MCQs / Physics / Superposition, Stationary Waves and Beats
Superposition, Stationary Waves and Beats MCQs — Physics (page 5) 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.
101. A tuning fork of 400 Hz and an unknown fork give 3 beats/s. The unknown frequency is: 1200 Hz 133 Hz 397 Hz or 403 Hz exactly 403 Hz Answer: C) 397 Hz or 403 Hz
Explanation: f = 400 pm 3. SLO P-11-D-15 · Understanding
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102. A node in a stationary wave is a point of: zero amplitude (no displacement) maximum amplitude maximum speed maximum energy flow Answer: A) zero amplitude (no displacement)
Explanation: Waves always cancel there. SLO P-11-D-11 · Understanding
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103. Destructive interference occurs when two waves meet: in antiphase (crest on trough) at the same speed only with zero amplitude in phase Answer: A) in antiphase (crest on trough)
Explanation: Amplitudes subtract. SLO P-11-D-08 · Understanding
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104. A sound wave and its exact inverted copy add to give: a wave of double frequency zero displacement everywhere a stationary wave double displacement Answer: B) zero displacement everywhere
Explanation: Complete destructive interference. SLO P-11-D-09 · Understanding
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105. We can hear people round a corner but cannot see them because: sound wavelengths are similar to doorway sizes but light wavelengths are tiny light cannot travel round corners at all sound travels faster than light sound is transverse Answer: A) sound wavelengths are similar to doorway sizes but light wavelengths are tiny
Explanation: Sound diffracts much more. SLO P-11-D-13 · Understanding
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106. Two tuning forks of 100 Hz and 102 Hz are sounded together. The number of beats heard per second is: 3 101 1.5 202 Answer: C) 1.5
Explanation: Beat frequency = f1 - f2 = 1.5 Hz. SLO P-11-D-15 · Understanding
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107. A pipe closed at one end produces only: even harmonics the fundamental only odd harmonics (1 : 3 : 5 …) all harmonics Answer: C) odd harmonics (1 : 3 : 5 …)
Explanation: Node at closed end, antinode at open end. SLO P-11-D-12 · Understanding
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108. A progressive wave differs from a stationary wave because in a progressive wave: all points have the same amplitude and energy is transferred no energy is transferred there are nodes and antinodes the wave does not move Answer: A) all points have the same amplitude and energy is transferred
Explanation: Key comparison. SLO P-11-D-11 · Understanding
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109. In a region of destructive interference, the wave energy: is destroyed is redistributed to regions of constructive interference is created elsewhere turns into heat only Answer: B) is redistributed to regions of constructive interference
Explanation: Energy is conserved. SLO P-11-D-08 · Understanding
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110. Why don't noise-cancelling headphones cancel the music you're listening to? the music is polarized the circuit only generates anti-noise for the outside sound picked up by the microphone music is a different kind of wave music travels faster Answer: B) the circuit only generates anti-noise for the outside sound picked up by the microphone
Explanation: The music signal is added separately. SLO P-11-D-09 · Understanding
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111. Two separate light bulbs do not produce a visible interference pattern because they are: polarized too far apart too bright not coherent Answer: D) not coherent
Explanation: Their phase difference changes randomly. SLO P-11-D-14 · Understanding
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112. Diffraction of waves provides evidence that light is: longitudinal a particle only a sound wave a wave Answer: D) a wave
Explanation: Only waves diffract. SLO P-11-D-13 · Understanding
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113. A piano tuner tightens a string while sounding it with a 440 Hz fork. The beats slow from 4 Hz to 2 Hz. The string's frequency is now: 438 Hz or 442 Hz 444 Hz only 880 Hz exactly 440 Hz Answer: A) 438 Hz or 442 Hz
Explanation: Still 2 Hz from 440 Hz. SLO P-11-D-16 · Application
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114. Which condition is needed for a stable interference pattern? the sources must be very bright the waves must be longitudinal the sources must be different colours the sources must be coherent Answer: D) the sources must be coherent
Explanation: Constant phase difference. SLO P-11-D-08 · Understanding
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115. A string of length 0.5 m fixed at both ends carries waves at 200 m/s. Its fundamental frequency is: 800 Hz 200 Hz 400 Hz 100 Hz Answer: B) 200 Hz
Explanation: f1 = v2L = 2002(0.5) = 200 Hz. SLO P-11-D-12 · Understanding
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116. Two tuning forks of 330 Hz and 335 Hz are sounded together. The number of beats heard per second is: 665 5 332 10 Answer: B) 5
Explanation: Beat frequency = f1 - f2 = 5 Hz. SLO P-11-D-15 · Understanding
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117. Two coherent waves of amplitudes 4 mm and 4 mm meet. The maximum and minimum possible resultant amplitudes are: 8 mm and 4 mm 16 mm and 0 mm 4 mm and 4 mm 8 mm and 0 mm Answer: D) 8 mm and 0 mm
Explanation: In phase: 4 + 4 = 8 mm; in antiphase: 4 - 4 = 0 mm. SLO P-11-D-07 · Application
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118. Beats are an example of interference in: time (at one place) space only polarization diffraction Answer: A) time (at one place)
Explanation: Phase difference changes with time. SLO P-11-D-15 · Understanding
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119. For constructive interference, the phase difference is: any value 0, 2pi, 4pi, ldots pi2 pi, 3pi, 5pi, ldots Answer: B) 0, 2pi, 4pi, ldots
Explanation: Whole cycles. SLO P-11-D-08 · Understanding
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120. Two coherent sources in phase emit waves of wavelength lambda. At a point, the path difference is 0.5lambda. The interference there is: neither — there is no interference constructive constructive only for sound destructive Answer: D) destructive
Explanation: Path difference = nlambda gives constructive; (n + 12)lambda gives destructive. Here 0.5lambda ⇒ destructive. SLO P-11-D-07 · Application
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121. Two coherent sources in phase emit waves of wavelength lambda. At a point, the path difference is 2lambda. The interference there is: constructive only for sound neither — there is no interference destructive constructive Answer: D) constructive
Explanation: Path difference = nlambda gives constructive; (n + 12)lambda gives destructive. Here 2lambda ⇒ constructive. SLO P-11-D-07 · Application
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122. For the second harmonic of a string, the number of loops is: 1 2 3 4 Answer: B) 2
Explanation: L = lambda. SLO P-11-D-12 · Understanding
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123. Noise-cancelling headphones need a battery because: electronic circuits must process the noise and drive the speakers they use magnets only the sound waves need charging they heat the ear Answer: A) electronic circuits must process the noise and drive the speakers
Explanation: Active system. SLO P-11-D-09 · Understanding
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124. Walking past two loudspeakers playing the same note, you hear: a constant loudness silence everywhere a changing pitch alternate loud and quiet regions Answer: D) alternate loud and quiet regions
Explanation: Interference pattern. SLO P-11-D-08 · Understanding
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125. Microwaves from one transmitter passing through two slits act as: two incoherent sources two coherent sources longitudinal waves one incoherent source Answer: B) two coherent sources
Explanation: Double-slit with microwaves. SLO P-11-D-14 · Understanding
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