Stationary Waves — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Stationary Waves MCQs with step-by-step solutions (21 questions). Part of Superposition of Waves. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — 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.
Q2 — 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.
Q3 — 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.
Q4 — 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.
Q5 — Stationary Waves · medium · theory
In a stationary wave, the net transfer of energy along the medium is:
A. Equal to that of a progressive wave
B. Maximum at the antinodes
C. Maximum at the nodes
D. Zero ✓ Correct
Solution: Energy carried one way by the incident wave is carried back by the reflected wave, so the wave stores energy locally instead of transporting it.
Q6 — Stationary Waves · medium · theory
All particles lying between two consecutive nodes of a stationary wave vibrate:
A. With the same amplitude and phase
B. With continuously changing frequency
C. In opposite phase with equal amplitudes
D. In the same phase but with different amplitudes ✓ Correct
Solution: Within one loop every particle reaches its extreme position at the same instant; only the amplitude varies, being greatest at the antinode.
Q7 — 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 $2A$.
Q8 — 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.
Q9 — 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 $\dfrac{\lambda}{2} = \dfrac{0.8}{2} = 0.4\text{ m}$.
Q10 — 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 $\lambda = 2 \times 0.25 = 0.5\text{ m}$.
Q11 — Stationary Waves · hard · numerical
A stationary wave is given by $y = 2\sin(0.1\pi x)\cos(100\pi t)$, with $x$ in metre. Its wavelength is:
A. $20\text{ m}$ ✓ Correct
B. $10\text{ m}$
C. $2\text{ m}$
D. $0.1\text{ m}$
Solution: Comparing with $y = 2A\sin kx\cos\omega t$, $k = 0.1\pi$, so $\lambda = \dfrac{2\pi}{0.1\pi} = 20\text{ m}$.
Q12 — Stationary Waves · medium · numerical
In a stationary wave, a node and the adjacent antinode are $0.15\text{ m}$ apart. The wavelength is:
A. $0.6\text{ m}$ ✓ Correct
B. $0.3\text{ m}$
C. $1.2\text{ m}$
D. $0.15\text{ m}$
Solution: That separation is $\dfrac{\lambda}{4}$, so $\lambda = 4 \times 0.15 = 0.6\text{ m}$.
Q13 — 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: $v = n\lambda = 200 \times 1.7 = 340\text{ m/s}$.
Q14 — 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: $2A = 2 \times 0.03 = 0.06\text{ m}$.
Q15 — Stationary Waves · medium · numerical
In a stationary wave of wavelength $1.2\text{ m}$, the distance between the third and the fifth node is:
A. $1.2\text{ m}$ ✓ Correct
B. $0.3\text{ m}$
C. $2.4\text{ m}$
D. $0.6\text{ m}$
Solution: There are two node intervals between them, each $\dfrac{\lambda}{2} = 0.6\text{ m}$, giving $1.2\text{ m}$ in all.
Q16 — 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: $n = \dfrac{v}{\lambda} = \dfrac{200}{0.4} = 500\text{ Hz}$.
Q17 — Stationary Waves · hard · numerical
In a stationary sound wave, consecutive nodes are $0.3\text{ m}$ apart. If the speed of sound is $330\text{ m/s}$, the frequency is:
A. $110\text{ Hz}$
B. $1100\text{ Hz}$
C. $275\text{ Hz}$
D. $550\text{ Hz}$ ✓ Correct
Solution: Node spacing gives $\lambda = 0.6\text{ m}$, so $n = \dfrac{330}{0.6} = 550\text{ Hz}$.
Q18 — Stationary Waves · hard · numerical
A stationary wave is given by $y = 5\cos\left(\dfrac{2\pi x}{3}\right)\sin(40\pi t)$, with $x$ in metre. The speed of the component waves is:
A. $60\text{ m/s}$ ✓ Correct
B. $40\text{ m/s}$
C. $20\text{ m/s}$
D. $120\text{ m/s}$
Solution: $k = \dfrac{2\pi}{3}$ gives $\lambda = 3\text{ m}$, and $\omega = 40\pi$ gives $n = 20\text{ Hz}$. So $v = n\lambda = 60\text{ m/s}$.
Q19 — Stationary Waves · hard · numerical
In a stationary wave of wavelength $0.8\text{ m}$, the distance from a node to the second antinode is:
A. $0.2\text{ m}$
B. $0.8\text{ m}$
C. $0.6\text{ m}$ ✓ Correct
D. $0.4\text{ m}$
Solution: The first antinode is at $\dfrac{\lambda}{4}$ and the second a further $\dfrac{\lambda}{2}$ away, so the distance is $\dfrac{3\lambda}{4} = 0.6\text{ m}$.
Q20 — Stationary Waves · hard · numerical
A stationary wave of wavelength $0.5\text{ m}$ is set up along a $2\text{ m}$ length with nodes at both ends. The number of nodes is:
A. $5$
B. $9$ ✓ Correct
C. $4$
D. $8$
Solution: Nodes occur every $\dfrac{\lambda}{2} = 0.25\text{ m}$. Over $2\text{ m}$ there are $\dfrac{2}{0.25} = 8$ intervals, hence $9$ nodes counting both ends.
Q21 — Stationary Waves · medium · numerical
Two sound waves of frequency $300\text{ Hz}$ travelling in opposite directions at $330\text{ m/s}$ form a stationary wave. The distance between consecutive nodes is:
A. $2.2\text{ m}$
B. $1.1\text{ m}$
C. $0.55\text{ m}$ ✓ Correct
D. $0.275\text{ m}$
Solution: $\lambda = \dfrac{330}{300} = 1.1\text{ m}$, so node spacing is $\dfrac{\lambda}{2} = 0.55\text{ m}$.