Learn › MH-CET · Physics › Superposition of Waves
Superposition of Waves — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Superposition of Waves MCQs with step-by-step solutions covering Progressive Waves & Wave Equation, Superposition & Interference, Stationary Waves, Vibrations of Strings, Vibrations of Air Columns, Beats. Practise online on Prepizo — no login needed.
▶ Practise Superposition of Waves online (free)
Subtopics
Sample questions with solutions
Q1 — Progressive Waves & Wave Equation · easy · theory
A progressive wave travelling through a medium transfers:
A. Energy and momentum, but not matter ✓ Correct
B. Matter along with energy
C. Matter but not energy
D. Neither energy nor matter
Solution: The particles of the medium merely oscillate about their mean positions; the disturbance, and with it energy and momentum, travels onward.
Q2 — Progressive Waves & Wave Equation · easy · theory
Sound waves travelling through air are:
A. Stationary waves
B. Transverse, with particles vibrating perpendicular to propagation
C. Electromagnetic waves
D. Longitudinal, with particles vibrating along the direction of propagation ✓ Correct
Solution: A gas has no shear elasticity, so it cannot sustain transverse waves. Sound in air propagates as compressions and rarefactions along the direction of travel.
Q3 — Progressive Waves & Wave Equation · easy · theory
The relation between the speed $v$, frequency $n$ and wavelength $\lambda$ of a wave is:
A. $v = n^2\lambda$
B. $v = \dfrac{\lambda}{n}$
C. $v = \dfrac{n}{\lambda}$
D. $v = n\lambda$ ✓ Correct
Solution: In one period the wave advances exactly one wavelength, so $v = \dfrac{\lambda}{T} = n\lambda$.
Q4 — Progressive Waves & Wave Equation · easy · theory
The propagation constant (wave number) $k$ of a wave of wavelength $\lambda$ is:
A. $2\pi\lambda$
B. $\dfrac{2\pi}{\lambda}$ ✓ Correct
C. $\dfrac{1}{\lambda}$
D. $\dfrac{\lambda}{2\pi}$
Solution: The wave number counts the phase change per unit distance, which is $2\pi$ radian over one wavelength.
Q5 — Progressive Waves & Wave Equation · easy · numerical
A wave of frequency $500\text{ Hz}$ has a wavelength of $0.6\text{ m}$. Its speed is:
A. $500\text{ m/s}$
B. $300\text{ m/s}$ ✓ Correct
C. $833\text{ m/s}$
D. $0.0012\text{ m/s}$
Solution: $v = n\lambda = 500 \times 0.6 = 300\text{ m/s}$.
Q6 — Progressive Waves & Wave Equation · easy · numerical
A tuning fork of frequency $256\text{ Hz}$ produces sound of speed $340\text{ m/s}$. The wavelength is approximately:
A. $1.33\text{ m}$ ✓ Correct
B. $0.75\text{ m}$
C. $2.66\text{ m}$
D. $87040\text{ m}$
Solution: $\lambda = \dfrac{v}{n} = \dfrac{340}{256} \approx 1.33\text{ m}$.
Q7 — Progressive Waves & Wave Equation · easy · numerical
A wave of wavelength $2\text{ m}$ has a frequency of $170\text{ Hz}$. Its speed is:
A. $0.012\text{ m/s}$
B. $340\text{ m/s}$ ✓ Correct
C. $170\text{ m/s}$
D. $85\text{ m/s}$
Solution: $v = n\lambda = 170 \times 2 = 340\text{ m/s}$ — the speed of sound in air at ordinary temperature.
Q8 — Progressive Waves & Wave Equation · easy · numerical
The angular frequency of a wave of frequency $100\text{ Hz}$ is approximately:
A. $100\text{ rad/s}$
B. $1256\text{ rad/s}$
C. $314\text{ rad/s}$
D. $628\text{ rad/s}$ ✓ Correct
Solution: $\omega = 2\pi n = 2\pi \times 100 \approx 628\text{ rad/s}$.
Q9 — Superposition & Interference · easy · theory
The principle of superposition of waves states that the resultant displacement at a point is:
A. The product of the individual displacements
B. The vector sum of the displacements due to the individual waves ✓ Correct
C. Always the larger of the two displacements
D. The difference of the individual displacements
Solution: Because the wave equation is linear, each wave travels as though the other were absent and the displacements simply add.
Q10 — Superposition & Interference · easy · theory
Constructive interference between two waves occurs when the path difference is:
A. An odd multiple of half the wavelength
B. An odd multiple of a quarter wavelength
C. Always zero
D. An integral multiple of the wavelength ✓ Correct
Solution: A path difference of $n\lambda$ brings the waves into step, so their amplitudes add and the intensity is maximum.
Q11 — Superposition & Interference · easy · theory
Destructive interference between two waves occurs when the path difference is:
A. An integral multiple of a quarter wavelength
B. Always zero
C. An odd multiple of half the wavelength ✓ Correct
D. An integral multiple of the wavelength
Solution: A path difference of $(2n-1)\dfrac{\lambda}{2}$ puts the waves exactly out of step, so the amplitudes subtract.
Q12 — Superposition & Interference · easy · theory
Two waves of the same frequency travelling along the same line with zero phase difference produce a resultant of amplitude:
A. $|A_1 - A_2|$
B. Zero
C. $A_1 + A_2$ ✓ Correct
D. $\sqrt{A_1^2 + A_2^2}$
Solution: With $\cos\phi = 1$, the general result $\sqrt{A_1^2 + A_2^2 + 2A_1A_2\cos\phi}$ reduces to $A_1 + A_2$.
Q13 — Superposition & Interference · easy · theory
The intensity of a wave is proportional to:
A. The square of its amplitude ✓ Correct
B. Its amplitude
C. The inverse of its amplitude
D. The cube of its amplitude
Solution: Energy per unit volume in a wave varies as $A^2$, and intensity is the energy transported per unit area per unit time.
Q14 — Superposition & Interference · easy · numerical
Two waves of amplitudes $6\text{ cm}$ and $8\text{ cm}$ superpose in phase. The resultant amplitude is:
A. $10\text{ cm}$
B. $48\text{ cm}$
C. $14\text{ cm}$ ✓ Correct
D. $2\text{ cm}$
Solution: In-phase superposition adds the amplitudes: $A = 6 + 8 = 14\text{ cm}$.
Q15 — Stationary Waves · easy · theory
A stationary wave is produced by the superposition of:
A. Two identical waves travelling in opposite directions ✓ Correct
B. Two waves travelling in the same direction
C. A single wave reflected at a free end only
D. Two waves of slightly different frequencies
Solution: The incident and reflected waves, identical in amplitude and frequency but oppositely directed, combine to give a pattern of fixed nodes and antinodes.
Q16 — Stationary Waves · easy · theory
In a stationary wave, the points that always remain at rest are called:
A. Crests
B. Nodes ✓ Correct
C. Troughs
D. Antinodes
Solution: At a node the two component waves always arrive exactly out of phase, so the displacement there is permanently zero.
Q17 — Stationary Waves · easy · theory
The distance between two consecutive nodes in a stationary wave of wavelength $\lambda$ is:
A. $2\lambda$
B. $\lambda$
C. $\dfrac{\lambda}{2}$ ✓ Correct
D. $\dfrac{\lambda}{4}$
Solution: Nodes recur every half wavelength, and so do antinodes; the two sets interleave.
Q18 — Stationary Waves · easy · theory
The distance between a node and the adjacent antinode is:
A. $\lambda$
B. $\dfrac{\lambda}{2}$
C. $\dfrac{\lambda}{4}$ ✓ Correct
D. $\dfrac{\lambda}{8}$
Solution: Antinodes sit midway between neighbouring nodes, which are $\dfrac{\lambda}{2}$ apart.
Q19 — Stationary Waves · easy · theory
If each of the two superposing waves has amplitude $A$, the amplitude of vibration at an antinode is:
A. $2A$ ✓ Correct
B. $\dfrac{A}{2}$
C. $4A$
D. $A$
Solution: At an antinode the waves are always in step, so their amplitudes add to give $2A$.
Q20 — Stationary Waves · easy · theory
The amplitude of vibration of a particle situated exactly at a node is:
A. Equal to $A$
B. Maximum
C. Always zero ✓ Correct
D. Equal to $2A$
Solution: Complete destructive interference occurs permanently at a node, so that particle never moves.
Q21 — Stationary Waves · easy · numerical
In a stationary wave of wavelength $0.8\text{ m}$, the distance between two consecutive antinodes is:
A. $0.8\text{ m}$
B. $0.4\text{ m}$ ✓ Correct
C. $0.2\text{ m}$
D. $1.6\text{ m}$
Solution: Consecutive antinodes are separated by $\dfrac{\lambda}{2} = \dfrac{0.8}{2} = 0.4\text{ m}$.
Q22 — Stationary Waves · easy · numerical
Consecutive nodes in a stationary wave are $0.25\text{ m}$ apart. The wavelength is:
A. $0.5\text{ m}$ ✓ Correct
B. $1.0\text{ m}$
C. $0.25\text{ m}$
D. $0.125\text{ m}$
Solution: Node spacing is $\dfrac{\lambda}{2}$, so $\lambda = 2 \times 0.25 = 0.5\text{ m}$.
Q23 — Stationary Waves · easy · numerical
A stationary wave of frequency $200\text{ Hz}$ has a wavelength of $1.7\text{ m}$. The speed of the component waves is:
A. $118\text{ m/s}$
B. $400\text{ m/s}$
C. $340\text{ m/s}$ ✓ Correct
D. $170\text{ m/s}$
Solution: $v = n\lambda = 200 \times 1.7 = 340\text{ m/s}$.
Q24 — Stationary Waves · easy · numerical
Two waves each of amplitude $0.03\text{ m}$ form a stationary wave. The amplitude at an antinode is:
A. $0.06\text{ m}$ ✓ Correct
B. $0.03\text{ m}$
C. $0.015\text{ m}$
D. Zero
Solution: At an antinode the amplitudes add: $2A = 2 \times 0.03 = 0.06\text{ m}$.
Q25 — Stationary Waves · easy · numerical
A stationary wave has a wavelength of $0.4\text{ m}$ and the component waves travel at $200\text{ m/s}$. The frequency is:
A. $250\text{ Hz}$
B. $80\text{ Hz}$
C. $800\text{ Hz}$
D. $500\text{ Hz}$ ✓ Correct
Solution: $n = \dfrac{v}{\lambda} = \dfrac{200}{0.4} = 500\text{ Hz}$.
Q26 — Vibrations of Strings · easy · theory
The fundamental frequency of a stretched string of length $L$, tension $T$ and linear density $\mu$ is:
A. $2L\sqrt{\dfrac{T}{\mu}}$
B. $\dfrac{1}{2L}\sqrt{\dfrac{\mu}{T}}$
C. $\dfrac{1}{L}\sqrt{\dfrac{T}{\mu}}$
D. $\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}$ ✓ Correct
Solution: In the fundamental mode $\lambda = 2L$, and the wave speed on the string is $\sqrt{\dfrac{T}{\mu}}$, so $n = \dfrac{v}{2L}$.
Q27 — Vibrations of Strings · easy · theory
The law of length for a vibrating string states that, at constant tension and linear density, the fundamental frequency is:
A. Independent of the length
B. Proportional to the square of the length
C. Directly proportional to the vibrating length
D. Inversely proportional to the vibrating length ✓ Correct
Solution: From $n = \dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}$, shortening the string raises the pitch — which is how a guitarist frets a note.
Q28 — Vibrations of Strings · easy · theory
The law of tension for a vibrating string states that the fundamental frequency is proportional to:
A. $T^2$
B. $T$
C. $\dfrac{1}{T}$
D. $\sqrt{T}$ ✓ Correct
Solution: Only the square root of the tension enters $n = \dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}$, so the tension must be quadrupled to double the pitch.
Q29 — Vibrations of Strings · easy · theory
The instrument used to verify the laws of a vibrating string is the:
A. Barometer
B. Potentiometer
C. Sonometer ✓ Correct
D. Resonance tube
Solution: A sonometer lets the length, tension and material of a stretched wire be varied independently while the frequency is compared against a tuning fork.
Q30 — Vibrations of Strings · easy · theory
In the fundamental mode of a string fixed at both ends, the ends are:
A. Antinodes, with a node at the midpoint
B. Nodes, with an antinode at the midpoint ✓ Correct
C. Both antinodes
D. Points of maximum velocity
Solution: The fixed ends cannot move, so they must be nodes; the single loop between them has its antinode at the centre.