Wave Speed
Wave speed depends on the medium. For periodic waves use v = fλ; for string waves use v = √(T/μ); for sound in fluid use v = √(B/ρ).
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Wave Speed and Superposition
Class 11 Physics notes covering wave speed, wave velocity, principle of superposition, interference, resultant displacement, numericals and PYQs.
CBSENEETJEE MainJEE AdvancedIBICSEIGCSEA-Level
Superposition
When waves overlap, add displacement at each point: y = y1 + y2. This is the foundation of interference, beats, standing waves and wave optics.
Interference
Same-phase waves produce constructive interference; opposite-phase waves produce destructive interference. Always check phase before adding amplitudes.
Concept
Introduction
Wave speed and superposition form the bridge between simple wave description and real wave behaviour. Wave speed tells how fast the disturbance travels, while superposition tells what happens when two or more waves meet at the same place.
In coaching-class language, always keep three separate ideas in mind: the wave pattern travels with wave velocity, the medium particle vibrates with particle velocity, and when two disturbances overlap the actual displacement is the algebraic sum of individual displacements.
Real-life example: when two water ripples cross, each ripple continues after crossing, but during overlap the water level at every point is the sum of the two ripple displacements. That is superposition in front of your eyes.
Examination perspective: CBSE asks definitions and formulas, NEET checks quick use of v = fλ and interference conditions, while JEE often mixes signs, direction, phase and resultant amplitude.
Exam FocusMemory trick: speed tells how fast a wave goes; superposition tells what waves do when they meet.
Exam FocusConceptual trap: waves pass through each other, but displacement during overlap changes.
Concept
Speed of Wave
Wave speed is the distance travelled by a wave disturbance per unit time. For any periodic wave, the most used formula is v = fλ. Here v is wave speed, f is frequency and λ is wavelength.
Physical meaning: in one time period T, a crest advances by one wavelength λ. Therefore v = λ/T = fλ. The source decides frequency, the medium decides speed, and wavelength adjusts according to v = fλ.
For a wave on a stretched string, v = √(T/μ). T is tension in the string and μ is mass per unit length. Greater tension makes the wave faster; greater linear density makes it slower.
For longitudinal waves in a fluid, v = √(B/ρ). B is bulk modulus and ρ is density. Greater elasticity increases speed, while greater density decreases speed.
Real-life example: a tight guitar string transmits waves faster and gives a higher pitch for a fixed length; a heavier string transmits waves more slowly.
Common mistake: saying wave speed depends on amplitude. In ordinary linear waves, amplitude changes energy, not wave speed.
Exam FocusSpeed depends on medium properties.
Exam FocusFor string waves: more tension means larger speed.
Exam FocusFor sound: larger bulk modulus means faster sound.
Concept
Wave Velocity
Wave velocity is wave speed with direction. In one-dimensional wave motion along a string, positive wave velocity means the phase pattern moves in positive x-direction, and negative wave velocity means it moves in negative x-direction.
Mathematically, for y = A sin(ωt − kx), constant phase gives x increasing with t, so the wave travels in positive x-direction with speed v = ω/k.
Particle velocity is different. Particle velocity is dy/dt, the speed of a point of the medium. Wave velocity is the speed of the crest, compression or phase.
Real-life example: a stadium wave travels around the stadium, but each person only stands and sits locally. The wave velocity is around the stadium; the particle velocity is each person's up-down motion.
JEE conceptual trap: maximum particle velocity can be Aω, but wave velocity is ω/k. They can have different values and different directions.
Exam FocusWave velocity = phase velocity.
Exam FocusParticle velocity = local vibration velocity.
Exam FocusDo not use dy/dt for wave speed.
Concept
Principle of Superposition
The principle of superposition states that when two or more waves overlap in a medium, the resultant displacement of any particle is equal to the algebraic sum of the displacements produced by individual waves at that point.
Mathematical meaning: y = y1 + y2. If one wave displaces a particle upward by 3 mm and another displaces it downward by 1 mm, resultant displacement is 2 mm upward.
Physical meaning: each wave produces its own displacement, and the medium particle shows the combined effect. After crossing, waves continue almost unchanged in a linear medium.
Vector and algebraic addition: displacement is directional. Upward displacement may be taken positive and downward negative. For waves along the same line, add signs carefully.
Real-life example: two pulses on a rope overlap. If both are upward, the rope rises more; if one is upward and one downward, the rope may partly or completely cancel.
Exam FocusSuperposition works for linear waves.
Exam FocusAlways add displacements, not amplitudes blindly.
Exam FocusMemory trick: resultant y is the total y.
Concept
Interference Basic Introduction
Interference is the sustained effect produced by superposition of two coherent waves. In basic Class 11 wave study, we focus on constructive and destructive interference.
Constructive interference occurs when waves meet in the same phase. Resultant amplitude is maximum: A = A1 + A2.
Destructive interference occurs when waves meet in opposite phase. Resultant amplitude is minimum: A = |A1 − A2|. If A1 = A2, complete cancellation occurs.
Real-life example: noise-cancelling headphones produce a sound wave nearly opposite in phase to external noise, reducing resultant sound reaching the ear.
Conceptual trap: destructive interference does not destroy energy in the universe; energy redistributes in space or transforms depending on the system.
Exam FocusSame phase gives maximum amplitude.
Exam FocusOpposite phase gives minimum amplitude.
Exam FocusInterference needs phase relation.
Concept
Resultant Displacement
Resultant displacement is the actual displacement of a particle when multiple waves act at the same point. It is calculated point by point and instant by instant.
If y1 = A1 sin(ωt − kx) and y2 = A2 sin(ωt − kx), then the resultant can have amplitude A1 + A2 when waves are in phase.
Amplitude changes during superposition because displacements add. Energy depends on amplitude squared, so changing amplitude changes local energy distribution.
Real-life example: in a concert hall, sound may be loud at some seats and weak at others because waves from speakers interfere differently at different points.
Common mistake: adding intensities or amplitudes without checking phase. In wave displacement questions, first add displacement functions.
Exam FocusResultant displacement is not always resultant amplitude.
Exam FocusA graph must be added point by point.
Exam FocusEnergy follows amplitude squared, not amplitude linearly.
Concept
Applications
Noise cancellation uses destructive interference. A microphone detects external noise and electronics generate an opposite-phase wave, reducing the resultant displacement of air near the ear.
Sound systems use superposition in speaker arrays. Correct spacing and phase make sound stronger in desired regions and weaker in unwanted regions.
Music depends on superposition of harmonics. A violin, flute and guitar may play the same fundamental frequency, but different harmonic mixtures give different quality or timbre.
Wave engineering uses interference and superposition in vibration control, bridges, buildings, ultrasound, seismology and antenna design.
Radio communication and signal transmission use superposition when multiple signals share space. Receivers select the required frequency and filter out unwanted components.
Echo and reflection problems use wave speed and time delay. If sound travels to a wall and returns, distance to wall is vt/2, not vt.
Exam FocusReal applications are mostly controlled superposition.
Exam FocusEcho questions need round trip distance.
Exam FocusCommunication systems depend on waves adding and filtering.
Formula Sheet
Premium Formula Boxes
v = fλv is wave speed, f is frequency and λ is wavelength. Use this whenever frequency and wavelength are given.
v = √(T/μ)T is tension and μ is linear mass density. More tension increases speed; more mass per metre decreases speed.
v = √(B/ρ)B is bulk modulus and ρ is density. Elastic medium transmits faster; denser medium resists motion.
y = y1 + y2Add displacement functions algebraically with signs. Do not add only magnitudes.
A = A1 + A2Used when phase difference is 0, 2π, 4π and so on.
A = |A1 − A2|Used when phase difference is π, 3π, 5π and so on.
NCERT Style
Wave Speed and Superposition Diagrams
These SVG diagrams use black wave lines, red labels and red arrows in a clean board-work style.
Wave Speed on String
Wave Velocity Representation
Superposition Principle
Constructive Interference
Destructive Interference
Resultant Displacement
Need Personal Help in Superposition?
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Solved Bank
93 Diverse Solved Numericals
This bank is intentionally curated to avoid data-only duplicate questions. It includes NEET, JEE Main and JEE Advanced style numericals across wave speed, string tension, density, elasticity, echo, sonar, graph reading, phase/path difference, constructive interference, destructive interference, resultant displacement, energy ratio and conceptual superposition.
Numerical 1NEET Easy
A ripple tank shows 5 crests crossing a mark every second. Distance between consecutive crests is 0.12 m. Find wave speed.
Show Solution
Given: f = 5 Hz, λ = 0.12 m
Formula: v = fλ
Solution: v = 5 × 0.12 = 0.60 m s−1
Final Answer: 0.60 m s−1
Numerical 2NEET Easy
A sound wave travels at 340 m s−1 and has wavelength 0.85 m. Find frequency.
Show Solution
Given: v = 340 m s−1, λ = 0.85 m
Formula: f = v/λ
Solution: f = 340/0.85 = 400 Hz
Final Answer: 400 Hz
Numerical 3NEET Easy
A water wave has speed 1.8 m s−1 and frequency 3 Hz. Find wavelength.
Show Solution
Given: v = 1.8 m s−1, f = 3 Hz
Formula: λ = v/f
Solution: λ = 1.8/3 = 0.60 m
Final Answer: 0.60 m
Numerical 4NEET Easy
A crest takes 0.04 s to move to the position of the next crest. If wavelength is 2 m, find speed.
Show Solution
Given: T = 0.04 s, λ = 2 m
Formula: v = λ/T
Solution: v = 2/0.04 = 50 m s−1
Final Answer: 50 m s−1
Numerical 5NEET Easy
A sound echo is heard after 1.2 s. If sound speed is 340 m s−1, find distance of the wall.
Show Solution
Given: t = 1.2 s, v = 340 m s−1
Formula: d = vt/2
Solution: d = 340 × 1.2/2 = 204 m
Final Answer: 204 m
Numerical 6NEET Easy
An ultrasonic pulse returns from a fish after 0.08 s in water. If speed is 1500 m s−1, find depth.
Show Solution
Given: t = 0.08 s, v = 1500 m s−1
Formula: depth = vt/2
Solution: depth = 1500 × 0.08/2 = 60 m
Final Answer: 60 m
Numerical 7NEET Easy
A string is under tension 25 N and has linear density 0.01 kg m−1. Find wave speed.
Show Solution
Given: T = 25 N, μ = 0.01 kg m−1
Formula: v = √(T/μ)
Solution: v = √(25/0.01) = 50 m s−1
Final Answer: 50 m s−1
Numerical 8NEET Easy
Two upward pulses produce displacements 3 mm and 5 mm at the same point. Find resultant displacement.
Show Solution
Given: y1 = +3 mm, y2 = +5 mm
Formula: y = y1 + y2
Solution: y = 3 + 5 = 8 mm
Final Answer: 8 mm upward
Numerical 9NEET Easy
One pulse displaces a point 6 mm upward and another displaces it 2 mm downward. Find resultant displacement.
Show Solution
Given: y1 = +6 mm, y2 = −2 mm
Formula: y = y1 + y2
Solution: y = 6 − 2 = 4 mm
Final Answer: 4 mm upward
Numerical 10NEET Easy
Two equal pulses of amplitude 4 cm meet in opposite phase. Find resultant amplitude.
Show Solution
Given: A1 = 4 cm, A2 = 4 cm, opposite phase
Formula: A = |A1 − A2|
Solution: A = |4 − 4| = 0
Final Answer: 0 cm
Numerical 11NEET Medium
A string wave speed is 40 m s−1. If tension is made four times and μ is unchanged, find new speed.
Show Solution
Given: v = 40 m s−1, T' = 4T
Formula: v ∝ √T
Solution: v' = 40 × √4 = 80 m s−1
Final Answer: 80 m s−1
Numerical 12NEET Medium
A string wave speed is 60 m s−1. If linear density becomes 9 times at same tension, find new speed.
Show Solution
Given: v = 60 m s−1, μ' = 9μ
Formula: v ∝ 1/√μ
Solution: v' = 60/3 = 20 m s−1
Final Answer: 20 m s−1
Numerical 13NEET Medium
A fluid has bulk modulus 2.25 × 109 Pa and density 1000 kg m−3. Find sound speed.
Show Solution
Given: B = 2.25 × 109 Pa, ρ = 1000 kg m−3
Formula: v = √(B/ρ)
Solution: v = √(2.25 × 106) = 1500 m s−1
Final Answer: 1500 m s−1
Numerical 14NEET Medium
A wave in medium A has speed 300 m s−1. In medium B wavelength becomes 1.5 times while frequency is unchanged. Find speed in B.
Show Solution
Given: vA = 300 m s−1, λB = 1.5λA
Formula: v = fλ
Solution: vB = 1.5vA = 450 m s−1
Final Answer: 450 m s−1
Numerical 15NEET Medium
A graph shows two consecutive crests separated by 0.75 m and 12 crests pass a point in 3 s. Find speed.
Show Solution
Given: λ = 0.75 m, f = 12/3 = 4 Hz
Formula: v = fλ
Solution: v = 4 × 0.75 = 3 m s−1
Final Answer: 3 m s−1
Numerical 16NEET Medium
A sound of frequency 500 Hz travels from air to water. Frequency remains 500 Hz and speed becomes 1500 m s−1. Find wavelength in water.
Show Solution
Given: f = 500 Hz, v = 1500 m s−1
Formula: λ = v/f
Solution: λ = 1500/500 = 3 m
Final Answer: 3 m
Numerical 17NEET Medium
Two waves of amplitudes 7 cm and 2 cm meet in same phase. Find resultant amplitude.
Show Solution
Given: A1 = 7 cm, A2 = 2 cm
Formula: A = A1 + A2
Solution: A = 7 + 2 = 9 cm
Final Answer: 9 cm
Numerical 18NEET Medium
Two waves of amplitudes 7 cm and 2 cm meet in opposite phase. Find resultant amplitude.
Show Solution
Given: A1 = 7 cm, A2 = 2 cm
Formula: A = |A1 − A2|
Solution: A = |7 − 2| = 5 cm
Final Answer: 5 cm
Numerical 19NEET Medium
A path difference is 3λ. Decide interference type.
Show Solution
Given: Δx = 3λ
Formula: Constructive when Δx = nλ
Solution: 3λ is an integral multiple of λ, so interference is constructive.
Final Answer: Constructive interference
Numerical 20NEET Medium
A path difference is 5λ/2. Decide interference type.
Show Solution
Given: Δx = 5λ/2
Formula: Destructive when Δx = (n + 1/2)λ
Solution: 5λ/2 = 2.5λ, a half-integral multiple; destructive.
Final Answer: Destructive interference
Numerical 21NEET Medium
Two waves produce y1 = 4 sinθ and y2 = 4 sinθ at a point. Find resultant displacement expression.
Show Solution
Given: y1 = 4 sinθ, y2 = 4 sinθ
Formula: y = y1 + y2
Solution: y = 8 sinθ
Final Answer: 8 sinθ
Numerical 22NEET Medium
Two waves produce y1 = 5 sinθ and y2 = −2 sinθ. Find resultant.
Show Solution
Given: y1 = 5 sinθ, y2 = −2 sinθ
Formula: y = y1 + y2
Solution: y = 3 sinθ
Final Answer: 3 sinθ
Numerical 23NEET Medium
If amplitude of a wave is doubled, by what factor does energy become in the same medium?
Show Solution
Given: A' = 2A
Formula: Energy ∝ A2
Solution: E'/E = (2A/A)2 = 4
Final Answer: 4 times
Numerical 24NEET Medium
A speaker system creates opposite phase sound of equal amplitude to a noise. What is resultant amplitude?
Show Solution
Given: Anoise = Acancel, phase difference = π
Formula: A = |A1 − A2|
Solution: A = 0
Final Answer: Zero resultant amplitude ideally
Numerical 25NEET Medium
A wave travels 24 m in 0.6 s on a rope. Find speed and distance covered in the next 2 s.
Show Solution
Given: s = 24 m, t = 0.6 s
Formula: v = s/t
Solution: v = 24/0.6 = 40 m s−1; distance in 2 s = 80 m
Final Answer: 40 m s−1, 80 m
Numerical 26NEET Medium
A boat receives 10 water crests in 5 s. The distance between first and third crest is 4 m. Find wave speed.
Show Solution
Given: f = 10/5 = 2 Hz; distance from first to third crest = 2λ = 4 m
Formula: λ = 2 m, v = fλ
Solution: v = 2 × 2 = 4 m s−1
Final Answer: 4 m s−1
Numerical 27NEET Difficult
A wave speed on a string is 100 m s−1 at tension 40 N. Find μ.
Show Solution
Given: v = 100 m s−1, T = 40 N
Formula: v = √(T/μ), so μ = T/v2
Solution: μ = 40/10000 = 0.004 kg m−1
Final Answer: 0.004 kg m−1
Numerical 28NEET Difficult
Find tension required for speed 80 m s−1 on a string of μ = 0.02 kg m−1.
Show Solution
Given: v = 80 m s−1, μ = 0.02 kg m−1
Formula: T = μv2
Solution: T = 0.02 × 802 = 128 N
Final Answer: 128 N
Numerical 29NEET Difficult
A fluid has sound speed 1200 m s−1 and density 900 kg m−3. Find bulk modulus.
Show Solution
Given: v = 1200 m s−1, ρ = 900 kg m−3
Formula: B = ρv2
Solution: B = 900 × 12002 = 1.296 × 109 Pa
Final Answer: 1.296 × 109 Pa
Numerical 30NEET Difficult
Two sources produce waves of amplitudes 6 cm and 8 cm. State maximum and minimum resultant amplitudes.
Show Solution
Given: A1 = 6 cm, A2 = 8 cm
Formula: Amax = A1 + A2, Amin = |A1 − A2|
Solution: Amax = 14 cm; Amin = 2 cm
Final Answer: 14 cm and 2 cm
Numerical 31JEE Main Concept
For a travelling wave y = A sin(ωt − kx), what is wave velocity in terms of ω and k?
Show Solution
Given: Equation y = A sin(ωt − kx)
Formula: v = ω/k
Solution: Wave velocity is phase speed, so v = ω/k.
Final Answer: ω/k
Numerical 32JEE Main Concept
A wave is y = 0.02 sin(80t − 4x). Find wave velocity.
Show Solution
Given: ω = 80 rad s−1, k = 4 rad m−1
Formula: v = ω/k
Solution: v = 80/4 = 20 m s−1
Final Answer: 20 m s−1
Numerical 33JEE Main Concept
A string wave has equation y = 0.01 sin(50t − 2x). Find frequency and wavelength.
Show Solution
Given: ω = 50, k = 2
Formula: f = ω/2π, λ = 2π/k
Solution: f = 25/π Hz; λ = π m
Final Answer: 25/π Hz, π m
Numerical 34JEE Main Concept
A pulse on a string travels 6 m right while a particle of the string completes one oscillation. What is wavelength?
Show Solution
Given: One oscillation time = T; crest advances one λ in T
Formula: distance in one T = λ
Solution: λ = 6 m
Final Answer: 6 m
Numerical 35JEE Main Formula
If tension is increased by 21% on a string, find percentage increase in wave speed approximately.
Show Solution
Given: T' = 1.21T
Formula: v ∝ √T
Solution: v'/v = √1.21 = 1.1, so increase = 10%
Final Answer: 10%
Numerical 36JEE Main Formula
If linear density decreases by 36% at same tension, find speed ratio.
Show Solution
Given: μ' = 0.64μ
Formula: v ∝ 1/√μ
Solution: v'/v = 1/√0.64 = 1/0.8 = 1.25
Final Answer: 1.25
Numerical 37JEE Main Formula
A 2 m string has mass 0.04 kg and tension 100 N. Find wave speed.
Show Solution
Given: m = 0.04 kg, L = 2 m, T = 100 N
Formula: μ = m/L, v = √(T/μ)
Solution: μ = 0.02 kg m−1; v = √(100/0.02) = 70.71 m s−1
Final Answer: 70.71 m s−1
Numerical 38JEE Main Formula
A string of μ = 0.05 kg m−1 supports speed 30 m s−1. Find tension.
Show Solution
Given: μ = 0.05, v = 30
Formula: T = μv2
Solution: T = 0.05 × 900 = 45 N
Final Answer: 45 N
Numerical 39JEE Main Formula
Two strings have same tension. Their linear densities are in ratio 1:4. Find speed ratio.
Show Solution
Given: μ1:μ2 = 1:4
Formula: v ∝ 1/√μ
Solution: v1:v2 = √4:√1 = 2:1
Final Answer: 2:1
Numerical 40JEE Main Formula
Two strings have same μ. Their tensions are in ratio 9:16. Find wave speed ratio.
Show Solution
Given: T1:T2 = 9:16
Formula: v ∝ √T
Solution: v1:v2 = 3:4
Final Answer: 3:4
Numerical 41JEE Main Graph
A y-x graph has crest spacing 0.4 m. A y-t graph at one point has crest spacing 0.02 s. Find wave speed.
Show Solution
Given: λ = 0.4 m, T = 0.02 s
Formula: v = λ/T
Solution: v = 0.4/0.02 = 20 m s−1
Final Answer: 20 m s−1
Numerical 42JEE Main Graph
A displacement-time graph repeats every 5 ms. If wavelength is 1.5 m, find speed.
Show Solution
Given: T = 5 ms = 0.005 s, λ = 1.5 m
Formula: v = λ/T
Solution: v = 1.5/0.005 = 300 m s−1
Final Answer: 300 m s−1
Numerical 43JEE Main Echo
A bat emits ultrasound and receives reflected signal after 0.03 s from an insect. Speed of sound is 340 m s−1. Find insect distance.
Show Solution
Given: t = 0.03 s, v = 340 m s−1
Formula: d = vt/2
Solution: d = 340 × 0.03/2 = 5.1 m
Final Answer: 5.1 m
Numerical 44JEE Main Echo
A sonar signal takes 0.4 s for return from sea bed. Speed in water is 1500 m s−1. Find depth.
Show Solution
Given: t = 0.4 s, v = 1500 m s−1
Formula: d = vt/2
Solution: d = 1500 × 0.4/2 = 300 m
Final Answer: 300 m
Numerical 45JEE Main Interference
Two waves have amplitudes 10 units and 6 units. Find possible range of resultant amplitude.
Show Solution
Given: A1 = 10, A2 = 6
Formula: A ranges from |A1 − A2| to A1 + A2
Solution: Minimum = 4, maximum = 16
Final Answer: 4 to 16 units
Numerical 46JEE Main Interference
Two waves of amplitudes 5 cm and 12 cm interfere. Can resultant amplitude be 3 cm?
Show Solution
Given: A1 = 5 cm, A2 = 12 cm
Formula: Allowed range: |A1 − A2| to A1 + A2
Solution: Range is 7 cm to 17 cm. 3 cm is not possible.
Final Answer: No
Numerical 47JEE Main Interference
Two coherent waves arrive with phase difference 2π. If each amplitude is 4 cm, find resultant amplitude.
Show Solution
Given: A1 = A2 = 4 cm, Δφ = 2π
Formula: Constructive: A = A1 + A2
Solution: A = 8 cm
Final Answer: 8 cm
Numerical 48JEE Main Interference
Two coherent waves arrive with phase difference π. If each amplitude is 4 cm, find resultant amplitude.
Show Solution
Given: A1 = A2 = 4 cm, Δφ = π
Formula: Destructive: A = |A1 − A2|
Solution: A = 0
Final Answer: 0
Numerical 49JEE Main Interference
A path difference of 7λ/2 is observed. Identify interference type.
Show Solution
Given: Δx = 7λ/2 = 3.5λ
Formula: Half-integral multiple gives destructive interference
Solution: 3.5λ is half-integral, so destructive.
Final Answer: Destructive
Numerical 50JEE Main Interference
A path difference of 6λ is observed. Identify interference type.
Show Solution
Given: Δx = 6λ
Formula: Integral multiple gives constructive interference
Solution: 6λ is integral multiple, so constructive.
Final Answer: Constructive
Numerical 51JEE Main Displacement
At an instant y1 = 3 cm, y2 = −4 cm and y3 = 2 cm. Find resultant displacement.
Show Solution
Given: 3 cm, −4 cm, 2 cm
Formula: y = y1 + y2 + y3
Solution: y = 3 − 4 + 2 = 1 cm
Final Answer: 1 cm
Numerical 52JEE Main Displacement
Two waves are y1 = 2 sinθ and y2 = 2 cosθ. Find resultant amplitude.
Show Solution
Given: Equal perpendicular phase components
Formula: R = √(22 + 22)
Solution: R = 2√2
Final Answer: 2√2
Numerical 53JEE Main Displacement
Two waves are y1 = 3 sinθ and y2 = 4 cosθ. Find resultant amplitude.
Show Solution
Given: Components 3 and 4 in quadrature
Formula: R = √(32 + 42)
Solution: R = 5
Final Answer: 5
Numerical 54JEE Main Medium
In a liquid, bulk modulus is made 4 times while density stays same. Find speed ratio.
Show Solution
Given: B' = 4B, ρ unchanged
Formula: v ∝ √B
Solution: v'/v = 2
Final Answer: 2
Numerical 55JEE Main Medium
In a gas model, density becomes 4 times while elasticity stays same. Find speed ratio.
Show Solution
Given: ρ' = 4ρ, B unchanged
Formula: v ∝ 1/√ρ
Solution: v'/v = 1/2
Final Answer: 1/2
Numerical 56JEE Main Mixed
A string wave has speed 60 m s−1 and frequency 120 Hz. Find distance between two nearest particles in same phase.
Show Solution
Given: v = 60, f = 120
Formula: λ = v/f
Solution: λ = 60/120 = 0.5 m
Final Answer: 0.5 m
Numerical 57JEE Main Mixed
A string wave has speed 60 m s−1 and frequency 120 Hz. Find distance between nearest opposite phase particles.
Show Solution
Given: λ = 0.5 m
Formula: Opposite phase separation = λ/2
Solution: distance = 0.25 m
Final Answer: 0.25 m
Numerical 58JEE Main Mixed
Two loudspeakers produce same frequency. At a point path difference is 0.75 m and wavelength is 0.5 m. Identify interference.
Show Solution
Given: Δx = 0.75 m, λ = 0.5 m
Formula: Δx/λ = 1.5
Solution: 1.5λ is half-integral, so destructive.
Final Answer: Destructive
Numerical 59JEE Main Mixed
At another point path difference is 1.0 m and wavelength is 0.5 m. Identify interference.
Show Solution
Given: Δx = 1.0 m, λ = 0.5 m
Formula: Δx/λ = 2
Solution: 2λ is integral, so constructive.
Final Answer: Constructive
Numerical 60JEE Advanced Conceptual
A wave pulse on a string becomes faster after the string is tightened, but frequency of a driven periodic wave remains fixed. Which quantity adjusts?
Show Solution
Given: Frequency fixed by source, speed changes due to tension
Formula: v = fλ
Solution: Since f is fixed and v changes, wavelength adjusts.
Final Answer: Wavelength
Numerical 61JEE Advanced Conceptual
For two waves y1 = A sin(ωt − kx) and y2 = A sin(ωt − kx + π), find resultant.
Show Solution
Given: Equal amplitudes, phase difference π
Formula: sin(θ + π) = −sinθ
Solution: y = A sinθ − A sinθ = 0
Final Answer: 0
Numerical 62JEE Advanced Conceptual
For y1 = A sinθ and y2 = A sin(θ + 2π), find resultant amplitude.
Show Solution
Given: Phase difference 2π
Formula: Same phase
Solution: AR = A + A = 2A
Final Answer: 2A
Numerical 63JEE Advanced Conceptual
Two waves of amplitudes A and 2A meet with opposite phase. Find resultant amplitude.
Show Solution
Given: A1 = A, A2 = 2A
Formula: AR = |A1 − A2|
Solution: AR = A
Final Answer: A
Numerical 64JEE Advanced Conceptual
Two waves of amplitudes A and 2A meet with same phase. Find intensity ratio of resultant to first wave, assuming intensity ∝ amplitude2.
Show Solution
Given: AR = 3A
Formula: I ∝ A2
Solution: IR/I1 = (3A/A)2 = 9
Final Answer: 9
Numerical 65JEE Advanced Graph
A resultant graph is obtained by adding two identical in-phase sine waves. If each original amplitude is 3 cm, what is resultant crest height from mean?
Show Solution
Given: A = 3 cm each, same phase
Formula: AR = A1 + A2
Solution: AR = 6 cm
Final Answer: 6 cm
Numerical 66JEE Advanced Graph
A graph shows a crest of +5 cm crossing with a trough of −3 cm at the same x. Find instantaneous resultant displacement.
Show Solution
Given: +5 cm and −3 cm
Formula: y = y1 + y2
Solution: y = 5 − 3 = 2 cm
Final Answer: +2 cm
Numerical 67JEE Advanced Graph
Two opposite pulses have unequal amplitudes 9 cm and 4 cm. What pulse remains at maximum overlap?
Show Solution
Given: A1 = 9 cm, A2 = 4 cm opposite
Formula: A = |A1 − A2|
Solution: A = 5 cm in direction of larger pulse
Final Answer: 5 cm
Numerical 68JEE Advanced String
A string of length 5 m and mass 0.10 kg is stretched by 80 N. Find wave speed.
Show Solution
Given: L = 5 m, m = 0.10 kg, T = 80 N
Formula: μ = m/L, v = √(T/μ)
Solution: μ = 0.02; v = √(80/0.02) = √4000 = 63.25 m s−1
Final Answer: 63.25 m s−1
Numerical 69JEE Advanced String
A string's tension is changed from 50 N to 200 N. Initial wave speed is 30 m s−1. Find final speed.
Show Solution
Given: T changes by factor 4
Formula: v ∝ √T
Solution: v' = 30 × 2 = 60 m s−1
Final Answer: 60 m s−1
Numerical 70JEE Advanced String
A string is replaced by another with one-fourth linear density under same tension. Initial speed is 40 m s−1. Find new speed.
Show Solution
Given: μ' = μ/4
Formula: v ∝ 1/√μ
Solution: v' = 40 × 2 = 80 m s−1
Final Answer: 80 m s−1
Numerical 71JEE Advanced Fluid
A medium has density 1.2 kg m−3 and sound speed 330 m s−1. Find effective bulk modulus.
Show Solution
Given: ρ = 1.2, v = 330
Formula: B = ρv2
Solution: B = 1.2 × 3302 = 130680 Pa
Final Answer: 1.31 × 105 Pa
Numerical 72JEE Advanced Fluid
Two fluids have same density but bulk moduli in ratio 16:25. Find sound speed ratio.
Show Solution
Given: B1:B2 = 16:25
Formula: v ∝ √B
Solution: v1:v2 = 4:5
Final Answer: 4:5
Numerical 73JEE Advanced Fluid
Two fluids have same bulk modulus but densities in ratio 9:4. Find sound speed ratio.
Show Solution
Given: ρ1:ρ2 = 9:4
Formula: v ∝ 1/√ρ
Solution: v1:v2 = 1/3 : 1/2 = 2:3
Final Answer: 2:3
Numerical 74JEE Advanced Phase
Two coherent waves have phase difference π/3 and equal amplitude A. Find resultant amplitude.
Show Solution
Given: A, A, φ = π/3
Formula: R = 2A cos(φ/2)
Solution: R = 2A cos(π/6) = √3 A
Final Answer: √3 A
Numerical 75JEE Advanced Phase
Two coherent waves have phase difference 2π/3 and equal amplitude A. Find resultant amplitude.
Show Solution
Given: A, A, φ = 2π/3
Formula: R = 2A cos(φ/2)
Solution: R = 2A cos(π/3) = A
Final Answer: A
Numerical 76JEE Advanced Phase
Two coherent waves of equal amplitude A produce resultant amplitude √2 A. Find phase difference.
Show Solution
Given: R = √2 A
Formula: R = 2A cos(φ/2)
Solution: cos(φ/2) = 1/√2, so φ/2 = π/4
Final Answer: φ = π/2
Numerical 77JEE Advanced Phase
Two waves of amplitudes 3A and 4A are in quadrature. Find resultant amplitude.
Show Solution
Given: A1 = 3A, A2 = 4A, φ = π/2
Formula: R = √(A12 + A22)
Solution: R = √(9A2 + 16A2) = 5A
Final Answer: 5A
Numerical 78JEE Advanced Phase
Two waves of amplitudes 5 cm and 5 cm have phase difference 120°. Find resultant amplitude.
Show Solution
Given: A = 5 cm, φ = 120°
Formula: R = 2A cos(φ/2)
Solution: R = 10 cos60° = 5 cm
Final Answer: 5 cm
Numerical 79JEE Advanced Path
Two in-phase speakers have wavelength 0.8 m. At a point path difference is 1.2 m. Identify interference.
Show Solution
Given: Δx = 1.2 m, λ = 0.8 m
Formula: Δx/λ = 1.5
Solution: Half-integral multiple, so destructive.
Final Answer: Destructive
Numerical 80JEE Advanced Path
For wavelength 0.6 m, find the smallest non-zero path difference for destructive interference.
Show Solution
Given: Destructive: Δx = λ/2
Formula: Δx = 0.6/2
Solution: Δx = 0.3 m
Final Answer: 0.3 m
Numerical 81JEE Advanced Path
For wavelength 0.6 m, find the smallest non-zero path difference for constructive interference.
Show Solution
Given: Constructive: Δx = λ
Formula: Δx = 0.6 m
Solution: smallest non-zero = 0.6 m
Final Answer: 0.6 m
Numerical 82JEE Advanced Energy
Two equal waves interfere constructively. Compare resultant energy density with one wave if energy ∝ A2.
Show Solution
Given: AR = 2A
Formula: E ∝ A2
Solution: ER/E = (2A/A)2 = 4
Final Answer: 4 times one wave
Numerical 83JEE Advanced Energy
Two equal waves interfere destructively at a point. What is resultant displacement there?
Show Solution
Given: Equal amplitudes, opposite phase
Formula: y = y1 + y2
Solution: They cancel at that point, so y = 0.
Final Answer: 0
Numerical 84JEE Advanced Superposition
At a point, three waves have displacements +2A, −3A and +5A. Find resultant.
Show Solution
Given: +2A, −3A, +5A
Formula: y = sum of displacements
Solution: y = 2A − 3A + 5A = 4A
Final Answer: 4A
Numerical 85JEE Advanced Superposition
Two pulses overlap for 0.02 s while moving through each other. What happens after overlap in a linear string?
Show Solution
Given: Linear medium
Formula: Principle of superposition
Solution: They regain their original shapes and continue moving.
Final Answer: They pass through unchanged ideally
Numerical 86JEE Advanced Superposition
A pulse travels right at 10 m s−1 and another travels left at 15 m s−1. Initial separation is 5 m. When do they meet?
Show Solution
Given: Relative speed = 10 + 15 = 25 m s−1
Formula: t = separation/relative speed
Solution: t = 5/25 = 0.2 s
Final Answer: 0.2 s
Numerical 87JEE Advanced Superposition
Two pulses travelling in opposite directions have widths 0.3 m and 0.2 m and relative speed 10 m s−1. Estimate overlap time.
Show Solution
Given: Total width = 0.5 m, relative speed = 10
Formula: time = total width/relative speed
Solution: t = 0.5/10 = 0.05 s
Final Answer: 0.05 s
Numerical 88JEE Advanced Mixed
A detector receives direct sound and reflected sound with path difference 1.7 m. If frequency is 100 Hz and speed is 340 m s−1, identify interference.
Show Solution
Given: v = 340, f = 100, Δx = 1.7 m
Formula: λ = v/f
Solution: λ = 3.4 m; Δx = λ/2, so destructive.
Final Answer: Destructive
Numerical 89JEE Advanced Mixed
At another detector, path difference is 3.4 m for the same sound. Identify interference.
Show Solution
Given: λ = 3.4 m, Δx = 3.4 m
Formula: Constructive when Δx = nλ
Solution: Path difference is one wavelength, so constructive.
Final Answer: Constructive
Numerical 90JEE Advanced Mixed
A string supports a wave with speed 120 m s−1. If the source frequency is 60 Hz, find distance between nearest same-phase and opposite-phase points.
Show Solution
Given: v = 120, f = 60
Formula: λ = v/f
Solution: λ = 2 m; same phase = 2 m, opposite phase = 1 m
Final Answer: 2 m and 1 m
Numerical 91JEE Advanced Mixed
An interference minimum occurs when path difference is 0.45 m. If it is the first minimum, find wavelength.
Show Solution
Given: First minimum: Δx = λ/2
Formula: λ = 2Δx
Solution: λ = 0.90 m
Final Answer: 0.90 m
Numerical 92JEE Advanced Mixed
A first constructive maximum after central maximum occurs at path difference 0.75 m. Find wavelength.
Show Solution
Given: First non-zero constructive: Δx = λ
Formula: λ = 0.75 m
Solution: 0.75 m
Final Answer: 0.75 m
Numerical 93JEE Advanced Mixed
Two waves have possible resultant amplitudes between 2 cm and 14 cm. Find individual amplitudes.
Show Solution
Given: Amax = 14, Amin = 2
Formula: A1 + A2 = 14; A1 − A2 = 2
Solution: Solving gives A1 = 8 cm and A2 = 6 cm
Final Answer: 8 cm and 6 cm
Question Bank
PYQs and Exam Questions
Questions cover CBSE, NEET, JEE Main, JEE Advanced, IB, ICSE, IGCSE, A-Level, assertion-reason, true-false, case-study, reasoning and conceptual practice.
Question 1CBSE
Define wave speed and write its SI unit.
Show Answer
Wave speed is distance travelled by wave per unit time. SI unit is m s−1.
Question 2CBSE
Write the relation between wave speed, frequency and wavelength.
Show Answer
v = fλ.
Question 3CBSE
State the principle of superposition.
Show Answer
When waves overlap, resultant displacement is algebraic sum of individual displacements.
Question 4CBSE
What is resultant displacement?
Show Answer
Actual displacement of a particle due to overlapping waves.
Question 5CBSE
What is constructive interference?
Show Answer
Interference in which waves meet in same phase and amplitude becomes maximum.
Question 6CBSE
What is destructive interference?
Show Answer
Interference in which waves meet in opposite phase and amplitude becomes minimum.
Question 7NEET
A wave has f = 100 Hz and λ = 2 m. Find speed.
Show Answer
v = fλ = 200 m s−1.
Question 8NEET
If frequency doubles in same medium, what happens to wavelength?
Show Answer
Wavelength becomes half.
Question 9NEET
Two equal amplitudes meet in opposite phase. What is resultant amplitude?
Show Answer
Zero.
Question 10NEET
Two waves of amplitudes 3 cm and 4 cm meet in same phase. Find resultant amplitude.
Show Answer
7 cm.
Question 11NEET
Two waves of amplitudes 9 cm and 5 cm meet in opposite phase. Find resultant amplitude.
Show Answer
4 cm.
Question 12NEET
Does wave speed depend on amplitude in a linear string?
Show Answer
No.
Question 13JEE Main
For a string wave, write speed formula.
Show Answer
v = √(T/μ).
Question 14JEE Main
If tension becomes four times, what happens to string wave speed?
Show Answer
Speed becomes two times.
Question 15JEE Main
If linear density becomes four times, what happens to string wave speed?
Show Answer
Speed becomes half.
Question 16JEE Main
For sound in fluid, write formula using bulk modulus.
Show Answer
v = √(B/ρ).
Question 17JEE Main
Explain particle velocity versus wave velocity.
Show Answer
Particle velocity is dy/dt of medium particle; wave velocity is speed of phase propagation.
Question 18JEE Main
What is y if y1 = 2 mm and y2 = −5 mm?
Show Answer
y = −3 mm.
Question 19JEE Advanced
Why does superposition require algebraic addition?
Show Answer
Displacement has sign/direction, so upward and downward displacements must be added with signs.
Question 20JEE Advanced
State the condition for maximum constructive interference.
Show Answer
Phase difference 2nπ or path difference nλ.
Question 21JEE Advanced
State the condition for destructive interference.
Show Answer
Phase difference (2n + 1)π or path difference (n + 1/2)λ.
Question 22JEE Advanced
For amplitudes A1 and A2, maximum resultant amplitude?
Show Answer
A1 + A2.
Question 23JEE Advanced
For amplitudes A1 and A2, minimum resultant amplitude?
Show Answer
|A1 − A2|.
Question 24JEE Advanced
Can two waves pass through each other after overlap?
Show Answer
Yes, in a linear medium they emerge practically unchanged.
Question 25IB
What does v = fλ mean physically?
Show Answer
One complete wavelength passes a point every time period, giving speed equal to frequency times wavelength.
Question 26IB
Why is echo distance vt/2?
Show Answer
Sound travels to the reflector and back, so total distance is twice wall distance.
Question 27IB
How does noise cancellation use interference?
Show Answer
It produces an opposite-phase sound wave to reduce resultant displacement.
Question 28IB
Why do some seats in a hall hear louder sound?
Show Answer
Constructive interference and room reflections can increase resultant amplitude.
Question 29ICSE
What happens when two crests meet?
Show Answer
Constructive interference; amplitude increases.
Question 30ICSE
What happens when a crest and equal trough meet?
Show Answer
Complete destructive interference.
Question 31ICSE
What is the unit of linear mass density?
Show Answer
kg m−1.
Question 32ICSE
What is bulk modulus related to?
Show Answer
Elasticity of a medium under compression.
Question 33IGCSE
A pulse on a tighter string travels faster or slower?
Show Answer
Faster.
Question 34IGCSE
A pulse on a heavier string travels faster or slower?
Show Answer
Slower.
Question 35IGCSE
What must be added in superposition: amplitudes or displacements?
Show Answer
Displacements are added point by point.
Question 36IGCSE
Does destructive interference mean energy is lost?
Show Answer
No, energy is redistributed.
Question 37A-Level
Give the phase condition for constructive interference.
Show Answer
Δφ = 2nπ.
Question 38A-Level
Give the phase condition for destructive interference.
Show Answer
Δφ = (2n + 1)π.
Question 39A-Level
What is coherent interference?
Show Answer
Interference of waves with constant phase difference and same frequency.
Question 40A-Level
Why is resultant intensity not simply A1 + A2?
Show Answer
Intensity depends on amplitude squared and phase relation.
Question 41Assertion-Reason
Assertion: Increasing string tension increases wave speed. Reason: v = √(T/μ).
Show Answer
Both are true and the reason correctly explains the assertion.
Question 42Assertion-Reason
Assertion: Increasing linear density increases speed. Reason: v = √(T/μ).
Show Answer
Assertion is false; reason is true.
Question 43Assertion-Reason
Assertion: Equal opposite pulses can cancel momentarily. Reason: Resultant displacement is algebraic sum.
Show Answer
Both are true and the reason correctly explains the assertion.
Question 44Assertion-Reason
Assertion: Destructive interference destroys energy. Reason: Resultant displacement can be zero.
Show Answer
Assertion is false; reason is true.
Question 45True/False
Wave velocity and particle velocity are always the same.
Show Answer
False.
Question 46True/False
For same phase waves, resultant amplitude is A1 + A2.
Show Answer
True.
Question 47True/False
For opposite phase waves, resultant amplitude is |A1 − A2|.
Show Answer
True.
Question 48True/False
Wave speed on a string is proportional to √T.
Show Answer
True.
Question 49Case Study
A boy hears an echo 2 s after shouting. If sound speed is 340 m s−1, find wall distance.
Show Answer
Distance = vt/2 = 340 × 2/2 = 340 m.
Question 50Case Study
Two loudspeakers emit same frequency sound in phase. What happens on the central line?
Show Answer
Constructive interference occurs.
Question 51Case Study
Noise-cancelling headphones reduce a steady hum. Which interference is used?
Show Answer
Destructive interference.
Question 52Case Study
A guitar string is tightened. What happens to wave speed?
Show Answer
Wave speed increases.
Question 53Reasoning
Why must signs be used in y = y1 + y2?
Show Answer
Because displacement can be upward/positive or downward/negative.
Question 54Reasoning
Why can waves cross without permanent change?
Show Answer
In a linear medium, each wave obeys the same equation and superposes temporarily.
Question 55Conceptual
If displacement is zero during destructive interference, is amplitude always zero everywhere?
Show Answer
No, cancellation may occur at specific points or instants depending on phase and amplitudes.
Question 56Conceptual
Why is v = fλ not enough for string speed dependence?
Show Answer
It relates wave variables; v = √(T/μ) shows medium dependence for strings.
Question 57Conceptual
What is the most common sign mistake in superposition?
Show Answer
Adding magnitudes even when one displacement is negative.
Question 58Conceptual
Why does a denser medium often reduce mechanical wave speed?
Show Answer
Greater inertia resists rapid motion, lowering speed when elasticity is unchanged.
Case Study
Case Study Practice
Case Study 1: Echo and Distance
A student shouts near a cliff and hears the echo after 1.5 s. Take speed of sound as 340 m s−1.
Question 1: Why is distance calculated using vt/2?
Because sound travels to the cliff and returns.
Question 2: Find distance of cliff.
d = 340 × 1.5 / 2 = 255 m.
Question 3: Which topic is used?
Wave speed and reflection of sound.
Case Study 2: Noise Cancellation
A headphone detects external noise and sends an opposite-phase sound into the ear.
Question 1: Which interference occurs?
Destructive interference.
Question 2: What is added in the ear canal?
Displacements of pressure waves.
Question 3: Does energy vanish magically?
No, the wave field and energy distribution are changed by the device.
Case Study 3: Loudspeakers
Two identical loudspeakers play the same tone in phase.
Question 1: Where is sound loudest?
At points where waves arrive in phase.
Question 2: What happens when path difference is λ/2?
Destructive interference occurs.
Question 3: Which formula gives wave speed?
v = fλ.
Case Study 4: Musical Instruments
A string instrument is tuned by changing tension.
Question 1: What happens to wave speed when tension increases?
Speed increases.
Question 2: Which formula explains it?
v = √(T/μ).
Question 3: What does μ mean?
Mass per unit length of the string.
Case Study 5: Wave Transmission
A signal wave travels along a stretched cable.
Question 1: What property affects speed if tension is fixed?
Linear mass density.
Question 2: What happens if cable is heavier per metre?
Wave speed decreases.
Question 3: Why is phase important in signals?
Overlapping signals can interfere depending on phase.
Mistakes
Common Mistakes
- Confusing particle velocity and wave velocity: particle velocity is local vibration; wave velocity is propagation speed.
- Wrong interference conditions: same phase gives maximum amplitude; opposite phase gives minimum amplitude.
- Sign mistakes in superposition: upward and downward displacements must be added with opposite signs.
- Confusing displacement with amplitude: displacement changes with time and position; amplitude is maximum possible displacement.
- Using echo distance as vt: echo is a round trip, so one-way distance is vt/2.
Revision
Quick Revision Notes
- v = fλ connects speed, frequency and wavelength.
- On a string, v = √(T/μ).
- In a fluid, sound speed is v = √(B/ρ).
- Wave velocity is different from particle velocity.
- Superposition means y = y1 + y2.
- Constructive interference gives A = A1 + A2.
- Destructive interference gives A = |A1 − A2|.
- Same phase means phase difference 2nπ.
- Opposite phase means phase difference (2n + 1)π.
- Echo distance is vt/2.
- Noise cancellation uses destructive interference.
- Always add displacement, not just amplitude.
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