MCQs / Physics / Superposition, Stationary Waves and Beats
Superposition, Stationary Waves and Beats MCQs — Physics (page 4) 150 multiple-choice questions on Superposition, Stationary Waves and Beats from Physics, each with the correct answer and, where available, a teacher's explanation. Read through them to revise the chapter, then practise the same questions one by one to test yourself.
76. Two identical waves interfere constructively. The resultant intensity is: 4 times that of one wave zero 2 times the same Answer: A) 4 times that of one wave
Explanation: Amplitude doubles; I propto A^2. SLO P-11-D-08 · Understanding
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77. Noise cancellation works best for: sudden, high-frequency sounds speech only music only steady, low-frequency sounds such as engine hum Answer: D) steady, low-frequency sounds such as engine hum
Explanation: Predictable, long-wavelength noise is easier to cancel. SLO P-11-D-09 · Understanding
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78. A pipe of length 0.85 m is closed at one end (v = 340 m/s, end correction ignored). Its fundamental frequency is: 300 Hz 50 Hz 100 Hz 200 Hz Answer: C) 100 Hz
Explanation: f1 = v4L = 100 Hz. SLO P-11-D-12 · Understanding
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79. If the anti-noise wave has a smaller amplitude than the noise, the result is: complete silence louder noise partial cancellation (quieter noise) no sound at all Answer: C) partial cancellation (quieter noise)
Explanation: Resultant amplitude = An - Aa. SLO P-11-D-09 · Understanding
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80. Increasing the tension in a guitar string makes its frequency: increase stay the same zero decrease Answer: A) increase
Explanation: v = T/mu increases. SLO P-11-D-16 · Application
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81. A stationary wave does not: transfer energy along the medium have a frequency have nodes have antinodes Answer: A) transfer energy along the medium
Explanation: Energy is stored, not transported. SLO P-11-D-11 · Understanding
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82. When two notes are exactly in tune, the beat frequency is: infinite equal to the note frequency maximum zero Answer: D) zero
Explanation: No difference in frequency. SLO P-11-D-15 · Understanding
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83. A harbour wall gap 5 m wide is approached by sea waves of wavelength 5 m. The waves inside the harbour will: spread out widely double in height stop travel straight with no spreading Answer: A) spread out widely
Explanation: Gap ≈ wavelength. SLO P-11-D-13 · Understanding
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84. Two coherent waves of amplitudes 6 mm and 2 mm meet. The maximum and minimum possible resultant amplitudes are: 6 mm and 2 mm 12 mm and 4 mm 8 mm and 4 mm 6.32 mm and 0 mm Answer: C) 8 mm and 4 mm
Explanation: In phase: 6 + 2 = 8 mm; in antiphase: 6 - 2 = 4 mm. SLO P-11-D-07 · Application
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85. Constructive interference occurs when two waves meet: at right angles in phase (crest on crest) in antiphase (crest on trough) with different frequencies Answer: B) in phase (crest on crest)
Explanation: Amplitudes add. SLO P-11-D-08 · Understanding
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86. Two sources are coherent if they have: the same frequency and a constant phase difference different frequencies the same amplitude only random phases Answer: A) the same frequency and a constant phase difference
Explanation: Needed for a stable interference pattern. SLO P-11-D-14 · Understanding
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87. In a stretched-string experiment, increasing the frequency of the vibrator (same tension) produces: a travelling wave only fewer loops no change more loops on the string Answer: D) more loops on the string
Explanation: Higher harmonics fit more half-wavelengths. SLO P-11-D-10 · Application
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88. A closed tube resonates first at 20 cm and next at 60 cm. The wavelength is: 40 cm 80 cm 20 cm 120 cm Answer: B) 80 cm
Explanation: Successive resonances differ by lambda2 = 40 cm. SLO P-11-D-10 · Application
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89. Bright fringes in Young's double-slit experiment are due to: polarization destructive interference reflection constructive interference Answer: D) constructive interference
Explanation: Path difference nlambda. SLO P-11-D-08 · Understanding
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90. The distance between a node and the next antinode is: lambda lambda2 lambda4 2lambda Answer: C) lambda4
Explanation: Half of node–node spacing. SLO P-11-D-11 · Understanding
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91. At a point, one wave alone gives a displacement of 5 cm and another alone gives -3 cm. When both arrive, the displacement is: 8 cm 2 cm -15 cm 5.83 cm Answer: B) 2 cm
Explanation: By superposition, displacements add: 5 + (-3) = 2 cm. SLO P-11-D-07 · Application
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92. All particles between two adjacent nodes of a stationary wave vibrate: with a phase difference of 90^circ in antiphase in phase with each other randomly Answer: C) in phase with each other
Explanation: They move up and down together. SLO P-11-D-11 · Understanding
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93. Two coherent sources in phase emit waves of wavelength lambda. At a point, the path difference is 1.5lambda. The interference there is: neither — there is no interference constructive destructive constructive only for sound Answer: C) destructive
Explanation: Path difference = nlambda gives constructive; (n + 12)lambda gives destructive. Here 1.5lambda ⇒ destructive. SLO P-11-D-07 · Application
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94. The amplitude of a stationary wave: varies from zero at nodes to a maximum at antinodes increases along the string is the same everywhere is zero everywhere Answer: A) varies from zero at nodes to a maximum at antinodes
Explanation: Unlike a progressive wave. SLO P-11-D-11 · Understanding
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95. Twice per cycle, a stretched string carrying a stationary wave is: broken momentarily straight (all displacements zero) still at nodes only at maximum displacement everywhere Answer: B) momentarily straight (all displacements zero)
Explanation: Graphically, the two travelling waves cancel at that instant. SLO P-11-D-11 · Understanding
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96. Guitar players can use 'harmonics' and beats to tune adjacent strings because: harmonics have no frequency harmonics are louder matching harmonics give zero beats when in tune beats make strings longer Answer: C) matching harmonics give zero beats when in tune
Explanation: Compare equal frequencies. SLO P-11-D-16 · Application
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97. At the fixed end of a stretched string there is always: an antinode a node zero tension maximum amplitude Answer: B) a node
Explanation: The end cannot move. SLO P-11-D-11 · Understanding
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98. Two coherent waves of amplitudes 3 mm and 5 mm meet. The maximum and minimum possible resultant amplitudes are: 5 mm and 3 mm 8 mm and 2 mm 5.83 mm and 0 mm 15 mm and 2 mm Answer: B) 8 mm and 2 mm
Explanation: In phase: 3 + 5 = 8 mm; in antiphase: 3 - 5 = 2 mm. SLO P-11-D-07 · Application
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99. In a resonance tube (air column) experiment, a loud sound is heard when: the tuning fork stops the tube is empty a stationary wave forms in the air column (resonance) the water boils Answer: C) a stationary wave forms in the air column (resonance)
Explanation: Length matches a harmonic. SLO P-11-D-10 · Application
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100. In the microwave experiment, the detector is moved along the line between transmitter and reflector. It shows: a constant reading only minima alternating maxima (antinodes) and minima (nodes) only maxima Answer: C) alternating maxima (antinodes) and minima (nodes)
Explanation: Stationary wave pattern. SLO P-11-D-10 · Application
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