: For

D ≫ d

the two paths are nearly parallel and

Δ x = S_{2} Q = d sin θ \approx \frac{d x}{D}

. Maxima where

Δ x = nλ

, minima where

Δ x = (2 n - 1) λ /2

(Solved Example 7).

On the perpendicular bisector of $S_{1} S_{2}$ ( $x = 0$ ), $Δ x = 0$ : always a maximum.
$n$ -th maximum at $x_{n} = \frac{nλ D}{d}$ ; $n$ -th minimum at $x_{n} = \frac{( 2 n - 1 ) λ D}{2 d}$ .
Along the line $S_{1} S_{2}$ extended (beyond either source), $Δ x = d$ everywhere.
At angle $θ$ from the line $S_{1} S_{2}$ (far away), $Δ x = d cos θ$ ; counting how many values of $nλ$ fit between $- d$ and $d$ gives the number of maxima around the sources.

Figure 6:

Δ x = 3 λ cos θ = 0, \pm λ, \pm 2 λ, \pm 3 λ

gives

cos θ = 0, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm 1

: 12 maxima on the circle (labels show

∣Δ x ∣

; Solved Example 9).

Exam Trick

Path to phase first. Every interference question reduces to one number, the phase difference $ϕ = \frac{2 π}{λ} Δ x$ (add $π$ for each inverting reflection). Even multiple of $π$ : maximum; odd multiple: minimum. On the line $S_{1} S_{2}$ extended beyond both sources, $Δ x = d$ at every point, so that whole line is equally loud.

Quick Recall: tap to check

Two waves of intensities

I

and

4 I

interfere. What are

I_{m a x}

and

I_{m i n}

(I + 2 I)^{2} = 9 I

and

(2 I - I)^{2} = I

What path difference gives destructive interference?

An odd multiple of

λ /2

λ /2, 3 λ /2, \dots

Two identical sources in phase: is the midpoint between them loud or quiet?

Loud: the path difference is zero.

Figure 7: Always convert path to phase first (

λ \leftrightarrow 2 π

), add

π

for an inverting reflection, then compare with

2 nπ

and

(2 n + 1) π

JEE Advanced

Sources that are not in phase. If $S_{2}$ leads $S_{1}$ by $ϕ_{0}$ , the phase difference at $P$ is $ϕ = ϕ_{0} + \frac{2 π}{λ} (S_{1} P - S_{2} P)$ and the whole pattern shifts. With $ϕ_{0} = π$ the central line (equal paths) becomes a minimum. A displacement wave reflected from a rigid wall or a denser string gains the same built-in $π$ , which is why such a path is counted as $Δ x + \frac{λ}{2}$ (for the pressure wave at a rigid wall there is no phase change).

5. Quincke's Tube

Quincke's tube demonstrates interference of sound and measures its speed. It has two U-tubes: $A$ fixed, and $B$ which slides in and out of $A$ . A tuning fork or source of known frequency $f$ sounds at opening $P$ , and a detector at opening $Q$ receives the sound arriving by the two paths.

Figure 8: Quincke's tube. Moving

B

between two successive maxima changes the path difference by

2 x = λ

, so

v = f λ = 2 f x

Adjust $B$ for a maximum at $Q$ . If the path lengths via $A$ and $B$ are $l_{1}$ and $l_{2}$ : $l_{2} - l_{1} = N λ$ .
Pull $B$ out by $x$ until the next maximum. Both arms of $B$ lengthen, so $l_{2}^{'} = l_{2} + 2 x$ and $l_{2}^{'} - l_{1} = (N + 1) λ$ .
Subtract: $2 x = λ$ , so $x = \frac{λ}{2}$ and
$v = f λ = 2 f x$

The distance between a maximum and the next minimum is $λ /4$ of tube movement.

6. Reflection of Waves at a Boundary

6.1 Fixed end and free end

Fixed end. When a pulse reaches a fixed end, the string pulls up on the wall; by Newton's third law the wall pulls the string down, first down and then up as the pulse arrives. This produces a reflected pulse that is inverted but otherwise identical: a phase change of $π$ .

Free end. Let the end be a light ring sliding on a smooth vertical rod: the rod keeps the tension but exerts no transverse force. The ring rises to a maximum, comes momentarily to rest with the string stretched, and is pulled back down, sending back a pulse on the same side as the incident one: no phase change.

Figure 9: Image method. Fixed end: an inverted image (dashed) keeps the end at

y = 0

, so the pulse returns inverted (phase change

π

). Free end: an upright image; the end momentarily rises to twice the height and the pulse returns upright (no phase change). Time runs downward.

Both cases are superpositions of the real pulse and an imaginary image pulse moving the other way from behind the boundary. An inverted image keeps the fixed end at zero displacement at all times; an upright image makes the free end's displacement momentarily double the pulse height.

Fixed end (rigid boundary)

Reflected wave inverted; phase change $π$ ; path change $λ /2$ . The end is always a node (displacement zero).

Free end (open boundary)

Reflected wave upright; no phase change. The end is an antinode (displacement maximum).

For sound, a rigid wall (closed end of a pipe) is a displacement node, so the displacement wave is inverted but the pressure wave comes back unchanged: a compression returns as a compression. At an open end the pressure wave is inverted. These rules build the air-column modes in the Stationary Waves concept.

6.2 Reflection and transmission between two strings

When the end is neither fixed nor free, for example a light string joined to a heavy one under the same tension $T$ , part of the wave is reflected and part is transmitted. The frequency stays the same (set by the source); the speed changes from $v_{1} = T / μ_{1}$ to $v_{2} = T / μ_{2}$ , and so does the wavelength.

Figure 10: Light to heavy string (

v_{2} < v_{1}

): the reflected pulse is inverted, the transmitted one upright. Heavy to light (

v_{2} > v_{1}

): both upright. The limits

v_{2} \to 0

(fixed end) and

v_{2} \to \infty

(free end) give

A_{r} = - A_{i}

and

A_{r} = + A_{i}

. Sketch for

v_{2} = v_{1} /2

A_{t} = \frac{2}{3} A_{i}

A_{r} = - \frac{1}{3} A_{i}

Energy: incident power $=$ reflected power $+$ transmitted power. With $P = 2 π^{2} f^{2} A^{2} μv$ and $μ = T / v^{2}$ , each power is proportional to $A^{2} / v$ :
$\frac{A _{i}^{2}}{v _{1}} = \frac{A _{r}^{2}}{v _{1}} + \frac{A _{t}^{2}}{v _{2}}$
Continuity: the string does not break or kink at the junction $P$ , so $A_{i} + A_{r} = A_{t}$ .
Factor the first equation: $(A_{i} - A_{r}) (A_{i} + A_{r}) = \frac{v _{1}}{v _{2}} A_{t}^{2}$ , so $A_{i} - A_{r} = \frac{v _{1}}{v _{2}} A_{t}$ .
Solve with step 2:
$A_{t} = \frac{2 v _{2}}{v _{1} + v _{2}} A_{i}, A_{r} = \frac{v _{2} - v _{1}}{v _{1} + v _{2}} A_{i}$

Case	Reflected wave	Transmitted wave
Light to heavy string ( $v_{2} < v_{1}$ )	Inverted ( $A_{r} < 0$ ), smaller	Upright, smaller amplitude, shorter $λ$
Heavy to light string ( $v_{2} > v_{1}$ )	Upright ( $A_{r} > 0$ )	Upright, larger amplitude than $A_{i}$ , longer $λ$
Fixed end ( $v_{2} \to 0$ )	$A_{r} = - A_{i}$ : fully inverted	None
Free end ( $v_{2} \to \infty$ )	$A_{r} = + A_{i}$ : fully upright	None (end moves $2 A_{i}$ )

Key idea

Going into a denser (slower) medium flips the reflected wave; the transmitted wave is never inverted.

Quick Recall: tap to check

A pulse reflects from a rigid end. What is the phase change?

π

: the reflected pulse is inverted (the end is a displacement node).

Two waves of intensities

I_{1}

and

I_{2}

interfere. What is the average intensity over the pattern?

I_{1} + I_{2}

: energy is only redistributed.

A pulse goes from a light string to a heavy one. Is the transmitted pulse inverted?

No. The transmitted pulse is always upright; only the reflected one is inverted.

Figure 11: Revision map: the principle, interference and its conditions, intensity, Quincke's tube and reflection at a boundary.

7. Solved Examples

Solved Example 1

Waves from two sources of the same frequency, travelling in the same direction, have intensities in the ratio

4 : 1

. Find the ratio of the maximum to the minimum intensity when they interfere.

Solution:

$\frac{I _{m a x}}{I _{m i n}} = (\frac{I _{1} + I _{2}}{I _{1} - I _{2}})^{2} = (\frac{I _{1} / I _{2} + 1}{I _{1} / I _{2} - 1})^{2} = (\frac{2 + 1}{2 - 1})^{2} = 9$ .

Answer: $9 : 1$ .

Solved Example 2

A triangular pulse moving at

2 cm s^{- 1}

on a rope approaches an end where the rope is free to slide on a vertical pole. The pulse is

1 cm

high; its trailing edge rises over

2 cm

, its leading edge falls over

1 cm

, and its front is

1 cm

from the pole. (a) Draw the pulse at

\frac{1}{2} s

intervals until it is completely reflected. (b) What is the particle speed on the trailing edge at the instant shown?

Solution:

(a) Reflection at a free end is the superposition of the real pulse and an identical, upright image pulse moving in the opposite direction from behind the pole. Each travels $1 cm$ every $\frac{1}{2} s$ . Adding them on the rope ( $x \leq 0$ ) gives the shapes below: the end rises to $2 cm$ at $t = 1 s$ , and by $t = 2 s$ the whole pulse has reflected, upright, with its steep edge leading as it moves away.

Figure 12: String shape

y = f (x - v t) + f (- x - v t)

(real pulse plus an upright image). The end rises to twice the pulse height at

t = 1 s

; at

t = 2 s

the pulse is fully reflected, upright and reversed in shape.

(b) Particle speed $= v \times ∣ slope ∣$ . The trailing edge rises $1 cm$ over $2 cm$ , so the slope is $\frac{1}{2}$ and $∣ v_{p} ∣ = 2 \times \frac{1}{2} = 1 cm s^{- 1}$ (moving down, since $v_{p} = - v \partial y / \partial x$ ).

Answer: (b) $1 cm s^{- 1}$ .

Solved Example 3

A rectangular pulse (height

2 cm

, from

x = - 2

0 cm

) and a triangular pulse (height

2 cm

, from

x = 1

3 cm

, peak at

2 cm

) approach each other at

0.5 cm s^{- 1}

. Sketch the resultant at

t = 2 s

Solution:

In $2 s$ each pulse moves $1 cm$ : the rectangle now spans $- 1$ to $1 cm$ , the triangle $0$ to $2 cm$ with its peak at $1 cm$ . They overlap between $0$ and $1 cm$ , where the heights add.

Figure 13: At

t = 2 s

each pulse has moved

1 cm

. Resultant (solid) = rectangle + triangle (dashed): height

2

, rising to

4 cm

x = 1

, then the triangle alone down to zero at

x = 2

(Solved Example 3).

Answer: height $2 cm$ from $- 1$ to $0$ ; rising linearly to $4 cm$ at $x = 1$ ; dropping to $2 cm$ just after $x = 1$ (the rectangle ends); falling to zero at $x = 2 cm$ .

Solved Example 4

In a large room a person receives sound directly from a source

120 m

away. The person also receives sound from the same source reflected from the

25 m

high ceiling at a point halfway between them. For which wavelengths will the two waves interfere constructively?

Solution:

For the reflection $∠ i = ∠ r$ , so the reflection point $C$ is midway and $S C = C P = 6 0^{2} + 2 5^{2} = 65 m$ . Reflected path $= 130 m$ ; path difference $Δ x = 130 - 120 = 10 m$ .

A rigid ceiling reflects the pressure wave with no phase change, so constructive interference needs $Δ x = nλ$ : $λ = \frac{10}{n} m$ .

Answer: $λ = 10 m, 5 m, 3.33 m, \dots$ ( $10/ n$ m).

Solved Example 5

A sound signal enters a tube and reaches a receiver by two paths: a straight path and a semicircular path of radius

20.0 cm

joining the same two points. The source frequency can be varied from

1000

4000 Hz

. At which frequencies are intensity maxima detected? (

v = 340 m s^{- 1}

)

Solution:

Straight path: $l_{1} = 2 \times 20 = 40 cm$ . Semicircular path: $l_{2} = π \times 20 \approx 62.83 cm$ . Path difference $Δ l \approx 22.83 cm = 0.2283 m$ .

Maxima when $Δ l = N λ = N \frac{v}{f}$ : $f = N \frac{340}{0.2283} \approx N \times 1489 Hz$ .

Answer: $N = 1$ and $2$ : about $1489 Hz$ and $2978 Hz$ ( $N = 3$ gives $4467 Hz$ , outside the range).

Solved Example 6

Two sources

S_{1}

and

S_{2}

2.0 m

apart, vibrate as

y_{1} = 0.03 sin π t

and

y_{2} = 0.02 sin π t

(SI units) and send out waves of speed

1.5 m s^{- 1}

. Find the amplitude of the resultant motion of a particle on the line

S_{1} S_{2}

(a) to the right of

S_{2}

(b) to the left of

S_{1}

Solution:

$f = \frac{ω}{2 π} = 0.5 Hz$ , $λ = \frac{v}{f} = 3.0 m$ .

(a) For any point $P_{2}$ to the right of $S_{2}$ : $Δ x = S_{1} P_{2} - S_{2} P_{2} = 2 m$ , $ϕ = \frac{2 π}{3} \times 2 = \frac{4 π}{3}$ .

R = 0.0 3^{2} + 0.0 2^{2} + 2 (0.03) (0.02) cos \frac{4 π}{3} = 0.0007 \approx 0.0265 m

(b) To the left of $S_{1}$ the path difference is again $2 m$ : $R \approx 0.0265 m$ .

(c) Midway, $Δ x = 0$ : constructive, $R = 0.03 + 0.02 = 0.05 m$ .

Solved Example 7

Two point sources of sound are a distance

d

apart. A detector moves on a straight line parallel to the line joining the sources, at a distance

D ≫ d

. It starts at the point equidistant from both sources. Find its displacement when it detects the

n

-th maximum and the

n

-th minimum.

Solution:

At displacement $x$ from the central point, the path difference is $Δ x = S_{2} Q \approx d sin θ \approx d tan θ = \frac{d x}{D}$ (Figure 5). The central point itself is a maximum.

$n$ -th maximum: $\frac{d x}{D} = nλ$ , so $x = \frac{nλ D}{d}$ .

$n$ -th minimum: $\frac{d x}{D} = (2 n - 1) \frac{λ}{2}$ , so $x = \frac{( 2 n - 1 ) λ D}{2 d}$ .

Answer: $x_{m a x} = \frac{nλ D}{d}$ , $x_{m i n} = \frac{( 2 n - 1 ) λ D}{2 d}$ . This path-difference formula is used again and again.

Solved Example 8

Two coherent sources

S_{1}

and

S_{2}

emit sound of wavelength

λ

in phase. They are

2 λ

apart on a line perpendicular to a screen,

S_{1}

being at distance

D ≫ λ

from the screen.

O

is where the line

S_{2} S_{1}

meets the screen. Find the distance

x = O P

of the nearest point

P

where the intensity equals that at

O

Solution:

At $O$ the path difference is $S_{2} O - S_{1} O = 2 λ$ : a maximum. Moving away from $O$ the path difference falls; the next point of equal (maximum) intensity has $Δ x = λ$ .

At $P$ , with $θ$ the angle between the line of the sources and the direction to $P$ : $Δ x = d cos θ = 2 λ cos θ = λ$ , so $cos θ = \frac{1}{2}$ and $θ = 6 0^{\circ}$ .

Answer: $x = D tan 6 0^{\circ} = 3 D$ .

Solved Example 9

Two coherent sources

S_{1}

and

S_{2}

3 λ

apart, emit sound of wavelength

λ

in phase. A circular wire of large radius lies in their plane with its centre at the midpoint of

S_{1} S_{2}

. At how many points on the wire does constructive interference occur?

Solution:

For a point at angle $θ$ from the line $S_{1} S_{2}$ , $Δ x = 3 λ cos θ$ . Maxima need $Δ x = nλ$ with $∣ n ∣ \leq 3$ : $cos θ = 0, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm 1$ .

$cos θ = \pm 1$ gives 2 points (on the line); $cos θ = 0$ gives 2 points; each of $\pm \frac{1}{3}$ and $\pm \frac{2}{3}$ gives 2 points, 8 in all.

Answer: 12 points (Figure 6): $θ = 0^{\circ}, 18 0^{\circ}$ ; $\pm 9 0^{\circ}$ ; $cos^{- 1} (\pm \frac{1}{3}) \approx 70. 5^{\circ}, 109. 5^{\circ}$ and $cos^{- 1} (\pm \frac{2}{3}) \approx 48. 2^{\circ}, 131. 8^{\circ}$ above and below the line.

Solved Example 10

Two waves of the same frequency with amplitudes

3 mm

and

4 mm

meet with a phase difference of

π /2

. Find the resultant amplitude. How does the resultant intensity compare with

I_{1} + I_{2}

Solution:

$A^{2} = 3^{2} + 4^{2} + 2 (3) (4) cos \frac{π}{2} = 25$ , so $A = 5 mm$ .

Since $cos (π /2) = 0$ , $I = I_{1} + I_{2}$ exactly.

Answer: $5 mm$ ; the intensities simply add.

Solved Example 11

Two coherent waves have amplitudes in the ratio

3 : 1

. Find

I_{m a x} / I_{m i n}

Solution:

$\frac{I _{m a x}}{I _{m i n}} = (\frac{3 + 1}{3 - 1})^{2} = 4$ .

Answer: $4 : 1$ .

Solved Example 12

In a Quincke's tube experiment with a

1000 Hz

source, the sliding tube has to be pulled out by

16.5 cm

between two successive maxima. Find the speed of sound.

Solution:

$x = \frac{λ}{2}$ , so $λ = 33 cm$ and $v = f λ = 2 f x = 2 \times 1000 \times 0.165 = 330 m s^{- 1}$ .

Answer: $330 m s^{- 1}$ .

Solved Example 13

A wave of amplitude

3 mm

travels at

20 m s^{- 1}

on a string joined to a heavier string in which waves travel at

10 m s^{- 1}

(same tension). Find the amplitudes of the reflected and transmitted waves, and check energy conservation.

Solution:

$A_{t} = \frac{2 v _{2}}{v _{1} + v _{2}} A_{i} = \frac{2 ( 10 )}{30} (3) = 2 mm$ ; $A_{r} = \frac{v _{2} - v _{1}}{v _{1} + v _{2}} A_{i} = \frac{- 10}{30} (3) = - 1 mm$ (inverted).

Check $\frac{A ^{2}}{v}$ : incident $\frac{9}{20} = 0.45$ ; reflected $\frac{1}{20} = 0.05$ ; transmitted $\frac{4}{10} = 0.40$ . $0.05 + 0.40 = 0.45$ .

Answer: reflected $1 mm$ (inverted), transmitted $2 mm$ (upright).

Solved Example 14

Two waves

y_{1} = a sin (ω t - k x)

and

y_{2} = a cos (ω t - k x)

superpose. The amplitude of the resultant is
(A)

a

(B)

a 2

(C)

2 a

(D)

0

Solution:

$cos (ω t - k x) = sin (ω t - k x + π /2)$ , so $ϕ = π /2$ and $A = a^{2} + a^{2} = a 2$ .

Answer: (B).

Solved Example 15

Two coherent sound waves with intensities in the ratio

9 : 1

interfere. The ratio of maximum to minimum intensity is
(A)

9 : 1

(B)

3 : 1

(C)

4 : 1

(D)

16 : 1

Solution:

Amplitude ratio $r = 9 = 3$ , so $\frac{I _{m a x}}{I _{m i n}} = (\frac{r + 1}{r - 1})^{2} = (\frac{4}{2})^{2} = 4$ .

Answer: (C).

Practice Questions

Two coherent sources have intensities $I$ and $9 I$ . Find $I_{m a x} : I_{m i n}$ .Answer: $(1 + 3)^{2} : (3 - 1)^{2} = 4 : 1$ .
For two interfering waves $I_{m a x} / I_{m i n} = 25$ . Find the ratio of their amplitudes and intensities.Answer: $A_{1} : A_{2} = 3 : 2$ ; $I_{1} : I_{2} = 9 : 4$ .
Two equal waves of intensity $I_{0}$ meet with path difference $0.5 m$ ; $λ = 2 m$ . Find the resultant intensity.Answer: $ϕ = π /2$ : $4 I_{0} cos^{2} (π /4) = 2 I_{0}$ .
A wave passes from a string where $v_{1} = 10 m s^{- 1}$ to one where $v_{2} = 20 m s^{- 1}$ . Find $A_{t}$ and $A_{r}$ in terms of $A_{i}$ .Answer: $A_{t} = \frac{4}{3} A_{i}$ , $A_{r} = + \frac{1}{3} A_{i}$ (upright).
In Quincke's tube with a $500 Hz$ fork, the tube moves $34 cm$ between successive minima. Find $v$ .Answer: $v = 2 f x = 340 m s^{- 1}$ .
A pulse is sent towards a fixed end. Describe the reflected pulse.Answer: Same shape and size, inverted (phase change $π$ ).

Common Mistakes to Avoid

Watch out

Adding intensities directly when waves are coherent. Add amplitudes with the phase: $A^{2} = A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} cos ϕ$ .
Using $Δ x = (2 n + 1) λ$ for destructive interference. It is an odd multiple of $λ /2$ .
Writing $I_{m a x} / I_{m i n} = (I_{1} + I_{2}) / (I_{1} - I_{2})$ . Use square roots: $(\frac{I _{1} + I _{2}}{I _{1} - I _{2}})^{2}$ .
Forgetting that in Quincke's tube the path changes by $2 x$ , not $x$ .
Inverting the transmitted wave. Only the reflected wave can be inverted (when going into a slower, denser string).
Changing the frequency at a boundary. $f$ stays the same; $v$ and $λ$ change.
Thinking energy is destroyed at destructive points. It is redistributed; the average intensity is still $I_{1} + I_{2}$ .
Expecting interference from two independent sources. They must be coherent (constant phase difference).

Frequently Asked Questions

What is the principle of superposition of waves?

When two or more waves meet at a point, the resultant displacement of the medium there is the vector sum of the displacements each wave would produce alone. Each wave travels on unchanged after the overlap. The principle holds for small displacements, which covers strings, sound and light in exam problems.

What are the conditions for constructive and destructive interference?

Constructive interference needs a phase difference of $2 nπ$ , or a path difference of a whole number of wavelengths, giving amplitude $A_{1} + A_{2}$ . Destructive interference needs a phase difference of $(2 n + 1) π$ , or a path difference of an odd number of half wavelengths, giving amplitude $∣ A_{1} - A_{2} ∣$ .

What are coherent sources?

Coherent sources emit waves of the same frequency with a constant phase difference, so the interference pattern stays steady. Two independent sources are practically never coherent. Coherent sources are made by splitting one wave into two paths, as in Quincke's tube, or by driving two loudspeakers from the same oscillator.

How do you find the ratio of maximum to minimum intensity?

Use $I_{m a x} / I_{m i n} = (\frac{I _{1} + I _{2}}{I _{1} - I _{2}})^{2}$ , or with the amplitude ratio $r$ , $(\frac{r + 1}{r - 1})^{2}$ . For intensities in the ratio 4 to 1, the amplitude ratio is 2 and the answer is 9 to 1.

Why is a pulse inverted on reflection from a fixed end?

At a fixed end the string pulls the wall up as the pulse arrives, and by Newton's third law the wall pulls the string down. This creates an inverted reflected pulse, a phase change of pi. At a free end there is no transverse force, so the pulse returns upright.

What happens when a wave goes from a light string to a heavy string?

Part of the wave is reflected and part transmitted. The transmitted wave is upright, slower, with a shorter wavelength and amplitude $2 v_{2} A_{i} / (v_{1} + v_{2})$ . The reflected wave is inverted, with amplitude $(v_{2} - v_{1}) A_{i} / (v_{1} + v_{2})$ . The frequency does not change.

How is superposition of waves tested in JEE Main?

JEE Main asks for resultant amplitude or intensity at a phase difference, the ratio $I_{m a x} / I_{m i n}$ , path-difference conditions for two sources, Quincke's tube, and amplitudes of reflected and transmitted waves at a string junction. Remember $A^{2} = A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} cos ϕ$ .

What does JEE Advanced ask about superposition and reflection?

JEE Advanced sets pulse-sketching problems (overlapping pulses, reflection from fixed and free ends using image pulses), counting maxima around two sources, interference with reflected sound, and energy checks at junctions of strings. Drawing the image pulse and tracking signs is the key skill.

Previous year questions on Superposition of Waves

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

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