Previous year questions: Waves

Waves30 min readLesson 8 of 8

Q1 (JEE Main 2026, 2 Apr Shift 1, 4 marks) A plane progressive wave follows y = 5 cos(200t - x/150), with x and y in centimetres and t in seconds. Work out the wave velocity. Answer: Compare the argument with the standard form omega t minus kx. The angular frequency is 200 radians per second and the wave number is 1/150 radians per centimetre. The wave speed is omega divided by k, which is 200 times 150, that is 30000 centimetres per second. Converting to metres per second gives 300 metres per second. The wave travels along the positive x direction because the signs of the x and t terms are opposite. The correct option is 300.

v = \frac{ω}{k} = \frac{200}{1/150} = 30000 cm s^{- 1} = 300 m s^{- 1}

Wave speed from the angular frequency and wave number

Q2 (JEE Main 2026, 5 Apr Shift 1, 4 marks) A transverse wave on a string follows y = 3 sin(36t + 0.018x + pi/4), with x and y measured in centimetres and t in seconds. Work out how far apart two successive crests are, rounded to the nearest whole number. Answer: The coefficient of x is the wave number k, so k equals 0.018 radians per centimetre. The distance between successive crests is the wavelength, which is 2 pi divided by k. Taking pi as 3.14 gives 6.28 divided by 0.018, which is about 348.9 centimetres. The nearest integer is 349, so the answer is 349 centimetres. The phase constant pi/4 shifts the wave along x but does not change the crest separation.

λ = \frac{2 π}{k} = \frac{2 ( 3.14 )}{0.018} \approx 348.9 cm \to 349 cm

Crest separation is one wavelength

Q3 (NCERT Exemplar, 1 mark) Water waves produced by a motor boat sailing in water are which kind of waves? Answer: On the surface of water the particles move in small circles, so the motion has a component along the direction of propagation and a component perpendicular to it. The waves are therefore both longitudinal and transverse. They are not purely one kind, because the surface particles neither move only in the direction of travel nor only across it. The correct choice is that they are both longitudinal and transverse. Source: NCERT Exemplar Class 11 Physics, chapter 15 Waves, MCQ 15.1.

Q4 (NCERT Exemplar, 1 mark) Sound waves of wavelength lambda travel in a medium with speed v and enter another medium where the speed is 2v. What is the wavelength in the second medium? Answer: When a wave crosses from one medium into another, its frequency is set by the source and does not change; the speed and the wavelength change together. The wavelength equals the speed divided by the frequency. With the speed doubled and the frequency unchanged, the wavelength doubles, becoming 2 lambda. The correct choice is 2 lambda. Source: NCERT Exemplar Class 11 Physics, chapter 15 Waves, MCQ 15.2.

Q5 (NCERT Exemplar, 1 mark) How does the speed of sound in air depend on temperature, pressure and humidity? Answer: The speed of sound in a gas equals the square root of gamma R T divided by the molar mass, so it rises with the absolute temperature. It does not depend on pressure at constant temperature, because pressure and density change together and their ratio is fixed. It does depend on humidity: moist air is less dense than dry air at the same temperature and pressure, so sound travels slightly faster in moist air. The correct choice is that the speed increases with an increase in humidity. Source: NCERT Exemplar Class 11 Physics, chapter 15 Waves, MCQ 15.3.

Q6 (NCERT Exemplar, 1 mark) A change in the temperature of the medium changes which property of the sound wave? Answer: The frequency of a sound wave is fixed by the source and does not change when the medium changes temperature. The speed changes with temperature, and the wavelength equals the speed divided by the frequency, so the wavelength changes. Amplitude and loudness depend on the strength of the source and on distance, not directly on the medium temperature. The correct choice is that the wavelength changes. Source: NCERT Exemplar Class 11 Physics, chapter 15 Waves, MCQ 15.4.

Q7 (NCERT Exemplar, 1 mark) When a longitudinal wave propagates through a medium, what quantities are transmitted? Answer: A longitudinal wave carries energy and momentum. In a sound wave the particles of the medium oscillate about their mean positions and do not travel with the wave, so no matter is transported from one place to another. Energy is transported because the wave sets the particles in motion, and momentum is carried in the direction of propagation. The correct choice is energy and momentum; matter is not transmitted. Source: NCERT Exemplar Class 11 Physics, chapter 15 Waves, MCQ 15.5.

Q8 (NCERT Exemplar, 2 marks) The displacement of an elastic wave is given by y = 3 sin(omega t) + 4 cos(omega t), with y in centimetres and t in seconds. Find the resultant amplitude. Answer: The sum of a sine and a cosine of the same argument is a single harmonic with amplitude equal to the square root of the sum of the squares of the coefficients. Here the coefficients are 3 and 4, so the resultant amplitude is the square root of 9 plus 16, which is the square root of 25, that is 5. The resultant can be written as 5 sin(omega t plus phi) with a phase constant. The amplitude is therefore 5 centimetres.

A = 3^{2} + 4^{2} = 5 cm

Amplitude of a sine plus cosine combination

Source: NCERT Exemplar Class 11 Physics, chapter 15 Waves, SA 15.21.

The two JEE items exercise the wave equation from both directions: one asks for the speed from the angular frequency and the wave number, and the other asks for the crest separation from the wave number. Together they show why the same pair of numbers, omega and k, is enough to describe a progressive wave completely. Q9 (NCERT Exemplar, 2 marks) A steel wire is 12 m long and has a mass of 2.10 kg. A tension of 2.06 x 10^4 N is applied to it. Calculate the speed of a transverse wave travelling on this wire. Answer: The linear mass density is the mass divided by the length, 2.10 divided by 12, which is 0.175 kilogram per metre. The wave speed is the square root of the tension divided by the linear density, so it is the square root of 2.06 times 10 to the fourth divided by 0.175. The quotient is about 1.177 times 10 to the fifth, whose square root is about 343 metres per second. The result comes out close to 343 metres per second, which is the familiar speed of sound in dry air when the temperature is 20 degrees Celsius.

v = \frac{T}{μ} = \frac{2.06 \times 1 0 ^{4}}{0.175} \approx 343 m s^{- 1}

Speed of the transverse wave on the steel wire

Source: NCERT Exemplar Class 11 Physics, chapter 15 Waves, SA 15.25.

Q10 (NCERT Exemplar, 3 marks) Take a 20 cm pipe with one end sealed. In air the sound speed is 330 m/s, and a 1237.5 Hz source is applied. Determine which natural mode of the pipe is resonantly excited. Answer: A pipe closed at one end has natural frequencies given by odd multiples of v over 4L, where L is 0.20 metre. The fundamental is 330 divided by 0.80, which is 412.5 hertz. The allowed frequencies are the odd multiples of this value: 412.5 hertz, 1237.5 hertz, 2062.5 hertz, and so on. The source frequency of 1237.5 hertz equals three times the fundamental, which is the third harmonic. The mode excited is therefore the third harmonic, called the first overtone for a closed pipe.

f_{n} = \frac{( 2 n - 1 ) v}{4 L} = (2 n - 1) (412.5) Hz 1237.5 = 3 \times 412.5

Odd harmonics of the closed pipe

Source: NCERT Exemplar Class 11 Physics, chapter 15 Waves, SA 15.26. The two JEE items at the start of the set show the two most common wave-equation tasks in this unit, and both are solved from the same pair of parameters: finding the wave speed from omega and k, and finding the crest separation from the wave number. The Exemplar items cover the conceptual side of the unit: the nature of water waves, the change of wavelength when sound enters a new medium, the factors that control the speed of sound, what a wave transports, what changes when the temperature changes, the amplitude of a sine plus cosine combination, the string speed formula, and the harmonic series of a closed pipe. Between them they cover every section of the chapter, so a student can use this page as a final revision set after the notes.