Class 11 Physics · NCERT Chapter 14
~33 min readWaves Class 11 Physics Notes
Complete, exam-ready notes on waves: transverse and longitudinal wave motion, the wave equation and superposition, the speed of sound with Newton's formula and Laplace correction, reflection, standing waves in strings and organ pipes, beats and the Doppler effect — every NCERT topic with MCQs, mark-wise questions and solved numericals for CBSE, JEE and NEET.
What is the Doppler effect in sound?
The apparent change in the frequency of a wave due to relative motion between the source and the observer: f′ = f(v ± v₀)/(v ∓ vₛ), using the upper sign when source or observer moves towards the other. The frequency rises on approach and falls on recession.
Wave Motion and Its Types
Wave
A wave is a disturbance that travels through a medium, transferring energy and momentum without the permanent transfer of matter — the particles of the medium oscillate about their mean positions while the disturbance moves on.
- Transverse waves: particles vibrate perpendicular to the direction of propagation (e.g. waves on a string, light). Can travel through solids and the surface of liquids.
- Longitudinal waves: particles vibrate parallel to the direction of propagation, forming compressions and rarefactions (e.g. sound). Can travel through solids, liquids and gases.
- Mechanical waves need a medium; electromagnetic waves (light, radio) do not.
- Categorised by wavelength: water ripples (tiny λ), sound (~15 m), light (~10⁻⁷ m).
Energy travels, matter does not
MCQ Questions
1 Mark Questions
- 1Define a wave. Distinguish between mechanical and electromagnetic waves.
- 2Give one example each of a transverse and a longitudinal wave.
- 3In what direction do particles vibrate in a longitudinal wave?
2 Mark Questions
- 1Distinguish between transverse and longitudinal waves with respect to particle displacement.
- 2Explain why sound cannot travel in vacuum but light can.
- 3Why can transverse waves travel in solids but generally not in gases?
3 Mark Questions
- 1A wave travels the length of a 20 m string in 5 s. If consecutive crests are 0.4 m apart, find the frequency and period of the wave.
- 2Describe how a particle at the surface of water vibrates as a ripple passes, and how that differs from the motion of the ripple itself.
Characteristics of a Wave
Wave parameters
Amplitude A is the maximum displacement of a particle from equilibrium. Wavelength λ is the distance between two successive crests (or any two corresponding points). The period T is the time for one complete oscillation and the frequency f = 1/T is the number of oscillations per second.
All particles of a wave oscillate with the same frequency and amplitude but with a continuously varying phase. The frequency is fixed by the source; the speed is fixed by the medium; the wavelength adjusts so that v = fλ. The unit of frequency is the hertz (Hz = s⁻¹).
- Tuning a radio dial selects the frequency of the transmitted signal.
- A 100 Hz sound in air at 20 °C (v ≈ 343 m/s) has wavelength about 3.43 m.
- Speed does not change with frequency of the wave in a given medium (no dispersion for sound in air).
MCQ Questions
1 Mark Questions
- 1Define wavelength and frequency of a wave.
- 2Write the relation connecting wave speed, frequency and wavelength.
- 3Define the hertz.
2 Mark Questions
- 1Derive the relation v = fλ starting from the definition of wave speed.
- 2A tuning fork of frequency 512 Hz produces a wave in air with speed 345 m/s. Find the wavelength.
- 3Why is the frequency of a wave fixed by its source and not by the medium?
3 Mark Questions
- 1A source emits a wave of frequency 500 Hz. If the wave speed in the medium is 1500 m/s, find the wavelength and the time period, and state how far the wavefront travels in 2 s.
- 2Explain the meaning of amplitude, frequency, wavelength and speed of a wave, and give their SI units.
The Speed of Sound
Sound travels through a medium as a longitudinal wave, with a speed set by the elastic and inertial properties of the medium: the faster the medium returns to equilibrium and the lighter it is, the faster sound moves.
Newton's formula and Laplace correction
Newton assumed the compressions and rarefactions of a sound wave occur isothermally, giving v = √(P/ρ) ≈ 280 m s⁻¹ for air — well below the measured 332 m s⁻¹. Laplace showed the process is adiabatic, replacing P by γP: v = √(γP/ρ) with γ = Cp/Cv ≈ 1.4, giving ≈ 332 m s⁻¹.
- Speed order: solids > liquids > gases (bulk modulus divided by density).
- In a gas, v ∝ √T — the speed rises with temperature (≈ 0.6 m/s per degree Celsius).
- At constant temperature, speed is independent of pressure of the gas.
- Moist air is denser-lighter: humidity increases the speed slightly because water-vapour molecules are light.
Air vs water vs steel
MCQ Questions
1 Mark Questions
- 1State Newton's formula for the speed of sound in a gas.
- 2What correction did Laplace make and why?
- 3In which medium is the speed of sound maximum — solid, liquid or gas?
2 Mark Questions
- 1Explain the Laplace correction and derive v = √(γP/ρ).
- 2The speed of sound in air at 0 °C is 331 m/s. Estimate its speed at 20 °C.
- 3Why does the speed of sound increase with humidity?
3 Mark Questions
- 1Derive an expression for the speed of sound in a gas using Newton's formula and apply the Laplace correction to obtain the standard result.
- 2Compare the speed of sound in solids, liquids and gases, explaining the physical reasons for the order.
- 3Show that the speed of sound in a gas is proportional to the square root of its absolute temperature.
Progressive Waves and Superposition
Progressive (travelling) wave
A wave that travels through a medium, carrying energy away from the source, with a shape that does not change as it moves. A harmonic progressive wave is described by y = A sin(kx − ωt), where k = 2π/λ is the wave number and ω = 2πf is the angular frequency.
The sign between kx and ωt fixes the direction: (kx − ωt) is a wave moving in the +x direction, (kx + ωt) in the −x direction. The phase (kx − ωt) is what travels; its constancy, kx − ωt = constant, gives dx/dt = ω/k = v.
Superposition principle
When two or more waves overlap in a region, the resultant displacement at every point is the vector (algebraic) sum of the displacements each wave would produce alone. This is the basis of interference, beats and standing waves.
- Phase difference Δφ and path difference Δx are related by Δφ = (2π/λ)Δx.
- A path difference of λ gives a phase difference of 2π (one full cycle).
- Phase and time difference: Δφ = ωΔt.
MCQ Questions
1 Mark Questions
- 1Define a progressive wave.
- 2State the superposition principle for waves.
- 3What is the wave number? Write its relation with wavelength.
2 Mark Questions
- 1Write the general equation of a one-dimensional progressive wave and identify each symbol.
- 2Show that the wave speed v = ω/k follows from the constancy of the phase.
- 3Derive the relation between phase difference and path difference.
3 Mark Questions
- 1A progressive wave is described by y = 0.05 sin π(200t − 2x) m. Find its (a) frequency, (b) wavelength and (c) speed of propagation.
- 2State and explain the principle of superposition of waves, and give three phenomena that it explains.
Reflection and Standing Waves
Reflection of a wave
When a wave hits a boundary it is reflected. At a rigid (fixed) boundary the reflected pulse is inverted — a phase change of π. At a free boundary it returns un-inverted with no phase change, because the boundary cannot exert a transverse force on the string.
When two identical harmonic waves travel in opposite directions along a string, their superposition gives a stationary (standing) wave. The pattern does not travel: energy is stored, not transported, and the medium divides into nodes (zero displacement) and antinodes (maximum displacement).
- Nodes occur where cos(kx) = 0; antinodes where cos(kx) = ±1.
- Consecutive nodes (or antinodes) are λ/2 apart.
- The standing wave does not transfer energy along the string.
- A string fixed at both ends forms a standing wave with nodes at both ends.
Phase change on reflection
MCQ Questions
1 Mark Questions
- 1Define a stationary or standing wave.
- 2What is a node? What is an antinode?
- 3What phase change does a wave suffer on reflection at a fixed boundary?
2 Mark Questions
- 1Show that consecutive nodes (or antinodes) of a stationary wave are λ/2 apart.
- 2Obtain the equation of a standing wave from two progressive waves of equal amplitude and frequency travelling in opposite directions.
- 3Explain why a stationary wave does not transport energy along the medium.
3 Mark Questions
- 1Describe the reflection of a travelling wave at a rigid boundary and at a free boundary, stating the phase change in each case with the reasoning.
- 2Two waves y₁ = A sin(kx − ωt) and y₂ = A sin(kx + ωt) superpose on a string. Obtain the resultant wave and identify the positions of nodes and antinodes.
Standing Waves in Strings and Organ Pipes
A stretched string fixed at both ends supports standing waves only at certain frequencies, called normal modes. The ends are nodes, so an integral number of half wavelengths must fit into the length: L = nλₙ/2, giving fₙ = nv/2L for n = 1, 2, 3, …
- Fundamental (first harmonic): f₁ = v/2L.
- All harmonics are present: f₂ = 2f₁, f₃ = 3f₁, … — the string can vibrate in any of these modes simultaneously.
- The wave speed on a string is v = √(T/μ), where T is the tension and μ the mass per unit length.
Organ pipes
An open pipe has displacement antinodes at both ends and produces all harmonics: fₙ = nv/2L. A closed pipe has a node at the closed end and an antinode at the open end, so it produces only the odd harmonics: fₙ = (2n − 1)v/4L.
Open vs closed pipes
MCQ Questions
1 Mark Questions
- 1Write the expression for the fundamental frequency of a stretched string.
- 2State the frequency formula for an open organ pipe and a closed organ pipe.
- 3Why does a closed organ pipe produce only odd harmonics?
2 Mark Questions
- 1Derive the expression for the frequencies of the normal modes of a string fixed at both ends.
- 2A string of length 1 m has a fundamental frequency of 110 Hz. Find its wavelength and the speed of waves on the string.
- 3Show that the fundamental frequency of a closed pipe is v/4L and that only odd harmonics are possible.
3 Mark Questions
- 1Compare the harmonics produced by an open organ pipe and a closed organ pipe of the same length.
- 2A stretched wire of length 0.90 m and mass 3.6 × 10⁻³ kg is under a tension of 90 N. Calculate its fundamental frequency and the frequencies of its first three harmonics.
- 3A closed organ pipe and an open organ pipe of the same length produce their fundamentals. Find the ratio of their frequencies.
Beats
Beats
When two sound waves of slightly different frequencies f₁ and f₂ reach a point together, the loudness alternately rises and falls. This periodic waxing and waning of intensity is called beats, and the number of intensity maxima heard per second equals the difference of the two frequencies.
Physically, the two waves fall alternately in phase (constructive — loud, sound reinforced) and out of phase (destructive — faint). The resultant amplitude varies slowly at the beat frequency while the wave oscillates rapidly at the average frequency. Beats are heard clearly only when the two frequencies are close (difference ≲ 10 Hz).
- Beats are a demonstration of the superposition principle.
- They are used to tune musical instruments: adjust a string until the beeps against a standard tuning fork vanish.
- A slightly loaded tuning fork shifts its frequency; a fork of known frequency can be calibrated by counting beats with another.
MCQ Questions
1 Mark Questions
- 1Define beats.
- 2Write the formula for the beat frequency when two waves of frequencies f₁ and f₂ superpose.
- 3Why are beats heard only when the two frequencies are close together?
2 Mark Questions
- 1Explain how beats arise from the principle of superposition.
- 2Two tuning forks A (300 Hz) and B produce 4 beats per second. What are the possible frequencies of B?
- 3How are beats used to tune musical instruments?
3 Mark Questions
- 1Derive an expression for the resultant intensity of two waves of slightly different frequencies and show that the intensity oscillates at the beat frequency.
- 2A fork of unknown frequency produces 5 beats/s with a 256 Hz fork. When the unknown fork is slightly loaded with wax, the beats drop to 3 beats/s. Find the unknown frequency and justify your answer.
The Doppler Effect
Doppler effect
The apparent change in the frequency of a wave when the source and the observer are in relative motion. When they approach each other the observed frequency increases; when they recede it decreases. For sound, v₀ and vₛ are measured relative to the medium.
- Observer moving towards a stationary source: f′ = f(v + v₀)/v.
- Source moving towards a stationary observer: f′ = fv/(v − vₛ).
- Both moving towards each other: f′ = f(v + v₀)/(v − vₛ).
- No shift when the relative velocity is zero, or when the motion is perpendicular to the line of sight (no radial component).
Sound vs light Doppler
Applications: radar and speed guns, Doppler echocardiography, tracking weather and storms, and measuring the recession of stars and galaxies in astronomy.
MCQ Questions
1 Mark Questions
- 1State the Doppler effect for sound.
- 2Write the expression for apparent frequency when the observer moves towards a stationary source.
- 3What happens to the apparent frequency as a source recedes from a stationary observer?
2 Mark Questions
- 1Derive the expression f′ = fv/(v − vₛ) for a source moving towards a stationary observer.
- 2A whistle of frequency 400 Hz is sounded by a train moving at 30 m/s. For a stationary observer towards whom the train moves, find the apparent frequency. (Speed of sound = 330 m/s.)
- 3Give two applications of the Doppler effect in technology.
3 Mark Questions
- 1Derive a general expression for the apparent frequency of sound when both the source and the observer are moving, and discuss the sign conventions.
- 2A source of sound of frequency 500 Hz moves towards a stationary observer at 20 m/s while the observer walks towards the source at 5 m/s. Find the apparent frequency heard. (Speed of sound = 340 m/s.)
NCERT Exercise Questions
From the textbook
Exercises (NCERT Q14.1–Q14.19)
- 14.1) 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?
- 14.2) 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⁻¹? (g = 9.8 m s⁻²)
- 14.3) 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 = 343 m s⁻¹?
- 14.4) Use the formula v = √(γP/ρ) to explain why the speed of sound in air (a) is independent of pressure, (b) increases with temperature, (c) increases with humidity.
- 14.5) 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, i.e. y = f(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 + vt)/x0] (c) 1/(x + vt)
- 14.6) 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⁻¹ and in water 1486 m s⁻¹.
- 14.7) 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⁻¹? The operating frequency of the scanner is 4.2 MHz.
- 14.8) A transverse harmonic wave on a string is described by y(x, t) = 3.0 sin (36t + 0.018x + π/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?
- 14.9) For the wave described in Exercise 14.8, plot the displacement (y) versus (t) graphs 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?
- 14.10) For the travelling harmonic wave y(x, t) = 2.0 cos 2π (10t − 0.0080 x + 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) λ/2, (d) 3λ/4.
- 14.11) The transverse displacement of a string (clamped at its both ends) is given by y(x, t) = 0.06 sin (2πx/3) cos (120πt), where x and y are in m and t in s. The length of the string is 1.5 m and its mass is 3.0 × 10⁻² kg. Answer the following: (a) Does the function represent a travelling wave or a stationary wave? (b) Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength, frequency, and speed of each wave? (c) Determine the tension in the string.
- 14.12) (i) For the wave on a string described in Exercise 15.11, do all the points on the string oscillate with the same (a) frequency, (b) phase, (c) amplitude? Explain your answers. (ii) What is the amplitude of a point 0.375 m away from one end?
- 14.13) Given below are some functions of x and t to represent the displacement (transverse or longitudinal) of an elastic wave. State which of these represent (i) a travelling wave, (ii) a stationary wave or (iii) none at all: (a) y = 2 cos (3x) sin (10t) (b) y = 2√(x − vt) (c) y = 3 sin (5x − 0.5t) + 4 cos (5x − 0.5t) (d) y = cos x sin t + cos 2x sin 2t
- 14.14) A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5 × 10⁻² kg and its linear mass density is 4.0 × 10⁻² kg m⁻¹. What is (a) the speed of a transverse wave on the string, and (b) the tension in the string?
- 14.15) A metre-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a tuning fork of frequency 340 Hz) when the tube length is 25.5 cm or 79.3 cm. Estimate the speed of sound in air at the temperature of the experiment. The edge effects may be neglected.
- 14.16) A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod are given to be 2.53 kHz. What is the speed of sound in steel?
- 14.17) A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source? Will the same source be in resonance with the pipe if both ends are open? (speed of sound in air is 340 m s⁻¹).
- 14.18) Two sitar strings A and B playing the note 'Ga' are slightly out of tune and produce beats of frequency 6 Hz. The tension in the string A is slightly reduced and the beat frequency is found to reduce to 3 Hz. If the original frequency of A is 324 Hz, what is the frequency of B?
- 14.19) Explain why (or how): (a) in a sound wave, a displacement node is a pressure antinode and vice versa, (b) bats can ascertain distances, directions, nature, and sizes of the obstacles without any 'eyes', (c) a violin note and sitar note may have the same frequency, yet we can distinguish between the two notes, (d) solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases, and (e) the shape of a pulse gets distorted during propagation in a dispersive medium.
Answers at a glance
- 1Q14.1 → The disturbance reaches the other end in 0.5 s.
- 2Q14.2 → The splash is heard about 8.7 s after the stone is dropped.
- 3Q14.3 → Required tension ≈ 2.06 × 10⁴ N.
- 4Q14.4 → (a) P cancels with ρ via ρ = PM/RT; (b) v ∝ √T; (c) humid air is lighter, so v increases.
- 5Q14.5 → Functions (b) and (c) can represent travelling waves (along −x); (a) cannot.
- 6Q14.6 → (a) 3.4 × 10⁻⁴ m (b) 1.5 × 10⁻³ m.
- 7Q14.7 → Wavelength in the tissue ≈ 4.0 × 10⁻⁴ m (0.40 mm).
- 8Q14.8 → (a) Travelling, 20 m s⁻¹ towards the left; (b) 3.0 cm, 5.73 Hz; (c) π/4; (d) ≈ 3.5 m.
- 9Q14.9 → All three graphs are sine curves of identical shape; the oscillatory motion differs only in phase, not in amplitude or frequency.
- 10Q14.10 → (a) 6.4π rad (b) 0.8π rad (c) π rad (d) 3π/2 rad.
- 11Q14.11 → (a) Stationary wave; (b) λ = 3 m, f = 60 Hz, v = 180 m s⁻¹ each way; (c) T = 648 N.
- 12Q14.12 → (i) Same frequency and phase for all points, but different amplitudes; (ii) ≈ 4.2 cm.
- 13Q14.13 → (b), (c): travelling waves; (a): stationary wave; (d): none at all.
- 14Q14.14 → (a) 78.75 m s⁻¹ (b) ≈ 248 N.
- 15Q14.15 → Speed of sound ≈ 366 m s⁻¹.
- 16Q14.16 → Speed of sound in steel ≈ 5.1 × 10³ m s⁻¹.
- 17Q14.17 → Fundamental (first harmonic) mode of the closed pipe, since 430 Hz ≈ 425 Hz; no resonance with both ends open (nearest harmonic is 850 Hz).
- 18Q14.18 → fB = 318 Hz.
- 19Q14.19 → (a) Node ⇔ pressure antinode; (b) echolocation; (c) timbre from different overtones; (d) rigidity lets solids support transverse waves; (e) components travel at different speeds in a dispersive medium.
CBSE Pattern Questions
How CBSE asks this chapter
MCQ Questions
1 Mark Questions
- 1Define wavelength and frequency of a progressive wave.
- 2Define beats and write the beat frequency in terms of the combined wave frequencies.
- 3What are nodes and antinodes in a stationary wave?
- 4State the Doppler effect for sound.
2 Mark Questions
- 1A wave travels with speed 330 m/s and frequency 660 Hz. Find its wavelength. (Ans: λ = v/f = 0.5 m)
- 2A string of length 2 m and mass 50 g is under a tension of 200 N. Find the speed of a transverse wave on it. (Ans: v = √(T/μ) = 89.4 m/s)
- 3Two tuning forks of frequencies 256 Hz and 260 Hz are sounded together. Find the beat frequency. (Ans: 4 Hz)
3 Mark Questions
- 1Derive the expression for the velocity of a transverse wave on a stretched string. (Ans: v = √(T/μ))
- 2Write and explain the equation of a plane progressive wave and derive relations among particle velocity, wave velocity and slope. (Ans: vp = −v × slope)
- 3Derive the conditions for a stationary wave on a string fixed at both ends and find the frequencies of the first three harmonics.
5 Mark Questions
- 1Describe the formation of stationary waves on a stretched string and derive the expressions for the allowed frequencies. Hence find the fundamental frequency and its overtones for a string of length L with tension T and linear density μ. (Ans: fn = [n/(2L)]√(T/μ))
- 2State and derive the Doppler effect for sound — obtain the observed frequency when a source moves towards a stationary observer and when the observer moves towards a stationary source.
PDF: Assignments & Tests
Printable study material
Free PDF Downloads & Assignments
Practice path: complete Assignment 1 first (topic-by-topic), then Assignment 2 (mixed difficulty with full solutions), then attempt the timed tests.
Covers all eight topics: wave motion and its types; characteristics of a wave; speed of sound; progressive waves and superposition; reflection and standing waves; standing waves in strings and organ pipes; beats; and the Doppler effect. Arranged topic by topic.
Higher-difficulty set with multi-topic integration problems (string waves, Doppler shift, organ pipes, standing waves) and full worked answers. Includes an answer key.
Full board-pattern test: Section A (MCQ/VSA 1M × 10), Section B (2M × 4), Section C (3M × 4), Section D (5M × 2) = 40 marks. Time: 90 minutes. Includes answer key.
Second full paper with different questions, same 40-mark / 90-minute blueprint. Includes answer key.
Quick Revision
Key formulas at a glance
Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.
Wave speed
frequency by source, speed by medium.
Speed of sound in gas
γ = 1.4 for air; ~343 m/s at 20 °C.
Progressive wave
k = 2π/λ, ω = 2πf; v = ω/k.
Phase–path relation
λ path difference ↔ 2π phase difference.
Standing wave
nodes every λ/2.
String / open pipe frequencies
n = 1, 2, 3, … (all harmonics).
Closed pipe frequencies
only odd harmonics.
Beats
intensity waxes and wanes at this rate.
Doppler effect
upper sign on approach; v₀, vₛ measured relative to the medium.
Exam Strategy
How this chapter is asked
High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.
- v = fλ; speed depends only on the medium, frequency only on the source.
- Newton's isothermal value ≈ 280 m/s was corrected by Laplace to v = √(γP/ρ) ≈ 332 m/s.
- Standing wave: nodes and antinodes λ/2 apart; string and open pipe give all harmonics, closed pipe only odd ones.
- Beats occur only when the two frequencies are close; beat frequency = |f₁ − f₂|.
- Doppler: frequency rises on approach, falls on recession; formula f′ = f(v ± v₀)/(v ∓ vₛ).
- Phase difference Δφ = (2π/λ)Δx.
Practice Numericals
Step-by-step worked examples
Standard exam-level problems solved with clear conceptual breakdowns.
Wavelength of a tuning fork's sound
Step-by-Step Solution
- 1λ = v/f = 345 ÷ 512 ≈ 0.674 m.
- 2The crests are about 67 cm apart in the air column.
Final Answer
≈ 0.674 m
Speed of sound using Laplace correction
Step-by-Step Solution
- 1v = √(γP/ρ) = √(1.4 × 1.013 × 10⁵ ÷ 1.29).
- 2γP/ρ = 1.418 × 10⁵ / 1.29 ≈ 1.099 × 10⁵.
- 3v ≈ √(1.099 × 10⁵) ≈ 331.5 m s⁻¹.
Final Answer
≈ 332 m s⁻¹
Beats between two tuning forks
Step-by-Step Solution
- 1Beat frequency = |260 − 256| = 4 Hz.
- 2Time period of beats = 1/4 = 0.25 s.
- 3The intensity of the combined sound rises and falls four times each second.
Final Answer
4 beats per second; period 0.25 s
FAQ
Frequently asked questions
Why did Newton's formula underestimate the speed of sound and how was it corrected?
Newton assumed the compressions and rarefactions were isothermal, giving v = √(P/ρ) ≈ 280 m/s. Laplace pointed out the changes are adiabatic and rapid, replacing P by γP, so v = √(γP/ρ) ≈ 332 m/s — matching experiment.
What is the difference between a node and an antinode?
In a standing wave a node is a point of permanent zero displacement where waves meet in opposite phase, while an antinode is a point of maximum displacement. Nodal and antinodal points alternate, spaced λ/2 apart.
Why does a closed organ pipe produce only odd harmonics?
A closed pipe has a displacement node at the closed end and an antinode at the open end, so its length fits an odd number of quarter wavelengths, giving fₙ = (2n − 1)v/4L — only odd harmonics are possible.
What are beats, and how are they used?
Beats are the periodic waxing and waning of loudness heard when two sources of slightly different frequency (say 256 Hz and 260 Hz) sound together; the beat frequency is |f₁ − f₂|. Musicians use beats to tune instruments by minimising the beat rate.
When does the Doppler effect apply and what changes on approach?
It applies whenever source or observer moves relative to the medium (for sound) carrying waves. On approach the apparent frequency rises (f′ = f(v + v₀)/(v − vₛ)) and on recession it falls — the pitch you hear from a siren changes as it passes you.
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