Important Questions Class 11 Physics Chapter 14: Waves

Waves are moving disturbances that transfer energy and information without transporting matter as a whole. Mechanical waves need a material medium, while electromagnetic waves can travel through vacuum.

Waves connect oscillations with energy transfer through elastic media such as strings, air columns, water and solids. Important Questions Class 11 Physics Chapter 14 help students practise transverse waves, longitudinal waves, progressive waves, displacement relation, wave speed, superposition, reflection, standing waves, harmonics, organ pipes and beats. The CBSE 2026 chapter also explains sound waves, normal modes, resonance and the difference between travelling and stationary wave patterns.

Key Takeaways

• Mechanical Waves: Mechanical waves need a medium and depend on elastic and inertial properties.
• Wave Speed: A progressive wave follows v = λν = ω/k.
• String Wave Speed: A stretched string supports transverse waves with speed v = √(T/μ).
• Beats: Beat frequency equals the difference between two close frequencies.

Important Questions Class 11 Physics Chapter 14 Structure 2026

Important Questions Class 11 Physics Chapter 14 with Answers

Waves carry energy through a medium or field without carrying matter as a whole.
Students should identify wave type, direction, boundary condition and frequency before solving.
These waves class 11 important questions follow the NCERT 2026 sequence.

1. What does Important Questions Class 11 Physics Chapter 14 mainly test?

Important Questions Class 11 Physics Chapter 14 mainly test wave types, progressive waves, wave speed, superposition, reflection, standing waves and beats. The chapter connects oscillations with energy transfer.

1. Wave Type Skill: Identify transverse and longitudinal waves.
2. Equation Skill: Use y = a sin(kx − ωt + φ).
3. Speed Skill: Use v = λν and v = √(T/μ).
4. Standing Wave Skill: Use node and antinode conditions.
5. Final Result: The chapter tests travelling and stationary wave behaviour.

2. What is a wave in Class 11 Physics?

A wave is a moving disturbance that transfers energy from one point to another. It does not transfer matter as a whole.

1. Disturbance: Moves through space or medium.
2. Energy: Transfers from source to receiver.
3. Matter: Medium particles only oscillate locally.
4. Final Result: A wave transfers energy without bulk matter flow.

3. Why does a cork on water move up and down when waves pass?

The cork moves up and down because water particles oscillate locally. The wave pattern travels outward, but water does not flow outward as a whole.

1. Wave Motion: Disturbance moves forward.
2. Cork Motion: Local oscillation.
3. Matter Flow: No bulk outward flow.
4. Final Result: Water waves carry disturbance, not water mass.

Transverse and Longitudinal Waves Class 11 Questions

Mechanical waves are classified by the direction of particle oscillation.
Transverse waves need shear rigidity, while longitudinal waves need compressibility.
These transverse and longitudinal waves class 11 questions cover examples, media and particle motion.

4. What is a transverse wave?

A transverse wave is a wave in which medium particles oscillate perpendicular to the direction of wave propagation. Waves on a stretched string are transverse.

1. Particle Motion: Perpendicular to wave direction.
2. Example: Wave on a string.
3. Medium Need: Shear rigidity.
4. Final Result: Transverse waves have perpendicular particle oscillation.

5. What is a longitudinal wave?

A longitudinal wave is a wave in which medium particles oscillate parallel to the direction of wave propagation. Sound in air is longitudinal.

1. Particle Motion: Parallel to wave direction.
2. Pattern: Compressions and rarefactions.
3. Example: Sound wave in air.
4. Final Result: Longitudinal waves have parallel particle oscillation.

6. Why can sound travel through air?

Sound travels through air because air can undergo compressions and rarefactions. These pressure-density disturbances move through the medium.

1. Compression: Higher pressure and density region.
2. Rarefaction: Lower pressure and density region.
3. Restoring Force: Pressure difference pushes air layers.
4. Final Result: Sound in air is a longitudinal mechanical wave.

7. Why can transverse waves not travel through gases?

Transverse waves cannot travel through gases because gases cannot sustain shear stress. Transverse waves need restoring force against shearing.

1. Gas Property: No shear rigidity.
2. Transverse Wave Need: Shearing strain support.
3. Result: No transverse propagation in gases.
4. Final Result: Gases support only longitudinal mechanical waves.

8. Why can solids support both transverse and longitudinal waves?

Solids support both waves because they resist shear and compression. They have shear modulus and bulk modulus.

1. Shear Resistance: Supports transverse waves.
2. Compression Resistance: Supports longitudinal waves.
3. Example: Steel can transmit both types.
4. Final Result: Solids support both transverse and longitudinal waves.

9. What type of wave is produced by a piston moving in an air-filled pipe?

A piston moving back and forth in an air-filled pipe produces a longitudinal wave. Air particles oscillate along the pipe.

1. Piston Motion: Forward and backward.
2. Air Disturbance: Compression and rarefaction.
3. Particle Motion: Along propagation direction.
4. Final Result: The wave is longitudinal.

10. Are water waves transverse or longitudinal?

Water waves have both transverse and longitudinal features. Water particles move up-down and back-forth.

1. Vertical Motion: Transverse component.
2. Horizontal Motion: Longitudinal component.
3. Ocean Waves: Combination motion.
4. Final Result: Water waves are a combination of wave motions.

Progressive Wave Class 11 Questions

A progressive wave moves from one part of a medium to another.
Each particle oscillates locally while the disturbance travels forward.
These progressive wave class 11 questions cover travelling wave meaning and direction.

11. What is a progressive wave?

A progressive wave is a wave that travels from one point of a medium to another. It carries energy through the medium.

1. Wave Pattern: Moves with time.
2. Medium Particles: Oscillate about equilibrium.
3. Energy Transfer: Occurs along wave direction.
4. Final Result: Progressive waves transfer energy through a medium.

12. What is the difference between progressive and stationary waves?

A progressive wave transfers energy through the medium, while a stationary wave has fixed nodes and antinodes. Stationary waves do not travel forward.

1. Progressive Wave: Pattern moves.
2. Stationary Wave: Pattern stays fixed.
3. Energy Transfer: Present in progressive wave.
4. Final Result: Progressive and stationary waves have different patterns.

13. What is a crest?

A crest is the point of maximum positive displacement in a wave. It is the highest point in a transverse wave profile.

1. Displacement: Maximum positive.
2. Wave Graph: Top point.
3. Neighbouring Feature: Trough follows after half wavelength.
4. Final Result: Crest is the highest displacement point.

14. What is a trough?

A trough is the point of maximum negative displacement in a wave. It is the lowest point in a transverse wave profile.

1. Displacement: Maximum negative.
2. Wave Graph: Bottom point.
3. Crest Separation: Crest to trough is half wavelength.
4. Final Result: Trough is the lowest displacement point.

15. How does a progressive wave differ from motion of matter?

A progressive wave transfers disturbance and energy, not matter as a whole. Medium particles only oscillate near their mean positions.

1. Wave: Carries energy.
2. Wind or Stream: Carries matter.
3. Sound Example: Air does not flow from speaker to ear.
4. Final Result: Wave motion is not bulk matter motion.

Displacement Relation in Progressive Wave Class 11

A travelling wave needs a function of position and time.
The phase decides displacement at each point and instant.
These displacement relation in progressive wave class 11 questions explain wave equation symbols.

16. What is the displacement relation for a progressive wave travelling in +x direction?

The displacement relation is y(x, t) = a sin(kx − ωt + φ). It represents a sinusoidal wave travelling along positive x-direction.

1. Amplitude: a.
2. Wave Number: k.
3. Angular Frequency: ω.
4. Phase Constant: φ.
5. Final Result: y = a sin(kx − ωt + φ).

17. What is the equation of a wave travelling in −x direction?

A wave travelling in −x direction is written as y(x, t) = a sin(kx + ωt + φ). The plus sign before ωt indicates negative direction.

1. Positive Direction Form: kx − ωt.
2. Negative Direction Form: kx + ωt.
3. Amplitude: a remains positive.
4. Final Result: kx + ωt represents travel along −x.

18. What is amplitude of a wave?

Amplitude is the maximum displacement of a medium particle from its equilibrium position. It is denoted by a.

1. Maximum Positive Displacement: +a.
2. Maximum Negative Displacement: −a.
3. Energy Link: Larger amplitude means larger energy.
4. Final Result: Amplitude is maximum particle displacement.

19. What is phase of a wave?

Phase is the argument of the sine or cosine function in a wave equation. It determines the state of oscillation.

1. For y = a sin(kx − ωt + φ): Phase is kx − ωt + φ.
2. At Fixed x: Phase changes with time.
3. At Fixed t: Phase changes with position.
4. Final Result: Phase decides displacement at a point.

20. What is wavelength?

Wavelength is the minimum distance between two points in the same phase. It equals the distance between two consecutive crests or troughs.

1. Symbol: λ.
2. Crest to Crest: One wavelength.
3. Trough to Trough: One wavelength.
4. Final Result: Wavelength is same-phase separation.

21. What is angular wave number?

Angular wave number is k = 2π/λ. It represents phase change per unit length.

1. Symbol: k.
2. Formula Used: k = 2π/λ.
3. SI Unit: rad m^-1 or m^-1.
4. Final Result: Wave number gives spatial phase rate.

22. What is angular frequency of a wave?

Angular frequency is ω = 2π/T = 2πν. It represents phase change per unit time.

1. Symbol: ω.
2. Time Period: T.
3. Frequency: ν.
4. Final Result: Angular frequency gives temporal phase rate.

23. A wave is y = 0.005 sin(80x − 3t). Find amplitude, wavelength and period.

The amplitude is 0.005 m, wavelength is 7.85 cm, and period is 2.09 s. Compare with y = a sin(kx − ωt).

1. Amplitude: a = 0.005 m.
2. Wave Number: k = 80 m^-1.
3. Wavelength: λ = 2π/80 = 0.0785 m.
4. Angular Frequency: ω = 3 s^-1.
5. Period: T = 2π/3 = 2.09 s.
6. Final Result: a = 5 mm, λ = 7.85 cm, T = 2.09 s.

Speed of Travelling Wave Class 11 Questions

Wave speed is the speed of a fixed phase point such as a crest.
It depends on the medium for mechanical waves.
These speed of travelling wave class 11 questions cover v = λν and v = ω/k.

24. What is the speed of a travelling wave?

The speed of a travelling wave is the speed with which a fixed phase point moves. It is given by v = λν.

1. Wavelength: λ.
2. Frequency: ν.
3. Formula Used: v = λν.
4. Final Result: Wave speed equals wavelength times frequency.

25. How is wave speed related to angular frequency and wave number?

Wave speed is v = ω/k. This comes from keeping the phase constant.

1. Phase: kx − ωt.
2. Fixed Phase Condition: kx − ωt = constant.
3. Speed: dx/dt = ω/k.
4. Final Result: v = ω/k.

26. Does mechanical wave speed depend on source speed?

Mechanical wave speed in a medium does not depend on source speed. It depends on elastic and inertial properties of the medium.

1. String Wave: Depends on tension and linear density.
2. Sound Wave: Depends on modulus and density.
3. Source: Decides frequency.
4. Final Result: Medium decides mechanical wave speed.

27. If wave speed is 340 m/s and frequency is 170 Hz, find wavelength.

The wavelength is 2 m. Use λ = v/ν.

1. Given Data:
   v = 340 m/s
   ν = 170 Hz
2. Formula Used: λ = v/ν.
3. Calculation:
   λ = 340/170
   λ = 2 m
4. Final Result: Wavelength = 2 m.

28. If wavelength is 0.5 m and frequency is 20 Hz, find wave speed.

The wave speed is 10 m/s. Use v = λν.

1. Given Data:
   λ = 0.5 m
   ν = 20 Hz
2. Formula Used: v = λν.
3. Calculation:
   v = 0.5 × 20
   v = 10 m/s
4. Final Result: Wave speed = 10 m/s.

Wave Speed on String Class 11 Questions

A stretched string supports transverse waves due to tension.
Its wave speed depends on tension and linear mass density.
These wave speed on string class 11 questions cover v = √(T/μ).

29. What is the speed of transverse wave on a stretched string?

The speed is v = √(T/μ). Here T is tension and μ is linear mass density.

1. Tension: T.
2. Linear Mass Density: μ = mass/length.
3. Formula Used: v = √(T/μ).
4. Final Result: String wave speed increases with tension.

30. How does tension affect wave speed on a string?

Wave speed increases when tension increases. The relation is v ∝ √T.

1. Formula: v = √(T/μ).
2. If T Becomes Four Times: v becomes two times.
3. Linear Density: Kept constant.
4. Final Result: Higher tension gives higher wave speed.

31. How does linear mass density affect wave speed?

Wave speed decreases when linear mass density increases. The relation is v ∝ 1/√μ.

1. Formula: v = √(T/μ).
2. If μ Becomes Four Times: v becomes half.
3. Tension: Kept constant.
4. Final Result: Heavier string gives lower wave speed.

32. A steel wire of length 0.72 m has mass 5.0 × 10^-3 kg and tension 60 N. Find wave speed.

The wave speed is about 93 m/s. Use v = √(T/μ).

1. Given Data:
   L = 0.72 m
   m = 5.0 × 10^-3 kg
   T = 60 N
2. Linear Density:
   μ = m/L = 5.0 × 10^-3/0.72
   μ = 6.94 × 10^-3 kg/m
3. Wave Speed:
   v = √(60/6.94 × 10^-3)
   v ≈ 93 m/s
4. Final Result: Wave speed ≈ 93 m/s.

33. A 20 m string has mass 2.5 kg and tension 200 N. Find time for disturbance to reach the other end.

The disturbance takes 2.5 s. First find wave speed using v = √(T/μ).

1. Given Data:
   L = 20 m
   m = 2.5 kg
   T = 200 N
2. Linear Density:
   μ = 2.5/20 = 0.125 kg/m
3. Speed:
   v = √(200/0.125) = 40 m/s
4. Time:
   t = L/v = 20/40 = 0.5 s
5. Final Result: Time = 0.5 s.

Speed of Sound Class 11 Questions

Sound is a longitudinal mechanical wave.
Its speed depends on elastic modulus and density of the medium.
These speed of sound class 11 questions cover Newton’s formula and Laplace correction.

34. What is the speed of sound in a fluid?

The speed of sound in a fluid is v = √(B/ρ). Here B is bulk modulus and ρ is density.

1. Bulk Modulus: B.
2. Density: ρ.
3. Formula Used: v = √(B/ρ).
4. Final Result: Sound speed depends on elasticity and density.

35. What is speed of longitudinal wave in a solid bar?

The speed is v = √(Y/ρ). Here Y is Young’s modulus and ρ is density.

1. Young’s Modulus: Y.
2. Density: ρ.
3. Formula Used: v = √(Y/ρ).
4. Final Result: Longitudinal wave speed in a bar is √(Y/ρ).

36. What is Newton’s formula for speed of sound in gas?

Newton’s formula is v = √(P/ρ). It assumes isothermal pressure changes in gas.

1. Pressure: P.
2. Density: ρ.
3. Assumption: Isothermal compression and rarefaction.
4. Final Result: Newton’s formula underestimates sound speed in air.

37. What is Laplace correction for speed of sound?

Laplace correction uses adiabatic pressure changes and gives v = √(γP/ρ). It matches measured sound speed in air.

1. Adiabatic Bulk Modulus: B = γP.
2. Specific Heat Ratio: γ = Cp/Cv.
3. Corrected Formula: v = √(γP/ρ).
4. Final Result: Laplace correction gives accurate sound speed.

38. Why are sound waves adiabatic in air?

Sound waves are adiabatic because compressions and rarefactions happen very fast. Heat does not get enough time to flow.

1. Pressure Variation: Rapid.
2. Heat Exchange: Negligible during one cycle.
3. Correct Model: Adiabatic process.
4. Final Result: Sound propagation uses adiabatic changes.

39. Why is sound faster in solids than gases?

Sound is faster in solids because solids have much larger elastic moduli. This effect dominates their larger density.

1. Solid Elasticity: High modulus.
2. Gas Elasticity: Lower modulus.
3. Formula: v = √(modulus/density).
4. Final Result: Higher elasticity makes sound faster in solids.

40. Estimate speed of sound in air at STP using Laplace correction.

The speed is about 331 m/s. Use v = √(γP/ρ).

1. Given Data:
   γ = 1.4
   P = 1.013 × 10^5 Pa
   ρ = 1.29 kg/m³
2. Formula Used: v = √(γP/ρ).
3. Calculation:
   v = √(1.4 × 1.013 × 10^5/1.29)
   v ≈ 331 m/s
4. Final Result: Speed of sound ≈ 331 m/s.

Superposition of Waves Class 11 Questions

When waves overlap, their displacements add algebraically.
This principle explains interference, standing waves and beats.
These superposition of waves class 11 questions cover constructive and destructive interference.

41. State the principle of superposition of waves.

The principle states that resultant displacement equals the algebraic sum of individual displacements. It applies when waves overlap in a medium.

1. Wave 1 Displacement: y1.
2. Wave 2 Displacement: y2.
3. Resultant: y = y1 + y2.
4. Final Result: Superposition adds wave displacements algebraically.

42. What is constructive interference?

Constructive interference occurs when waves meet in phase. The resultant amplitude becomes maximum.

1. Phase Difference: 0 or 2π.
2. Equal Amplitudes: Resultant amplitude = 2a.
3. Effect: Larger displacement.
4. Final Result: In-phase waves reinforce each other.

43. What is destructive interference?

Destructive interference occurs when waves meet in opposite phase. Equal-amplitude waves cancel completely.

1. Phase Difference: π.
2. Equal Amplitudes: Resultant amplitude = 0.
3. Effect: Zero displacement.
4. Final Result: Opposite-phase waves cancel each other.

44. What is resultant amplitude of two equal waves with phase difference φ?

The resultant amplitude is 2a cos(φ/2). Here a is the amplitude of each wave.

1. Wave 1: a sin(kx − ωt).
2. Wave 2: a sin(kx − ωt + φ).
3. Resultant Amplitude: 2a cos(φ/2).
4. Final Result: Amplitude depends on phase difference.

45. What happens when two equal opposite pulses overlap?

Their displacements cancel during overlap if they have equal and opposite shapes. After crossing, both pulses continue separately.

1. During Overlap: Net displacement can become zero.
2. After Crossing: Pulses retain identity.
3. Principle Used: Superposition.
4. Final Result: Waves pass through each other after overlap.

Reflection of Waves Class 11 Questions

Waves reflect when they meet a boundary.
The reflected phase depends on whether the boundary is rigid or open.
These reflection of waves class 11 questions explain phase reversal and boundary conditions.

46. What happens when a wave reflects from a rigid boundary?

A wave reflecting from a rigid boundary undergoes phase reversal. The phase change is π.

1. Rigid Boundary: Displacement must be zero.
2. Reflected Pulse: Inverted.
3. Phase Change: π or 180°.
4. Final Result: Rigid reflection produces phase reversal.

47. What happens when a wave reflects from an open boundary?

A wave reflecting from an open boundary has no phase reversal. The phase change is zero.

1. Open Boundary: Boundary can move freely.
2. Reflected Pulse: Same orientation.
3. Phase Change: 0.
4. Final Result: Open reflection gives no phase change.

48. What is reflection at a closed end in sound waves?

Reflection at a closed end gives a displacement node and pressure antinode. Air cannot move at the closed end.

1. Closed End: Displacement is zero.
2. Pressure Variation: Maximum.
3. Boundary Type: Rigid boundary.
4. Final Result: Closed end acts as a displacement node.

49. What is reflection at an open end in sound waves?

Reflection at an open end gives a displacement antinode and pressure node. Air displacement is maximum near the open end.

1. Open End: Air can move freely.
2. Displacement: Maximum.
3. Pressure Variation: Minimum.
4. Final Result: Open end acts as a displacement antinode.

Standing Waves Class 11 Questions

Standing waves form by superposition of two identical waves travelling in opposite directions.
They have fixed nodes and antinodes.
These standing waves class 11 questions cover stationary wave equation and normal modes.

50. What is a standing wave?

A standing wave is a wave pattern with fixed nodes and antinodes. It forms from two identical waves moving in opposite directions.

1. Wave 1: Travels one way.
2. Wave 2: Travels opposite way.
3. Result: Fixed amplitude pattern.
4. Final Result: Standing wave does not travel forward.

51. What is the equation of a standing wave on a string?

The standing wave equation is y(x, t) = 2a sin kx cos ωt. It forms from opposite travelling waves.

1. Forward Wave: a sin(kx − ωt).
2. Backward Wave: a sin(kx + ωt).
3. Superposition: y = 2a sin kx cos ωt.
4. Final Result: Standing wave has separate x and t factors.

52. What is a node?

A node is a point of zero displacement in a standing wave. It remains at rest.

1. Amplitude: Zero.
2. Location: Fixed.
3. String Example: Ends fixed at nodes.
4. Final Result: Node has zero amplitude.

53. What is an antinode?

An antinode is a point of maximum displacement in a standing wave. It oscillates with maximum amplitude.

1. Amplitude: Maximum.
2. Location: Fixed.
3. Between Nodes: Antinode lies midway between adjacent nodes.
4. Final Result: Antinode has maximum amplitude.

54. What is distance between two consecutive nodes?

The distance between two consecutive nodes is λ/2. The same distance separates consecutive antinodes.

1. Node Position: x = nλ/2.
2. Next Node: x = (n + 1)λ/2.
3. Separation: λ/2.
4. Final Result: Node-to-node distance is λ/2.

55. What are normal modes?

Normal modes are allowed vibration patterns of a bounded system. Boundary conditions decide their frequencies.

1. String Ends: Fixed ends become nodes.
2. Air Column Ends: Open or closed conditions apply.
3. Allowed Frequencies: Resonant frequencies.
4. Final Result: Normal modes are natural vibration patterns.

56. What are harmonics of a stretched string fixed at both ends?

A stretched string fixed at both ends has frequencies νn = nv/(2L). Here n = 1, 2, 3 and so on.

1. Length: L.
2. Wave Speed: v.
3. Allowed Wavelengths: λn = 2L/n.
4. Final Result: Fixed string supports all harmonics.

57. What is the fundamental frequency of a string fixed at both ends?

The fundamental frequency is ν1 = v/(2L). It is the lowest natural frequency.

1. First Harmonic: n = 1.
2. Wavelength: λ = 2L.
3. Formula: ν1 = v/2L.
4. Final Result: Fundamental frequency is v/(2L).

58. What frequencies occur in a pipe closed at one end?

A pipe closed at one end supports only odd harmonics. Its frequencies are v/4L, 3v/4L, 5v/4L and so on.

1. Closed End: Displacement node.
2. Open End: Displacement antinode.
3. Allowed Harmonics: Odd only.
4. Final Result: Closed pipe supports odd harmonics.

59. What frequencies occur in a pipe open at both ends?

A pipe open at both ends supports all harmonics. Its frequencies are νn = nv/(2L).

1. Both Ends: Displacement antinodes.
2. Allowed Harmonics: n = 1, 2, 3 and so on.
3. Fundamental: v/2L.
4. Final Result: Open pipe supports all harmonics.

60. A pipe 30 cm long is open at both ends. Which harmonic resonates with 1.1 kHz source if v = 330 m/s?

The source resonates with the second harmonic. The fundamental frequency is 550 Hz.

1. Given Data:
   L = 0.30 m
   v = 330 m/s
2. Fundamental Frequency:
   ν1 = v/(2L) = 330/0.60
   ν1 = 550 Hz
3. Source Frequency: 1100 Hz = 2 × 550 Hz.
4. Final Result: 1.1 kHz is the second harmonic.

61. Will the same 1.1 kHz source resonate if one end of the pipe is closed?

No, it will not resonate. A closed pipe of length 30 cm has fundamental 275 Hz and only odd harmonics.

1. Closed Pipe Fundamental: ν1 = v/(4L) = 330/1.2 = 275 Hz.
2. 1.1 kHz: 1100/275 = 4.
3. Allowed Harmonics: 1, 3, 5 and so on.
4. Final Result: Fourth harmonic is not allowed in a closed pipe.

Beats Class 11 Physics Questions

Beats occur when two sound waves of close frequencies overlap.
The loudness rises and falls at a frequency equal to the frequency difference.
These beats class 11 physics questions cover tuning and beat frequency.

62. What are beats?

Beats are periodic rises and falls in sound intensity. They occur when two close frequencies are heard together.

1. Wave Frequencies: Nearly equal.
2. Amplitude: Waxing and waning.
3. Audible Effect: Loud-soft-loud pattern.
4. Final Result: Beats come from interference of close frequencies.

63. What is beat frequency?

Beat frequency is the difference between the two frequencies. It is written as νbeat = |ν1 − ν2|.

1. Frequency 1: ν1.
2. Frequency 2: ν2.
3. Formula Used: νbeat = |ν1 − ν2|.
4. Final Result: Beat frequency equals frequency difference.

64. Why do musicians use beats while tuning instruments?

Musicians use beats because beat frequency becomes zero when two notes match. Fewer beats mean closer tuning.

1. Out of Tune: Beats are heard.
2. Tuning Change: Beat frequency changes.
3. Perfect Match: No beats.
4. Final Result: Zero beats indicate same frequency.

65. Two sitar strings produce 5 Hz beats. String A has frequency 427 Hz, and increasing B tension reduces beats to 3 Hz. Find original B frequency.

The original frequency of B is 422 Hz. Increasing tension increases string frequency.

1. Given Data:
   νA = 427 Hz
   Initial beat frequency = 5 Hz
2. Possibilities: νB = 422 Hz or 432 Hz.
3. Tension Increase: νB increases.
4. Beat Decreases: B must be below A originally.
5. Final Result: νB = 422 Hz.

NCERT Class 11 Physics Chapter 14 Questions

NCERT questions test wave type, wave speed, standing waves, harmonics and beats.
Students should inspect units because equations may use cm, m, Hz or radian values.
These NCERT Class 11 Physics Chapter 14 questions follow the 2026 exercise pattern.

66. A stone is dropped from a 300 m tower. When is splash heard at top if sound speed is 340 m/s?

The splash is heard after about 8.70 s. Add fall time and sound travel time.

1. Given Data:
   h = 300 m
   g = 9.8 m/s²
   sound speed = 340 m/s
2. Fall Time:
   h = 1/2gt²
   t = √(2h/g) = √(600/9.8)
   t = 7.82 s
3. Sound Time:
   ts = 300/340 = 0.88 s
4. Total Time: 7.82 + 0.88 = 8.70 s.
5. Final Result: Splash is heard after 8.70 s.

67. A steel wire length is 12 m and mass is 2.10 kg. What tension gives wave speed 343 m/s?

The required tension is about 2.06 × 10^4 N. Use T = μv².

1. Given Data:
   L = 12 m
   m = 2.10 kg
   v = 343 m/s
2. Linear Density:
   μ = 2.10/12 = 0.175 kg/m
3. Formula Used: T = μv².
4. Calculation:
   T = 0.175 × 343²
   T ≈ 2.06 × 10^4 N
5. Final Result: Tension ≈ 2.06 × 10^4 N.

68. A bat emits 1000 kHz sound in air. Find wavelength of reflected sound in air.

The reflected wavelength is 3.4 × 10^-4 m. Frequency remains 1000 kHz.

1. Given Data:
   ν = 1000 kHz = 1.0 × 10^6 Hz
   vair = 340 m/s
2. Formula Used: λ = v/ν.
3. Calculation:
   λ = 340/(1.0 × 10^6)
   λ = 3.4 × 10^-4 m
4. Final Result: Reflected wavelength = 0.34 mm.

69. Find wavelength of transmitted ultrasonic sound in water if speed is 1486 m/s.

The transmitted wavelength is 1.486 × 10^-3 m. Frequency remains unchanged at boundary.

1. Given Data:
   ν = 1.0 × 10^6 Hz
   vwater = 1486 m/s
2. Formula Used: λ = v/ν.
3. Calculation:
   λ = 1486/(1.0 × 10^6)
   λ = 1.486 × 10^-3 m
4. Final Result: Transmitted wavelength = 1.486 mm.

70. An ultrasonic scanner has frequency 4.2 MHz and sound speed 1.7 km/s in tissue. Find wavelength.

The wavelength is about 0.405 mm. Use λ = v/ν.

1. Given Data:
   v = 1.7 km/s = 1700 m/s
   ν = 4.2 MHz = 4.2 × 10^6 Hz
2. Formula Used: λ = v/ν.
3. Calculation:
   λ = 1700/(4.2 × 10^6)
   λ = 4.05 × 10^-4 m
4. Final Result: λ = 0.405 mm.

Class 11 Physics Chapter 14 Numericals

Numericals in Waves often require converting cm to m and kHz to Hz.
Students should decide whether the question involves travelling waves, standing waves or beats.
These Class 11 Physics Chapter 14 numericals support stepwise CBSE 2026 practice.

71. A wave is y = 3.0 sin(36t + 0.018x + π/4), where x and y are in cm. Find direction.

The wave travels in the negative x-direction. The phase has the form kx + ωt.

1. Given Equation: y = 3.0 sin(36t + 0.018x + π/4).
2. Standard Negative Direction: y = a sin(kx + ωt + φ).
3. Sign Check: +ωt with +kx.
4. Final Result: The wave travels towards negative x-direction.

72. For the same wave, find speed.

The speed is 2000 cm/s or 20 m/s. Use v = ω/k.

1. Angular Frequency: ω = 36 s^-1.
2. Wave Number: k = 0.018 cm^-1.
3. Formula Used: v = ω/k.
4. Calculation:
   v = 36/0.018
   v = 2000 cm/s
5. Final Result: Wave speed = 20 m/s.

73. For the same wave, find amplitude and frequency.

The amplitude is 3.0 cm, and frequency is 5.73 Hz. Use ν = ω/2π.

1. Amplitude: a = 3.0 cm.
2. Angular Frequency: ω = 36 s^-1.
3. Frequency:
   ν = 36/(2π)
   ν = 5.73 Hz
4. Final Result: Amplitude = 3.0 cm and ν = 5.73 Hz.

74. A string fixed at both ends has y = 0.06 sin(2πx/3) cos(120πt). Is it travelling or stationary?

It is a stationary wave. The x-dependent and t-dependent factors appear separately.

1. Given Form: y = 0.06 sin(2πx/3) cos(120πt).
2. Stationary Wave Form: y = A(x) cos ωt.
3. No kx − ωt Term: Pattern does not travel.
4. Final Result: The wave is stationary.

75. For the same wave, find wavelength, frequency and wave speed.

The wavelength is 3 m, frequency is 60 Hz, and wave speed is 180 m/s.

1. Wave Number: k = 2π/3.
2. Wavelength: λ = 2π/k = 3 m.
3. Angular Frequency: ω = 120π.
4. Frequency: ν = ω/2π = 60 Hz.
5. Speed: v = λν = 3 × 60 = 180 m/s.
6. Final Result: λ = 3 m, ν = 60 Hz, v = 180 m/s.

76. If string length is 1.5 m and mass is 3.0 × 10^-2 kg, find tension for wave speed 180 m/s.

The tension is 648 N. Use T = μv².

1. Given Data:
   L = 1.5 m
   m = 3.0 × 10^-2 kg
   v = 180 m/s
2. Linear Density:
   μ = m/L = 0.03/1.5
   μ = 0.02 kg/m
3. Tension:
   T = μv²
   T = 0.02 × 180²
   T = 648 N
4. Final Result: Tension = 648 N.

77. Two strings produce 6 Hz beats. A has frequency 324 Hz, and reducing A tension reduces beats to 3 Hz. Find B frequency.

The frequency of B is 318 Hz. Reducing tension reduces A frequency, and beat frequency also reduces.

1. Given Data:
   νA = 324 Hz
   Initial beats = 6 Hz
2. Possible B Frequencies: 318 Hz or 330 Hz.
3. Reducing A Tension: νA decreases.
4. Beats Reduce: B must be below A.
5. Final Result: νB = 318 Hz.

CBSE Class 11 Physics Chapter-Wise Important Questions