Class XI · Chapter 14
Waves — Exam Practice
Practice questions in CBSE, NEET, and JEE Main style covering wave speed, standing waves, beats, and the Doppler effect.
A note on these questions: these are original practice questions, written to match the exact style, format, and difficulty of CBSE Board, NEET, and JEE Main papers — not verbatim reproductions of specific past papers.
1CBSE Board Exam Style
Q1. Why can’t sound waves travel through a vacuum?
Answer
Sound is a mechanical (longitudinal) wave that propagates via particle collisions transmitting the disturbance through a medium. A vacuum has no particles to collide and transmit the wave, so sound cannot travel through it.
Sound is a mechanical (longitudinal) wave that propagates via particle collisions transmitting the disturbance through a medium. A vacuum has no particles to collide and transmit the wave, so sound cannot travel through it.
Derive the expression for the fundamental frequency of a stretched string fixed at both ends.
Answer
A string fixed at both ends must have a node at each end. The simplest (fundamental) standing wave pattern fits exactly half a wavelength between the two ends: L = λ/2 → λ = 2L.
A string fixed at both ends must have a node at each end. The simplest (fundamental) standing wave pattern fits exactly half a wavelength between the two ends: L = λ/2 → λ = 2L.
Using v = fλ: f₁ = v/λ = v/(2L), the fundamental frequency (general formula: fn = nv/2L for the nth harmonic).
Q3. (a) Explain the formation of beats, and derive the expression for beat frequency. (b) Two tuning forks of frequency 256 Hz and 260 Hz are sounded together. Find the beat frequency and the time interval between successive intensity maxima.
Answer
(a) Superposing two waves of slightly different frequency produces a resultant whose amplitude periodically rises and falls, as the two waves drift in and out of phase. This is heard as a slow throbbing in loudness — beats, with beat frequency fbeat = |f₁ − f₂|.
(a) Superposing two waves of slightly different frequency produces a resultant whose amplitude periodically rises and falls, as the two waves drift in and out of phase. This is heard as a slow throbbing in loudness — beats, with beat frequency fbeat = |f₁ − f₂|.
(b) Beat frequency = |260−256| = 4 Hz.
Time between successive maxima = 1/fbeat = 1/4 = 0.25 s.
2NEET Style (Single-Correct MCQ)
Q4. The speed of sound in air increases with:
Solution
v = √(γRT/M) — speed of sound in a gas increases with the square root of absolute temperature.
v = √(γRT/M) — speed of sound in a gas increases with the square root of absolute temperature.
Q5. In a pipe closed at one end, which harmonics are present?
Solution
The closed end must be a node and the open end an antinode — a constraint only satisfied by odd multiples of the fundamental frequency (f, 3f, 5f, …).
The closed end must be a node and the open end an antinode — a constraint only satisfied by odd multiples of the fundamental frequency (f, 3f, 5f, …).
Q6. According to the Doppler effect, the observed frequency of sound appears higher when the source:
Solution
An approaching source compresses successive wavefronts closer together in the direction of travel, raising the frequency an observer ahead of it perceives.
An approaching source compresses successive wavefronts closer together in the direction of travel, raising the frequency an observer ahead of it perceives.
Q7. For a wave described by y = A sin(kx − ωt), the wave travels in the:
Solution
The form (kx − ωt) indicates propagation in the positive x-direction; (kx + ωt) would indicate the negative x-direction.
The form (kx − ωt) indicates propagation in the positive x-direction; (kx + ωt) would indicate the negative x-direction.
3JEE Main Style
Q8. Find the speed of sound in a gas at pressure 10⁵ Pa, density 1.2 kg/m³, with γ = 1.4.
Solution
v = √(γP/ρ) = √[(1.4×10⁵)/1.2] = √116,667 ≈ 341.6 m/s.
v = √(γP/ρ) = √[(1.4×10⁵)/1.2] = √116,667 ≈ 341.6 m/s.
Q9. A sound source moves away from a stationary observer. The frequency the observer hears:
Solution
A receding source stretches the wavefronts apart in the observer’s direction, lowering the perceived frequency.
A receding source stretches the wavefronts apart in the observer’s direction, lowering the perceived frequency.
Q10. A string of length 0.5 m, fixed at both ends, vibrates in its 3rd harmonic. If the wave speed on the string is 200 m/s, find the frequency.
Solution
f₃ = 3v/(2L) = (3×200)/(2×0.5) = 600/1 = 600 Hz.
f₃ = 3v/(2L) = (3×200)/(2×0.5) = 600/1 = 600 Hz.
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