23Superposition of Waves

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SUPERPOSITION OF WAVES
▢ Reflection and Refraction of Waves:

Table 1: Reflection and Refraction Basics

Point
Answer
Boundary incidence
Part of incident wave returns back → reflection; part transmits → refraction
Frequency in reflection/refraction
Unchanged
Incident, reflected, refracted ray and normal
Same plane
Reflection law
Angle of incidence = angle of reflection
Refraction relation for sound
sin i / sin r = v₁ / v₂
Phase in refraction
Unchanged
Rarer/denser concept for wave
Depends on speed, not density
Water for sound
Rarer than air because vwater > vair
Water for light
Denser than air
❖ Applications:
  • •
    Stethoscope
  • •
    Speaking tubes
  • •
    Whispering gallery
  • •
    Visibility of sun before sunrise and after sunset due to refraction of light
▢ Reflection From Boundaries:

Table 1: Free / Rigid Boundary Reflection

Boundary
Phase
Nature of pulse / sound
Equation
Free / open end / rarer boundary
No phase change
Compression ↔ rarefaction
y = a sin(ωt − kx) → y = a sin(ωt + kx)
Rigid / denser boundary
Phase reversed
Compression → compression; rarefaction → rarefaction
y = a sin(ωt − kx) → y = a sin(ωt + kx + π)
▢ Principle of Superposition:
❖ Statement: Resultant displacement at any point due to two or more waves is algebraic sum of displacements produced by individual waves
❖ Formula: y = y₁ + y₂ + y₃ + y₄ + ...
❖ Resultant Depends On:
  • •
    Amplitude of waves
  • •
    Frequency of waves
  • •
    Direction of propagation
  • •
    Phase difference between waves

Table 1: Types of Superposition

Type
Result
Interference
Redistribution of intensity
Beats
Periodic rise and fall of intensity
Stationary wave
Nodes and antinodes
Lissajous figures
Resultant path due to perpendicular SHMs
❖ Validity Conditions:
  • •
    Amplitude a << wavelength λ
  • •
    Particle velocity dy/dt << wave velocity v
  • •
    Valid for mechanical and non-mechanical waves
  • •
    Valid for transverse and longitudinal waves
  • •
    Energy is not destroyed; only redistributed
▢ Interference of Sound:
❖ Definition: Superposition of two sound waves of same frequency and amplitude moving in same direction producing maxima and minima of intensity
❖ Constructive Interference: Waves meet in same phase → maximum intensity
❖ Destructive Interference: Waves meet in opposite phase → minimum intensity

Table 1: Conditions for Interference

Condition
Answer
Frequencies
ν₁ = ν₂
Wave numbers
n₁ = n₂
Wavelengths
λ₁ = λ₂
Phase difference
Constant
Coherent sources
Required

Table 2: Constructive vs Destructive Interference

Feature
Constructive
Destructive
Phase difference
Δφ = 2nπ
Δφ = (2n − 1)π or (2n + 1)π
Path difference
Δx = nλ
Δx = (2n − 1)λ/2 or (2n + 1)λ/2
Amplitude
Amax = a₁ + a₂
Amin = |a₁ − a₂|
Intensity
Imax = (√I₁ + √I₂)²
Imin = (√I₁ − √I₂)²
❖ Useful Relations:
  • •
    Amax + Amin = 2a₁ if a₁ > a₂
  • •
    Amax − Amin = 2a₂
  • •
    Imax + Imin = 2(I₁ + I₂)
  • •
    Imax − Imin = 4√(I₁I₂)
  • •
    If a₁/a₂ = α, Amax/Amin = (α + 1)/(α − 1)
  • •
    If a₁/a₂ = α, Imax/Imin = ((α + 1)/(α − 1))²
▢ Beats:
❖ Definition: Periodic rise and fall in intensity of sound due to superposition of two waves of nearly equal frequencies travelling in same direction

Table 1: Beat Formulae

Quantity
Formula / Result
Beat frequency
fbeat = |n₁ − n₂|
Beat period
T = 1 / |n₁ − n₂|
Detectable beats condition
|n₁ − n₂| ≤ 10 Hz
One waxing + one waning
1 beat
Time interval for noticeable beat
> 1/10 sec
Resultant displacement
y = 2a cos[π(n₁ − n₂)t] sin[2π((n₁+n₂)/2)t]
Resultant amplitude
A = 2a cos[π(n₁ − n₂)t]
Intensity
I = 4I₀ cos²[π(n₁ − n₂)t]

Table 2: Maxima and Minima in Beats

Point
Condition / Time
Maxima condition
cos[π(n₁ − n₂)t] = ±1
Maxima time
t = n / |n₁ − n₂|
Minima condition
cos[π(n₁ − n₂)t] = 0
Minima time
t = (2n − 1) / 2|n₁ − n₂|
First maxima
t = 0
First minima
t = 1 / 2|n₁ − n₂|
❖ Important Points:
  • •
    If beat frequency > 10 Hz, beats are not clearly heard
  • •
    Tuning fork and Quincke's tube show interference of sound
  • •
    If two waves have equal amplitude but different frequency, resultant amplitude varies from 0 to 2A
  • •
    For equal-intensity waves, maximum intensity during beat = 4I
  • •
    Loading a tuning fork decreases frequency
  • •
    Filing/cutting prongs increases frequency
▢ Tuning Fork:

Table 1: Tuning Fork Facts

Fact
Answer
Antinodes
4
Nodes
3
Material
Elinvar
Sound
Single frequency
Prongs
Transverse stationary waves
Stem
Longitudinal stationary waves
Prongs vibrate
Opposite phase
Frequency on loading
Decreases
Frequency on filing/cutting
Increases
Frequency relation
n = m t / (4√3 ρ L²) × √(Y/ρ)
❖ Vibration Note: When one prong is cut, centre of gravity changes; if stem is held rigidly, it vibrates with greater amplitude compared to two-prong fork
▢ Echo:
❖ Definition: Sound heard after reflection from obstacle

Table 1: Echo Conditions

Quantity
Formula / Value
Persistence of hearing
1/10 sec
Minimum obstacle distance
x = v/20
For air, v = 330 m/s
x = 16.5 m
Monosyllabic sound duration
0.2 sec
Sharp echo condition
Reflected sound should reach after at least 1/10 sec
▢ Reverberation:
❖ Definition: Persistence of sound due to multiple reflections in a room/hall
❖ Condition: If reflected sound reaches observer within 0.1 sec after original sound, sound is prolonged and separate echo is not heard
❖ Good Audibility: Reverberation time should be properly controlled
▢ Diffraction:
❖ Definition: Bending of wave around obstacle or edge of aperture

Table 1: Diffraction Facts

Fact
Answer
Condition
Wavelength comparable to aperture/obstacle size
Sound diffraction
Observed in daily life
Reason
Sound wavelength is of order metre
Light diffraction
Less common in daily life
Reason
Light wavelength is very small
Radio waves
Can be detected but light cannot behind large obstacle
Thin edge of blade/window/door
Light may diffract
▢ Interference and Diffraction Extra Facts:

Table 1: Read and Digest

Fact
Answer
Interference
Can take place in all waves
Coherent sources
Same frequency and constant initial phase difference
Incoherent sources
Cannot produce sustained interference
Two candles
Do not produce interference pattern
Maxima
Equally separated
Minima
Equally separated
Interference energy
Neither created nor destroyed, only redistributed
Longitudinal stationary wave
Pressure changes at antinodes
Hydrogen for sound
Acts like concave lens because sound speed is greater in H₂
Stethoscope / speaking tube / whispering gallery
Based on reflection of sound
▢ Lissajous Figures:
❖ Definition: Resultant path when two waves propagate along perpendicular axes with same frequency

Table 1: Lissajous Figure Conditions

Phase difference
Result
0
Straight line
π/2
Circle
π/4
Ellipse
2π/3 or 120°
Ellipse
▢ High-Yield Recall:

Table 1: Superposition One-Liners

Fact
Answer
Superposition principle
y = y₁ + y₂ + ...
Reflection at free end
No phase change
Reflection at rigid end
Phase reversal π
Constructive interference
Δφ = 2nπ, Δx = nλ
Destructive interference
Δφ = (2n − 1)π, Δx = (2n − 1)λ/2
Beat frequency
|n₁ − n₂|
Beat period
1/|n₁ − n₂|
Detectable beats
Frequency difference ≤ 10 Hz
Loading tuning fork
Frequency decreases
Filing tuning fork
Frequency increases
Echo minimum distance
v/20 = 16.5 m in air
Diffraction
Bending of wave around obstacle
Interference requires
Coherent sources
Lissajous straight line
Phase difference 0
Lissajous circle
Phase difference π/2
Q1.
A tuning fork A of unknown frequency gives 4 beats/sec when sounded with another fork B of frequency 256 Hz. The fork A is now loaded with a piece of wax and again 4 beats/sec are heard. The frequency of fork A is
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Q2.
Two tuning forks have frequencies 440 Hz and 444 Hz. The resultant frequency is
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Q3.
The waves of length 50 cm and 51 cm produce 12 beats per second. The velocity of sound is
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Q4.
Two tuning forks of frequencies 252 Hz and 256 Hz are sounded simultaneously. The number of beats heard per second is
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Q5.
Two tuning forks A and B sounded together produce 5 beats. The frequency of B is 512 Hz. When fork A is filed and sounded together, the beat frequency increases. The frequency of A is
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Q6.
When two tuning forks of frequency 484 Hz and 486 Hz are sounded together, what will be the beat frequency?
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Q7.
For production of beats the radio should tune with
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Q8.
Sound waves of wavelengths 5 m and 6 m produce 33 beats in 3 seconds. The velocity of sound is
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Q9.
The phenomenon of beats is due to
Q10.
When the prongs of a tuning fork are cut, its frequency
Q11.
A wave is reflected from a free boundary. The change of phase on reflection will be
Q12.
The minimum distance of a reflector to hear the echo of a sharp sound in terms of speed of sound v is
Q13.
The ratio of intensities of two interfering waves is 4:1. Then the ratio of maximum to minimum intensity is
Q14.
Two plane waves of same frequency having intensities I and 4I are travelling in the same direction. The resultant intensity at minima is
Q15.
Two waves having intensities I and 9I produce interference. If the resultant intensity at a point is 7I, the phase difference between the two waves is
Q16.
If the difference of frequencies of two sounding sources is more than 10, then the beats
Q17.
Beats are produced by two progressive waves of equal amplitude. Maximum intensity at waxing is x times the intensity of each wave. The value of x is
Q18.
Two tuning forks of frequencies 252 Hz and 256 Hz are sounded simultaneously. The number of beats heard per second are
Q19.
There are three sources of sound of equal intensities but with frequencies 400, 401 and 404 Hz. The number of beats per second is
Q20.
A tuning fork A of unknown frequency gives 4 beats/sec when sounded with another fork B of frequency 256 Hz. The fork A is now loaded with wax and again 4 beats/sec are heard. Then frequency of fork A after loading is
Q21.
Tuning fork X of frequency 258 Hz gives 8 beats/sec with tuning fork Y. When Y's prongs are cut a little and sounded again, the number of beats remains same. The frequency of Y before cutting the prongs is
Q22.
Two tuning forks have frequencies 450 Hz and 454 Hz. On sounding these forks together, the time interval between successive maximum intensities will be
Q23.
Ten tuning forks are arranged in increasing order of frequency in such a way that any two nearest forks produce 4 beats/sec. The highest frequency is twice that of the lowest. Possible lowest and highest frequencies in Hz are
Q24.
A tuning fork A of frequency 200 Hz is sounded with fork B. The number of beats per second is 5. By putting some wax on A, the number of beats increases to 8. The frequency of fork B is
Q25.
If two waves of same frequency and same amplitude superpose to produce a resultant disturbance of same amplitude, the waves differ in phase by
Q26.
A source of frequency n gives 5 beats/s when sounded with a source of frequency 200 Hz. The second harmonic 2n gives 10 beats/s when sounded with a source of frequency 420 Hz. Then n is equal to
Q27.
Two waves y₁ = 0.25 sin 320πt and y₂ = 0.25 sin 326πt are travelling in the same direction. The number of beats produced per second will be
Q28.
Two waves of equal amplitudes a each and equal frequency travel in same direction in a medium. The amplitude of resultant wave in the medium is
Q29.
Two periodic waves of intensities I₁ and I₂ travel in a medium simultaneously in same direction. Difference of maximum intensity and minimum intensity will be
Q30.
Two vibrating bodies have frequencies 252 Hz and 256 Hz. The number of beats produced per minute is
Q31.
Two waves y₁ = a sin[(2π/λ)(vt − x)] and y₂ = a cos[(2π/λ)(vt − x)] are superposed. The resultant wave has amplitude
Q32.
Beats are a result of
Q33.
When two tuning forks A and B are sounded together, x beats/sec are heard. When one prong of B is loaded with a little wax, number of beats/sec decreases. If frequency of A is n, then frequency of B will be
Q34.
Two sound waves of equal intensity produce beats. The maximum intensity of sound produced in beats will be
Q35.
25 tuning forks are arranged in series with decreasing frequency 3 beats/sec. If frequency of last tuning fork is octave of first tuning fork, the frequency of first tuning fork is
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Q36.
Two waves of lengths 50 cm and 51 cm produce 12 beats per second. The velocity of sound is
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Q37.
Two tuning forks of frequencies 250 Hz and 256 Hz produce beats. If a maximum is produced just now, after how much time will the minimum be produced at the same place?
Q38.
Consider 10 identical sources of sound all giving same frequency but having random phase angles. If average intensity of each source is I₀, the average resultant intensity due to all these 10 sources will be
Q39.
A set of 28 tuning forks is arranged in series of decreasing frequencies. Each fork gives 3 beats with succeeding one. First fork is octave of the last. Frequencies of first and 15th tuning forks are
Q40.
Two equations of progressive waves are y₁ = a sin(ωt − 0.1x) and y₂ = a sin[(ωt − 0.1x) + φ₀/2]. The resulting amplitude of the superposing waves is
Q41.
The two equations of waves are y₁ = 10 sin(3πt + π/3), y₂ = 5[sin3πt + √3 cos3πt]. The ratio of their amplitudes is
Q42.
Three sources of sound of equal frequencies having amplitudes 10 mm, 4 mm and 7 mm respectively are superposing with successive phase difference π/2. Resultant amplitude due to superposition of three waves is
Q43.
Two simple harmonic wave equations are represented as y₁ = 5 sin104πt and y₂ = 5 sin100πt. The resultant amplitude of two superposing waves at time 1/4 sec is
Q44.
Two tuning forks A and B when sounded together produce 4 beats/sec. Frequency of fork A is 256 Hz. If fork B is loaded, beat frequency increases. Frequency of fork B before loading is
Q45.
Two tuning forks A and B when sounded together produce 4 beats. Frequency of A is 256 Hz. If fork B is loaded, beat frequency increases to 6 beats. Frequency of B after loading is
Q46.
The amplitude of superposition of two waves y₁ = 5 sinωt and y₂ = 5 cosωt is
Q47.
If amplitude ratio of two sources producing interference is 3:5, the ratio of intensities at maxima and minima is
Q48.
Two sounding bodies produce progressive waves as y₁ = 4 sin400πt and y₂ = 3 sin404πt. An observer will hear
Q49.
Two coherent sound waves of equal frequency traverse two paths S₁P and S₂P to reach point P. If S₂P − S₁P = 5λ/2, then point P is at
Q50.
A tuning fork vibrating with a sonometer having 20 cm wire produces 5 beats/sec. The beat frequency does not change if the length of wire is changed to 21 cm. Frequency of the tuning fork is
Q51.
The motion of a particle is given by x = A sinωt + B cosωt. The motion of the particle is
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