MCQs / Physics / Superposition, Stationary Waves and Beats
Superposition, Stationary Waves and Beats MCQs — Physics (page 6) 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.
126. Two coherent sources can have: a constant non-zero phase difference different frequencies only zero phase difference a random phase difference Answer: A) a constant non-zero phase difference
Explanation: Phase difference must be constant, not necessarily zero. SLO P-11-D-14 · Understanding
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127. Two tuning forks of 440 Hz and 443 Hz are sounded together. The number of beats heard per second is: 442 6 3 883 Answer: C) 3
Explanation: Beat frequency = f1 - f2 = 3 Hz. SLO P-11-D-15 · Understanding
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128. The phase difference between the noise and the anti-noise should be: pi4 rad 2pi rad pi rad 0 rad Answer: C) pi rad
Explanation: Antiphase. SLO P-11-D-09 · Understanding
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129. Particles in adjacent loops of a stationary wave vibrate: in antiphase with 90^circ phase difference with no motion in phase Answer: A) in antiphase
Explanation: One loop goes up while the next goes down. SLO P-11-D-11 · Understanding
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130. Two identical waves interfere completely destructively. The resultant amplitude is: zero the same as one wave half double Answer: A) zero
Explanation: They cancel. SLO P-11-D-08 · Understanding
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131. When two sounds produce beats, the pitch heard is: zero the average of the two frequencies the sum of the frequencies the difference of the frequencies Answer: B) the average of the two frequencies
Explanation: The 'carrier' frequency. SLO P-11-D-15 · Understanding
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132. In the microwave experiment, the distance moved by the detector between two successive minima is 1.6 cm. The microwave frequency is (c = 3 10^8 m/s): 4.7 10^9 Hz 9.4 10^9 Hz 1.9 10^10 Hz 1.6 10^10 Hz Answer: B) 9.4 10^9 Hz
Explanation: lambda = 3.2 cm; f = c/lambda = 9.4 10^9 Hz. SLO P-11-D-10 · Application
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133. Two instruments play together with 6 beats in 2 seconds. Their frequency difference is: 6 Hz 12 Hz 3 Hz 2 Hz Answer: C) 3 Hz
Explanation: fb = 6/2. SLO P-11-D-16 · Application
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134. Behind a small obstacle in a ripple tank, waves: leave a perfectly sharp shadow speed up reflect only bend round into the 'shadow' region Answer: D) bend round into the 'shadow' region
Explanation: Diffraction round obstacles. SLO P-11-D-13 · Understanding
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135. Ordinary light from a filament lamp is incoherent because: it has only one frequency it is polarized atoms emit light randomly and independently it is very weak Answer: C) atoms emit light randomly and independently
Explanation: Short, random wave trains. SLO P-11-D-14 · Understanding
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136. A string of length 0.8 m fixed at both ends carries waves at 320 m/s. Its fundamental frequency is: 200 Hz 800 Hz 400 Hz 100 Hz Answer: A) 200 Hz
Explanation: f1 = v2L = 3202(0.8) = 200 Hz. SLO P-11-D-12 · Understanding
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137. The coherence of a source is important in: holography sunbathing photographs with flash ordinary room lighting Answer: A) holography
Explanation: Holograms need laser light. SLO P-11-D-14 · Understanding
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138. In a microwave stationary-wave experiment, adjacent nodes are 1.4 cm apart. The wavelength of the microwaves is: 5.6 cm 2.8 cm 11.2 cm 1.4 cm Answer: B) 2.8 cm
Explanation: Node to node = lambda2, so lambda = 2 1.4 = 2.8 cm. SLO P-11-D-10 · Application
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139. A string of length 1.2 m fixed at both ends vibrates in 3 loops. The wavelength is: 0.4 m 0.8 m 1.6 m 3.6 m Answer: B) 0.8 m
Explanation: Each loop is lambda2: lambda = 2Ln = 0.8 m. SLO P-11-D-10 · Application
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140. A pipe open at both ends produces: only odd harmonics only even harmonics all harmonics (1 : 2 : 3 …) no harmonics Answer: C) all harmonics (1 : 2 : 3 …)
Explanation: Antinodes at both ends. SLO P-11-D-12 · Understanding
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141. In Melde's experiment (vibrating string), the stationary wave is set up by: two separate vibrators wind sound from a loudspeaker waves reflected from the fixed end overlapping the incident waves Answer: D) waves reflected from the fixed end overlapping the incident waves
Explanation: Reflection at the pulley end. SLO P-11-D-10 · Application
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142. During diffraction, the wavelength of the waves: increases stays the same decreases becomes zero Answer: B) stays the same
Explanation: Speed and frequency are unchanged. SLO P-11-D-13 · Understanding
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143. In the example above, sticking wax on the unknown fork (lowering its frequency) increases the beats to 5/s. The unknown frequency was: 397 Hz 405 Hz 403 Hz 395 Hz Answer: A) 397 Hz
Explanation: Lowering 397 Hz moves it further from 400 Hz. SLO P-11-D-15 · Understanding
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144. Destructive interference in noise-cancelling headphones means the energy of the noise is: converted into electric charge converted into light destroyed completely not destroyed — energy is redistributed or not delivered to the ear Answer: D) not destroyed — energy is redistributed or not delivered to the ear
Explanation: Energy is conserved. SLO P-11-D-09 · Understanding
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145. In a graph showing two waves travelling in opposite directions, at the instant they are exactly in phase the resultant: is zero everywhere has the same amplitude as one wave is a straight line has twice the amplitude of each wave at the antinodes Answer: D) has twice the amplitude of each wave at the antinodes
Explanation: Maximum displacement instant. SLO P-11-D-11 · Understanding
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146. Which waves would diffract most through a 1 m doorway? visible light ultraviolet sound of wavelength 1 m X-rays Answer: C) sound of wavelength 1 m
Explanation: Wavelength ≈ gap. SLO P-11-D-13 · Understanding
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147. If the anti-noise wave were in phase with the noise, the listener would hear: louder noise no change silence a higher pitch Answer: A) louder noise
Explanation: Constructive interference. SLO P-11-D-09 · Understanding
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148. A string of length 1 m fixed at both ends carries waves at 100 m/s. Its fundamental frequency is: 50 Hz 200 Hz 25 Hz 100 Hz Answer: A) 50 Hz
Explanation: f1 = v2L = 1002(1) = 50 Hz. SLO P-11-D-12 · Understanding
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149. When light passes through a single narrow slit, the pattern on a screen is: a single sharp line a broad central bright band with weaker bands either side a uniform bright patch complete darkness Answer: B) a broad central bright band with weaker bands either side
Explanation: Single-slit diffraction. SLO P-11-D-13 · Understanding
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150. After two wave pulses pass through each other, they: reverse direction merge into one pulse are destroyed continue with their original shapes Answer: D) continue with their original shapes
Explanation: Superposition is temporary. SLO P-11-D-07 · Application
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