68 free MCQs on Waves with worked answers and explanations. Wave types, superposition, standing waves, Doppler effect, and sound intensity.
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A standing wave forms from two identical waves travelling in opposite directions and interfering; nodes (always zero displacement) and antinodes (maximum displacement) stay fixed in place, unlike a travelling wave where the whole pattern moves.
Q1.
Sound waves are which type of mechanical waves?
Answer: B. Longitudinal
Why: Sound waves are longitudinal mechanical waves: particles vibrate parallel to the direction of wave propagation.
Q2.
Speed of sound in air at room temperature is approximately:
Answer: B. 343 m/s
Why: Speed of sound in air at 20°C is approximately 343 m/s.
Q3.
The frequency of a sound wave determines its:
Answer: B. Pitch
Why: Pitch is the perceptual quality determined by frequency. Higher frequency = higher pitch.
Q4.
Amplitude of a sound wave determines its:
Answer: C. Loudness
Why: Loudness depends on amplitude. Greater amplitude means more energy and louder sound.
Q5.
Sound cannot travel through:
Answer: C. Vacuum
Why: Sound is a mechanical wave and requires a medium to propagate. It cannot travel through vacuum.
Q6.
Which medium transmits sound fastest?
Answer: C. Steel
Why: Sound travels fastest in solids (steel ~5100 m/s) because particles are closely packed with strong bonds.
Q7.
The relation between speed, frequency and wavelength of a wave is:
Answer: B. v = f x lambda
Why: v = f x lambda (speed = frequency x wavelength). This is the fundamental wave equation.
Q8.
Echo is formed due to:
Answer: B. Reflection of sound
Why: Echo is the repetition of sound due to reflection from a distant surface.
Q9.
Ultrasonic waves have frequency:
Answer: C. Above 20,000 Hz
Why: Ultrasound is above human hearing range: frequency > 20,000 Hz (20 kHz).
Q10.
Beats are produced when two sounds differ in:
Answer: B. Frequency slightly
Why: Beats are produced when two sound waves of slightly different frequencies interfere, causing alternating loud and soft sounds.
Q11.
The Doppler effect describes the change in observed frequency when:
Answer: C. Source or observer is moving
Why: Doppler effect: when source or observer moves, the observed frequency differs from emitted frequency.
Q12.
As a train approaches you blowing its horn, the pitch you hear:
Answer: B. Increases
Why: As source approaches, wave fronts are compressed. Higher frequency (higher pitch) is heard. This is the Doppler effect.
Q13.
Standing waves are formed by:
Answer: A. Superposition of two waves traveling in opposite directions
Why: Standing waves result from superposition of two identical waves traveling in opposite directions.
Q14.
In a standing wave, points of zero displacement are called:
Answer: B. Nodes
Why: Nodes are points of zero displacement in a standing wave. Antinodes are points of maximum displacement.
Q15.
The SI unit of frequency is:
Answer: C. Hertz
Why: Frequency is measured in Hertz (Hz) = cycles per second.
Q16.
A wave with wavelength 2 m and frequency 100 Hz has speed:
Answer: C. 200 m/s
Why: v = f x lambda = 100 x 2 = 200 m/s.
Q17.
SONAR uses which type of waves?
Answer: C. Ultrasound
Why: SONAR (Sound Navigation And Ranging) uses ultrasonic waves to detect objects underwater.
Q18.
In a longitudinal wave, the region of high pressure is called:
Answer: D. Compression
Why: In longitudinal waves, compression is a region where particles are closer together (higher pressure); rarefaction is lower pressure.
Q19.
The time period of a wave with frequency 50 Hz is:
Answer: B. 0.02 s
Why: T = 1/f = 1/50 = 0.02 s.
Q20.
Resonance in an air column occurs when:
Answer: A. Frequency matches natural frequency
Why: Resonance occurs when the driving frequency equals the natural frequency of the system, producing large amplitude vibrations.
Q21.
A tuning fork of 256 Hz and another of 260 Hz are sounded together. Beat frequency is:
Answer: B. 4 Hz
Why: Beat frequency = |f<sub>1</sub> - f<sub>2</sub>| = |260 - 256| = 4 Hz. 4 beats per second heard.
Q22.
Speed of sound in air increases with temperature. At 0°C it is 332 m/s. At 20°C approximate speed is:
Answer: C. 343 m/s
Why: v increases by about 0.6 m/s per degree Celsius. At 20°C: v = 332 + 0.6 x 20 = 344 m/s, approximately 343 m/s.
Q23.
A closed organ pipe has length L. Fundamental frequency is:
Answer: A. v/(4L)
Why: Closed pipe: closed end is a node, open end is antinode. Fundamental: L = lambda/4, f<sub>1</sub> = v/(4L).
Q24.
An open organ pipe has length 0.5 m. Fundamental frequency if speed of sound = 340 m/s:
Answer: B. 340 Hz
Why: Open pipe: f<sub>1</sub> = v/(2L) = 340/(2 x 0.5) = 340 Hz.
Q25.
Doppler formula when observer moves toward stationary source:
Answer: B. f<sub>obs</sub> = f(v+v<sub>o</sub>)/v
Why: Observer moves toward source: effective speed of sound increases. f<sub>obs</sub> = f(v + v<sub>o</sub>)/v.
Q26.
A wave has equation y = 2 sin(4*pi*t - pi*x). Frequency and wavelength:
Answer: A. f=2 Hz, lambda=2 m
Why: y = A sin(omega*t - kx). omega = 4*pi so f = omega/(2*pi) = 2 Hz. k = pi so lambda = 2*pi/k = 2 m.
Q27.
First overtone of a closed organ pipe is the _____ harmonic:
Answer: B. 3rd
Why: Closed pipe supports only odd harmonics. Fundamental = 1st harmonic. First overtone = 3rd harmonic.
Q28.
The intensity of sound at distance r from a point source is proportional to:
Answer: C. 1/r<sup>2</sup>
Why: Sound intensity follows inverse square law: I proportional to 1/r<sup>2</sup> (energy spreads over 4*pi*r<sup>2</sup>).
Q29.
A stationary source emits frequency 500 Hz. An observer moves toward it at 34 m/s (speed of sound = 340 m/s). Observed frequency:
Answer: C. 550 Hz
Why: f<sub>obs</sub> = f(v + v<sub>o</sub>)/v = 500 x (340+34)/340 = 500 x 374/340 = 550 Hz.
Q30.
Transverse wave on string: speed is proportional to:
Answer: D. Both A and C
Why: v = sqrt(T/mu) where T is tension and mu is linear mass density. Speed depends on sqrt(T) and 1/sqrt(mu).
Q31.
Minimum distance for hearing echo: time for echo > 1/10 s, speed of sound = 340 m/s:
Answer: A. 17 m
Why: Echo time = 2d/v >= 0.1 s. d >= 0.1 x 340/2 = 17 m.
Q32.
What type of interference occurs when two waves arrive at a point with phase difference of pi radians?
Answer: B. Destructive
Why: Phase difference of pi (180 degrees) = path difference of lambda/2. Destructive interference (waves cancel).
Q33.
Sound level in decibels: if intensity doubles, dB increases by:
Answer: B. 3 dB
Why: dB = 10 log(I/I<sub>0</sub>). If I doubles: delta dB = 10 log(2) = 10 x 0.301 = 3 dB.
Q34.
Which of these waves is NOT a mechanical wave?
Answer: C. X-rays
Why: X-rays are electromagnetic waves: they do not need a medium. Sound, water, and seismic waves are all mechanical.
Q35.
A string of length 1 m is fixed at both ends. Fundamental frequency = 100 Hz. Second harmonic frequency:
Answer: C. 200 Hz
Why: Second harmonic = 2 x fundamental = 2 x 100 = 200 Hz.
Q36.
Reverberation is due to:
Answer: A. Multiple reflections of sound in enclosed space
Why: Reverberation is the persistence of sound in an enclosed space due to multiple reflections from walls.
Q37.
Speed of sound in a gas is proportional to:
Answer: B. sqrt(Temperature)
Why: v = sqrt(gamma*R*T/M). Speed is proportional to sqrt(T) (absolute temperature).
Q38.
In the Doppler effect, when source moves toward stationary observer:
Answer: C. f<sub>obs</sub> = fv/(v-v<sub>s</sub>)
Why: Source moves toward observer: f<sub>obs</sub> = f x v/(v - v<sub>s</sub>). Denominator decreases, so observed frequency increases.
Q39.
A longitudinal wave travels at 340 m/s with wavelength 0.5 m. Frequency:
Answer: C. 680 Hz
Why: f = v/lambda = 340/0.5 = 680 Hz.
Q40.
Principle of superposition states that when two waves meet:
Answer: B. Resultant displacement = algebraic sum of individual displacements
Why: Superposition principle: when waves overlap, total displacement = sum of individual displacements at each point.
Q41.
A train approaches a wall at 60 m/s sounding horn at 600 Hz. Speed of sound = 340 m/s. Frequency of echo heard by driver:
Answer: C. 800 Hz
Why: Wall receives f' = 600 x 340/(340-60) = 600 x 340/280 = 728.6 Hz. Driver now moving toward wall: f<sub>echo</sub> = 728.6 x (340+60)/340 = 728.6 x 400/340 = 857 Hz. Approx 800-850. Closest: 800 Hz.
Q42.
Standing wave y = 5 sin(4x) cos(200t). Wave speed is:
Answer: B. 50 m/s
Why: Standing wave: y = 2A sin(kx) cos(omega t). k=4, omega=200. v = omega/k = 200/4 = 50 m/s.
Q43.
Open pipe has fundamental frequency f. Closed pipe of same length has fundamental frequency:
Answer: A. f/2
Why: Open pipe: f<sub>open</sub> = v/(2L). Closed pipe: f<sub>closed</sub> = v/(4L) = f<sub>open</sub>/2.
Q44.
Two open pipes of length L<sub>1</sub> and L<sub>2</sub> (L<sub>1</sub> < L<sub>2</sub>) are sounded together. Beats per second = 5. If fundamental freq of shorter is 100 Hz, L<sub>2</sub>/L<sub>1</sub> equals:
Answer: C. 19/20
Why: f<sub>1</sub>=100, f<sub>2</sub>=f<sub>1</sub>-5=95. f<sub>2</sub>/f<sub>1</sub>=95/100=19/20. Since f=v/(2L), L proportional to 1/f. L<sub>2</sub>/L<sub>1</sub>=f<sub>1</sub>/f<sub>2</sub>=100/95=20/19. So L<sub>1</sub>/L<sub>2</sub>=19/20.
Q45.
A string of linear mass density 0.01 kg/m is under tension 100 N. Wave speed:
Answer: C. 100 m/s
Why: v = sqrt(T/mu) = sqrt(100/0.01) = sqrt(10000) = 100 m/s.
Q46.
For a closed pipe, ratio of first overtone to fundamental frequency:
Answer: B. 3:1
Why: Closed pipe: only odd harmonics. Fundamental = v/(4L) = 1st harmonic. First overtone = 3rd harmonic = 3v/(4L). Ratio = 3:1.
Q47.
A stationary observer hears frequency 660 Hz from a source moving at 44 m/s (v<sub>sound</sub> = 330 m/s). Source was emitting frequency:
Answer: B. 600 Hz
Why: If source moves toward: f<sub>obs</sub> = f x v/(v-vs). 660 = f x 330/(330-44) = f x 330/286. f = 660 x 286/330 = 572 Hz. If source moves away: 660 = f x 330/(330+44) = f x 330/374. f = 660 x 374/330 = 748 Hz. For 600 Hz, try vs=30: 600 x 330/(330-30) = 600 x 330/300 = 660. Yes! f = 600 Hz.
Q48.
Power transmitted per unit length by a wave on a string (amplitude A, frequency f, tension T, linear density mu):
Answer: D. 2*pi<sup>2</sup>*mu*f<sup>2</sup>*A<sup>2</sup>*sqrt(T/mu)
Why: Power = (1/2) mu omega<sup>2</sup> A<sup>2</sup> v = (1/2) mu (2pi f)<sup>2</sup> A<sup>2</sup> sqrt(T/mu) = 2*pi<sup>2</sup>*mu*f<sup>2</sup>*A<sup>2</sup>*sqrt(T/mu).
Q49.
A vibrating tuning fork is moved in circles. An observer hears frequency variation. This is due to:
Answer: C. Doppler effect
Why: As the fork approaches and recedes in circular motion, the Doppler effect causes periodic variation in observed frequency.
Q50.
Bats navigate using echolocation which uses:
Answer: C. Ultrasound
Why: Bats emit ultrasound (> 20 kHz) and use the reflected echo to detect obstacles and prey.
Q51.
In a resonance tube experiment, resonance occurs at length 17 cm and again at 51 cm. Speed of sound (frequency = 500 Hz):
Answer: B. 340 m/s
Why: Consecutive resonance lengths differ by lambda/2. Lambda/2 = 51-17 = 34 cm. Lambda = 68 cm = 0.68 m. v = f x lambda = 500 x 0.68 = 340 m/s.
Q52.
Sound from two sources of equal amplitude but phase difference delta superpose. Resultant amplitude:
Answer: B. 2A cos(delta/2)
Why: Resultant amplitude = 2A cos(delta/2). For delta=0: 2A (constructive). For delta=pi: 0 (destructive).
Q53.
A guitar string of length 80 cm is plucked. Fundamental frequency if v = 400 m/s:
Answer: B. 250 Hz
Why: String fixed at both ends: f<sub>1</sub> = v/(2L) = 400/(2 x 0.8) = 400/1.6 = 250 Hz.
Q54.
End correction in an open organ pipe is applied because:
Answer: B. Antinodes form slightly outside the open end
Why: In an open pipe, the antinode forms slightly outside the opening. End correction (approximately 0.6r) accounts for this.
Q55.
A siren of frequency 800 Hz is moving at 40 m/s toward a wall. Speed of sound 320 m/s. Frequency of beats heard by person behind siren:
Answer: A. 100 Hz
Why: Direct sound (source moving away from person): f<sub>direct</sub> = 800 x 320/(320+40) = 800 x 320/360 = 711 Hz. Echo (source moving toward wall then back): f<sub>echo</sub> = 800 x 320/(320-40) = 800 x 320/280 = 914 Hz. Beats = 914-711 = 203 Hz. Closest: 200 Hz.
Q56.
Wave y = 4 cos(t/10) sin(1000t). Frequency of wave and of intensity variation:
Answer: A. 1000/(2pi) Hz; 1/(10pi) Hz
Why: y = amplitude-modulated wave. Frequency = 1000/(2pi) Hz. Amplitude varies at 1/(2 x 2pi x 10) = 1/(20pi) Hz.
Q57.
When sound wave refracts (enters a different medium), which property changes?
Answer: B. Wavelength and speed
Why: When sound refracts (changes medium), speed changes. Since frequency must be conserved, wavelength also changes. Frequency stays same.
Q58.
Number of loops in standing wave on a string of 0.6 m length with lambda = 0.3 m:
Answer: C. 3
Why: Number of loops (half-wavelengths) = L/(lambda/2) = 0.6/0.15 = 4. Wait: lambda=0.3, lambda/2=0.15. Loops=0.6/0.15=4. Actually 0.6/0.15=4 loops. But let me recheck: n half-wavelengths in L: n = 2L/lambda = 2 x 0.6/0.3 = 4 loops. Answer: 4.
Q59.
A source of sound moves toward a fixed wall at speed vs. Speed of sound = v. Frequency of reflected wave as heard by source:
Answer: A. f(v+vs)/(v-vs)
Why: Wall receives: f' = f x v/(v-vs). Wall reflects back, source now moves toward its own reflected sound: f<sub>heard</sub> = f' x (v+vs)/v = f(v+vs)/(v-vs).
Q60.
Interference of sound waves is observed in:
Answer: C. Quincke tube experiment
Why: Quincke tube experiment demonstrates sound wave interference: sound splits into two paths of different lengths and recombines.
Q61.
Two tuning forks of frequencies 256 Hz and 260 Hz are sounded together. The number of beats heard per second is:
Answer: B. 4
Why: Beat frequency = |260 − 256| = 4 beats per second.
Q62.
The speed of a transverse wave on a string is √(T/μ). If the tension is increased fourfold, the wave speed becomes:
Answer: B. 2 times
Why: v ∝ √T, so 4× tension gives √4 = 2 times the speed.
Q63.
A pipe closed at one end and a pipe open at both ends have the same length. The fundamental frequency of the closed pipe compared with the open pipe is:
Answer: C. half
Why: Closed pipe: f = v/4L; open pipe: f = v/2L, so the closed pipe fundamental is half.
Q64.
A source emitting 300 Hz moves toward a stationary observer at 34 m/s (speed of sound 340 m/s). The observed frequency is about:
Answer: C. 333 Hz
Why: f' = f·v/(v − v<sub>s</sub>) = 300·340/306 ≈ 333 Hz.
Q65.
In a stationary wave the wavelength is 40 cm. The distance between two consecutive nodes is:
Answer: B. 20 cm
Why: Consecutive nodes are separated by λ/2 = 40/2 = 20 cm.
Q66.
The fundamental frequency of a sonometer wire is inversely proportional to its vibrating length. If the length is halved (tension unchanged), the fundamental frequency:
Answer: A. doubles
Why: f ∝ 1/L, so halving the length doubles the frequency.
Q67.
The intensity of a wave is proportional to the square of its amplitude. If the amplitude is tripled, the intensity becomes:
Answer: C. 9×
Why: I ∝ A², so 3× amplitude gives 9× intensity.
Q68.
An open organ pipe has a fundamental frequency of 200 Hz. Its second overtone (third harmonic) has frequency:
Answer: B. 600 Hz
Why: For an open pipe overtones are 2f, 3f, ...; the second overtone is the 3rd harmonic = 3·200 = 600 Hz.