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Waves – Class 11 Physics notes, important questions and MCQs

NCERT Class 11 Physics, chapter 14. Notes, key terms, important questions with answers and MCQs.

Chapter overview

Waves carry energy without carrying matter. Mechanical waves can be transverse or longitudinal; superposition gives standing waves and beats.

Wave basics

v = fλ; y = A sin(kx − ωt), k = 2π/λ, ω = 2πf.
Speed on a string v = √(T/µ); speed of sound v = √(γP/ρ) (Laplace correction to Newton's formula).
Sound needs a medium; it is longitudinal in air.

Superposition

Standing waves on a string fixed at both ends: fₙ = nv/2L (all harmonics).
Closed pipe: odd harmonics only, f = (2n − 1)v/4L; open pipe: all harmonics, f = nv/2L.
Beats: beat frequency = |f₁ − f₂|; Doppler effect changes observed frequency with relative motion.

Key terms

Node: point of zero displacement in a standing wave
Beats: periodic rise and fall of loudness from two close frequencies
Doppler effect: change in observed frequency due to relative motion between source and observer

Formulas

v = fλ
v = √(T/µ)
f_beat = |f₁ − f₂|

Important questions with answers

Why is sound heard better on a cold night than in the afternoon? (Also: why do we hear a train's sound earlier through the rails?)2 marks

At night air near the ground is cooler, so sound bends back towards the ground (refraction) instead of escaping upward. Sound travels much faster in solids (rails) than in air, so it reaches us earlier through the rails.

Explain Laplace's correction to Newton's formula for the speed of sound in air.3 marks

Newton assumed isothermal compressions and used v = √(P/ρ), giving about 280 m/s, lower than the measured 332 m/s. Laplace said compressions and rarefactions are so fast that heat cannot flow, so the process is adiabatic: v = √(γP/ρ). With γ = 1.4 for air this gives about 331 m/s, matching experiment.

Derive the frequencies of normal modes of a string fixed at both ends. A 1 m long string has tension 100 N and linear mass density 0.01 kg/m; find the wave speed, the fundamental frequency and the first two overtones.5 marks

Standing waves form with nodes at both ends, so the length must hold a whole number of half-wavelengths: L = nλ/2, λₙ = 2L/n. Frequencies fₙ = nv/2L, n = 1, 2, 3, …, with v = √(T/µ). Here v = √(100/0.01) = 100 m/s. f₁ = 100/2 = 50 Hz; first overtone (2nd harmonic) 100 Hz; second overtone (3rd harmonic) 150 Hz.

Quick revision

Sound cannot travel in vacuum.
Closed pipe: only odd harmonics.
Speed of sound increases with temperature.

Exam tips

Distance between two consecutive nodes = λ/2.
In a string fixed at both ends and an open pipe, all harmonics are present; in a closed pipe only odd harmonics occur.

Practice MCQs

A wave of frequency 50 Hz and wavelength 2 m has speed
1. 100 m/s
2. 25 m/s
3. 52 m/s
4. 48 m/s
Show answer
Answer: (a) 100 m/s. fλ.
Sound waves in air are
1. transverse
2. stationary only
3. electromagnetic
4. longitudinal
Show answer
Answer: (d) longitudinal. Compressions and rarefactions.
Two tuning forks of 256 Hz and 260 Hz give beats of
1. 4 Hz
2. 516 Hz
3. 2 Hz
4. 258 Hz
Show answer
Answer: (a) 4 Hz. Difference.
Distance between consecutive nodes is
1. λ
2. λ/2
3. λ/4
4. 2λ
Show answer
Answer: (b) λ/2. Half wavelength.
The speed of sound in air is 340 m/s. The wavelength of a 680 Hz note is
1. 0.5 m
2. 2 m
3. 1 m
4. 0.25 m
Show answer
Answer: (a) 0.5 m. v/f.
A pipe closed at one end produces
1. only odd harmonics
2. all harmonics
3. only even harmonics
4. no harmonics
Show answer
Answer: (a) only odd harmonics. Node at closed end.
If tension in a string is made 4 times, wave speed becomes
1. ½
2. 4 times
3. 2 times
4. 16 times
Show answer
Answer: (c) 2 times. v ∝ √T.
The fundamental frequency of an open pipe of length 0.5 m (v = 340 m/s) is
1. 170 Hz
2. 85 Hz
3. 680 Hz
4. 340 Hz
Show answer
Answer: (d) 340 Hz. v/2L.

All chapters of Class 11 Physics

Written by Kundali IQ from standard, well-established facts to help with revision. This is not an official NCERT publication; always read your textbook too.