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Waves NCERT Solutions, Complete PDF Content and Important Questions
NCERT Solutions for Class 11 Physics Chapter 14 Waves PDF Download (2026) + Important Questions Welcome to one of the most dynamic and relatable chapters in physics! Whether it is the sound of your favourite song playing on a speaker, the ripples forming in a quiet pond when you drop a pebble, or the vibration of guitar strings, waves are everywhere. For students preparing for the CBSE Class 11 Board Exams in 2026, as well as competitive entrance tests like JEE Main and NEET, Class 11 Physics Chapter 14: Waves is incredibly important. This chapter forms the foundation for advanced concepts in Class 12, such as wave optics and alternating current. It shifts your focus from the motion of single particles to the collective, rhythmic motion of continuous media.
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ChapterWaves
SubjectPhysics
Class11
BoardCBSE
DifficultyModerate to High
Exam Weightage6 - 8 Marks
Learning Objectives
- Differentiate between mechanical and non-mechanical waves, as well as transverse and longitudinal waves. Write and interpret the standard mathematical equation for a progressive wave.
- Calculate the speed of a travelling wave in various media (e.g., stretched strings, gases).
- Understand the Principle of Superposition and how waves interfere.
- Analyze the formation of Standing Waves in strings and organ pipes (nodes and antinodes).
- Explain the phenomenon of Beats and.
- Calculate beat frequency.
Key Concepts, Definitions and Formulas
- Mechanical Waves: Waves that require a material medium for their propagation (e.g., sound waves, water waves)
- Transverse Waves: The particles of the medium oscillate perpendicular to the direction of wave propagation
- Longitudinal Waves: The particles of the medium oscillate parallel to the direction of wave propagation (e.g., sound waves in air)
- Equation of a Progressive Wave: y(x,t)=asin(kx- t+ ) (where a is amplitude, k is angular wave number, is angular frequency, and is initial phase)
- Angular wave number: k= lambda 2pi
- Angular frequency: = T 2pi =2pinu
- Wave speed: v= nulambda= k Speed of a
- Transverse Wave on a Stretched String: v= T (where T is tension, is linear mass density)
- Speed of Sound in a Gas (Newton-Laplace Formula): v= rho gamma P (where gamma is the ratio of specific heats, P is pressure, rho is density)
- Principle of Superposition: When two or more waves overlap, the resultant displacement is the algebraic sum of their individual displacements
- Standing Waves: Formed by the superposition of two identical waves travelling in opposite directions
- Points of maximum amplitude. Beat Frequency: nu beat = nu 1 -nu 2 To help you visualize how waves interact, interfere, and create standing waves, explore this interactive Wave Superposition Simulator below! Show me the visualisation
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Full NCERT Solutions
NCERT Q14.1: A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?
Answer: Given: Mass of the string, M=2.50 kg Tension, T=200 N Length of the string, L=20.0 m Step 1: Calculate linear mass density () = L M = 20.0 2.50 =0.125 kg/m Step 2: Calculate wave speed (v) v= T = 0.125 200 = 1600 =40 m/s Step 3: Calculate time taken (t) t= Speed Distance = v L = 40 20.0 =0.5 s Result: The disturbance takes 0.5 seconds to reach the other end.
NCERT Q14.2: A stone dropped from the top of a tower of height 300 m splashes into the water of a pond near the base of the tower. When is the splash heard at the top given that the speed of sound in air is 340 m/s? (g=9.8 m/s 2 )
Answer: Given: Height of tower, h=300 m Speed of sound, v=340 m/s Acceleration due to gravity, g=9.8 m/s 2 Initial velocity of stone, u=0 Step 1: Time taken by stone to reach the pond (t 1 ) Using the equation of motion: h=ut 1 + 2 1 gt 1 2 300=0+ 2 1 (9.8)t 1 2 300=4.9t 1 2 t 1 2 = 4.9 300 approx 61.22 t 1 = 61.22 approx 7.82 s Step 2: Time taken by sound to travel up to the top (t 2 ) t 2 = v h = 340 300 approx 0.88 s Step 3: Total time T=t 1 +t 2 =7.82+0.88=8.7 s Result: The splash is heard after 8.7 seconds.
NCERT Q14.3: A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the tension in the wire so that speed of a transverse wave on the wire equals the speed of sound in dry air at 20 C (which is 343 m/s)?
Answer: Given: Length, L=12.0 m Mass, M=2.10 kg Wave speed, v=343 m/s Step 1: Calculate linear mass density () = L M = 12.0 2.10 =0.175 kg/m Step 2: Calculate Tension (T) Formula: v= T T=v 2 T=(343) 2 x 0.175 T=117649 x 0.175 approx 20588.5 N Result: The required tension is approximately 2.06 x 10 4 N.
NCERT Q14.4: Use the formula v= rho gamma P to explain why the speed of sound in air: (a) is independent of pressure, (b) increases with temperature, (c) increases with humidity.
Answer: (a) According to the ideal gas equation, PV=nRT= M m RT rho P = M RT . Since rho P is constant at a constant temperature, a change in pressure causes a proportionate change in density. Hence, v is independent of pressure. (b) Since rho P = M RT , we can write v= M gamma RT . Therefore, v T . As temperature increases, the speed of sound increases. (c) Water vapor is lighter than dry air (molar mass of H 2 O is 18, while dry air is approx 28.8). As humidity increases, the effective density (rho) of the air decreases. Since v rho 1 , the speed of sound increases with humidity.
NCERT Q14.5: You have learnt that a travelling wave in one dimension is represented by a function y=f(x,t) where x and t must appear in the combination x-vt or x+vt. Is the converse true? Examine if the following functions for y can possibly represent a travelling wave: (a) (x-vt) 2 (b) log( x 0 x+vt ) (c) x+vt 1
Answer: The converse is not true. For a function to represent a travelling wave, it must be finite and well-defined for all values of x and t. (a) (x-vt) 2 tends to infinity as x or t tends to infinity. It is not finite. Not a wave. (b) log( x 0 x+vt ) tends to infinity as x or t tends to infinity, and is undefined for negative arguments. Not a wave. (c) x+vt 1 tends to infinity when x=-vt. It is not finite at this point. Not a wave.
NCERT Q14.6: A bat emits ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of (a) the reflected sound, (b) the transmitted sound? Speed of sound in air is 340 m/s and in water is 1486 m/s.
Answer: Given: Frequency, nu=1000 kHz=10 6 Hz Speed in air, v a =340 m/s Speed in water, v w =1486 m/s Note: Frequency remains unchanged during reflection and refraction. (a) Wavelength of reflected sound (in air): lambda a = nu v a = 10 6 340 =3.4 x 10 -4 m (b) Wavelength of transmitted sound (in water): lambda w = nu v w = 10 6 1486 =1.486 x 10 -3 m
NCERT Q14.7: A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is 1.7 km/s? The operating frequency of the scanner is 4.2 MHz.
Answer: Given: Speed of sound in tissue, v=1.7 km/s=1700 m/s Frequency, nu=4.2 MHz=4.2 x 10 6 Hz Formula: v= nulambda lambda= nu v lambda= 4.2 x 10 6 1700 approx 4.05 x 10 -4 m Result: The wavelength is approximately 4.05 x 10 -4 m.
NCERT Q14.8: A transverse harmonic wave on a string is described by y(x,t)=3.0sin(36t+0.018x+ pi/4) where x and y are in cm and t in s. The positive direction of x is from left to right. (a) Is this a travelling wave or a stationary wave? If it is travelling, what are the speed and direction of its propagation? (b) What are its amplitude and frequency? (c) What is the initial phase at the origin? (d) What is the least distance between two successive crests in the wave?
Answer: Comparing with the standard equation y(x,t)=asin( t+kx+ ): a=3.0 cm, =36 rad/s, k=0.018 cm -1 , =pi/4 rad. (a) It is a travelling wave. Since the sign between t and kx is positive, it is travelling from right to left (negative x-direction). Speed v= k = 0.018 36 =2000 cm/s=20 m/s. (b) Amplitude a=3.0 cm. Frequency nu= 2pi = 2 x 3.14 36 approx 5.73 Hz. (c) Initial phase =pi/4 rad. (d) The least distance between two successive crests is the wavelength lambda. lambda= k 2pi = 0.018 2 x 3.14 approx 349 cm or 3.49 m.
NCERT Q14.9: (For the wave described in Exercise 14.8, plot the displacement y versus t for x = 0, 2 and 4 cm. What are the shapes of these graphs? In which aspects does the oscillatory motion in travelling wave differ from one point to another: amplitude, frequency or phase?)
Answer: For x=0, y(0,t)=3.0sin(36t+ pi/4). For x=2, y(2,t)=3.0sin(36t+0.036+ pi/4). For x=4, y(4,t)=3.0sin(36t+0.072+ pi/4). The shapes of all these graphs are sinusoidal (sine waves). The oscillatory motion of a travelling wave has the same amplitude and same frequency at every point. The only aspect that differs from one point to another is the phase.
NCERT Q14.10: For the travelling harmonic wave y(x,t)=2.0cos2 pi(10 t-0.0080x+0.35) where x and y are in cm and t in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of (a) 4 m, (b) 0.5 m, (c) lambda/2, (d) 3lambda/4.
Answer: Equation: y(x,t)=2.0cos(20 pi t-0.016 pi x+0.70 pi). Here, k=0.016 pi cm -1 . Phase difference Delta= k Delta x. (a) Distance Delta x=4 m=400 cm. Delta=0.016pi x 400=6.4pi rad. (b) Distance Delta x=0.5 m=50 cm. Delta=0.016pi x 50=0.8pi rad. (c) Delta x= lambda/2. Since k= lambda 2pi , Delta= lambda 2pi x 2 lambda =pi rad. (d) Delta x=3 lambda/4. Delta= lambda 2pi x 4 3lambda = 2 3pi rad.
Extra Important Questions for 2026 Exams
Important Q1: The speed of sound in a gas is maximum for: A) Hydrogen B) Oxygen C) Nitrogen D) Carbon dioxide
Answer: A) Hydrogen Explanation: v= gamma RT/M . Speed is inversely proportional to the square root of the molar mass. Hydrogen has the lowest molar mass.
Important Q2: Two waves of identical frequency and amplitude interfere. If the phase difference is pi, the resultant amplitude is: A) 2A B) A 2 C) A D) Zero
Answer: D) Zero Explanation: Resultant amplitude R= A 2 +A 2 +2A 2 cos pi . Since cos pi=-1, R=0 (Destructive interference).
Important Q3: In a stationary wave, the distance between a node and the nearest antinode is: A) lambda B) lambda/2 C) lambda/4 D) 2lambda
Answer: C) lambda/4 Explanation: Nodes are points of zero amplitude, antinodes are points of max amplitude. The distance between a consecutive node and antinode is a quarter of a wavelength.
Important Q4: Differentiate between Transverse and Longitudinal waves.
Answer: In transverse waves, particles vibrate perpendicular to wave propagation (can be polarized, form crests and troughs). In longitudinal waves, particles vibrate parallel to wave propagation (cannot be polarized, form compressions and rarefactions).
Important Q5: State the principle of superposition of waves.
Answer: When two or more waves simultaneously pass through a medium, the net displacement of any particle is the algebraic sum of the individual displacements produced by each wave at that point. y=y 1 +y 2 + +y n .
Important Q6: What are Beats? How are they formed?
Answer: Beats are periodic variations in the intensity of sound caused by the superposition of two sound waves of slightly different frequencies travelling in the same direction. The beat frequency is the difference between the two frequencies.
Important Q7: Why are longitudinal waves not possible in a string?
Answer: A string only possesses elasticity of shape (it can be stretched) but lacks volume elasticity (bulk modulus). Longitudinal waves require volume elasticity to propagate compressions and rarefactions.
Important Q8: Discuss the formation of standing waves in a string fixed at both ends. Derive the expression for the fundamental frequency.
Answer: When a transverse wave travels along a string fixed at both ends, it reflects back. The incident and reflected waves superpose to form standing waves. The fixed ends must be nodes. Let length of string be L. For the fundamental mode (first harmonic), the string vibrates in one loop. Distance between two nodes = lambda/2. L= lambda/2 lambda=2 L. Speed of wave v= nulambda nu= lambda v = 2L v . Since v= T/ , the fundamental frequency is nu= 2L 1 T .
Important Q9: Derive the Newton-Laplace formula for the speed of sound in a gas.
Answer: Newton assumed sound travels in air under isothermal conditions. v= B/ rho . For an isothermal process, PV=constant, differentiating gives bulk modulus B=P. So, v= P/ rho . This gave a theoretical value of 280 m/s, which was incorrect. Laplace corrected this, stating compressions and rarefactions occur so rapidly that the process is adiabatic, not isothermal. For an adiabatic process, PV gamma =constant. Differentiating gives B= gamma P. Therefore, the corrected speed of sound is v= rho gamma P .
Important Q10: Two tuning forks A and B produce 4 beats per second. The frequency of A is 256 Hz. When B is slightly loaded with wax, the beat frequency becomes 2 beats per second. Find the original frequency of B.
Answer: Frequency of A, nu A =256 Hz. Beat frequency =4 Hz. Therefore, nu B could be 256+4=260 Hz OR 256-4=252 Hz. When B is loaded with wax, its mass increases, so its frequency decreases. Case 1: If original nu B =260, after loading it decreases (e.g., to 258). New beat frequency = 256-258 =2 Hz. This matches the given condition. Case 2: If original nu B =252, after loading it decreases (e.g., to 250). New beat frequency = 256-250 =6 Hz. This does not match. Hence, original frequency of B is 260 Hz. A student blows air across the open top of a glass bottle containing some water. A sound of a specific pitch is produced. The student then adds more water to the bottle and blows again.
Important Q11: What type of organ pipe does the bottle act as?
Answer: It acts as a closed organ pipe (open at the top where air is blown, and closed at the bottom by the water surface).
Important Q12: When the student adds more water, what happens to the length of the air column?
Answer: The length of the vibrating air column decreases.
Important Q13: How does adding water affect the pitch (frequency) of the sound produced?
Answer: For a closed pipe, fundamental frequency nu= 4L v . Since adding water decreases the length L, the frequency nu increases. The pitch of the sound becomes higher. (Directions: Select A if both are true and Reason explains Assertion. Select B if both are true but Reason does not explain Assertion. Select C if Assertion is true, Reason is false. Select D if Assertion is false, Reason is true.)
Important Q14: Assertion (A): Sound waves cannot travel through a vacuum. Reason (R): Sound waves are mechanical and require a material medium for propagation.
Answer: A. Both A and R are true, and R is the correct explanation. Sound propagates through the vibration of particles, which are absent in a vacuum.
Important Q15: Assertion (A): The speed of a transverse wave on a stretched string increases with an increase in tension. Reason (R): The speed of a transverse wave is directly proportional to the square root of tension.
Answer: A. Both A and R are true, and R correctly explains A. The formula is v= T/ .
FAQ Section
Is Waves Chapter 14 important for JEE/NEET 2026?
Absolutely. It is highly scoring, and the concepts of superposition and path difference overlap heavily with Class 12 Wave Optics, giving you a dual advantage in competitive exams.
Where can I download the Class 11 Physics NCERT PDF for Waves?
You can download the official, updated NCERT Physics textbook for free directly from the official NCERT website (ncert.nic.in) under the 'Publications' -> 'PDFs' section.
What is the difference between progressive and standing waves?
Progressive waves travel through a medium carrying energy from one point to another. Standing waves are trapped between boundaries, storing energy in nodes and antinodes without propagating it forward.
Why is the Doppler effect missing from these notes?
The Doppler effect in sound has been officially deleted from the latest rationalized CBSE Class 11 Physics syllabus. Focus on Superposition, Beats, and Standing Waves instead!
How do I find phase difference from path difference?
Use the golden formula: Phase Difference (Delta) = lambda 2pi x Path Difference (Delta x)
More Class 11 Physics Chapters
Ch 1: Units and MeasurementsCh 2: Motion in a Straight LineCh 3: Motion in a PlaneCh 4: Laws of MotionCh 5: Work, Energy and PowerCh 6: System of Particles and Rotational MotionCh 7: GravitationCh 8: Mechanical Properties of SolidsCh 9: Mechanical Properties of FluidsCh 10: Thermal Properties of MatterCh 11: ThermodynamicsCh 12: Kinetic TheoryCh 13: OscillationsCh 14: Waves