CBSE Class 9 Science Chapter 10 Sound Waves Characteristics And Applications MCQs Set 02

Answer: (B) Sound spreads outward in all directions as spherical waves from the source

Explanation:
A vibrating tuning fork produces compressions and rarefactions that spread in all directions through the surrounding medium, creating spherical waves. The chapter explicitly states that when the medium is not confined, vibrating particles collide with surrounding particles in all directions, causing sound to propagate as spherical waves that spread through the medium in all directions.

Answer: (B) Energy travels through the medium while individual particles only vibrate locally

Explanation:
The slinky model demonstrates that regions of compression and rarefaction travel along the spring while the marked turn does not travel with the disturbance but only vibrates about its mean position. This illustrates the fundamental property of sound waves: it is the energy and density disturbance that propagates, not the particles themselves. Each particle oscillates about its equilibrium position as the wave passes through.

Answer: (B) Sound waves require a material medium to propagate through

Explanation:
The vacuum bell jar experiment demonstrates that sound cannot propagate without a medium. As air is removed from the box, sound becomes progressively fainter until almost none is heard in a near vacuum, even though the speaker continues vibrating visibly. This proves sound waves are mechanical waves requiring a material medium for propagation.

Answer: (B) 680 Hz

Explanation:
The distance between consecutive compressions equals the wavelength, so λ = 0.5 m. Using the relationship v = λν, we get ν = v/λ = 340 m/s ÷ 0.5 m = 680 Hz.

Answer: (A) Wave X has a wavelength 4 times longer than Wave Y

Explanation:
• Wavelength is inversely proportional to frequency when speed is constant (v = λν)
• Since frequency of X is one-fourth that of Y (50 Hz vs 200 Hz)
• Wave X must have a wavelength 4 times longer than Wave Y to maintain the same wave speed
• Amplitude does not affect wavelength

Answer: (A) 2250 m

Explanation:
The pulse travels to the object and back, so the total distance traveled is 1500 m/s × 3 s = 4500 m. Since this is the round-trip distance, the actual distance to the object is 4500 m ÷ 2 = 2250 m.

Answer: (B) Reverberation that blends with the original sound without clear separation

Explanation:
Reverberation occurs when reflected sound arrives with a time gap less than 0.05 seconds, causing the reflected sound to blend with the original. Since 0.04 seconds is less than 0.05 seconds, the listener experiences reverberation rather than a distinct echo. Echoes require a minimum time separation of 0.1 seconds to be heard as separate sounds.

Answer: (B) Sound particles vibrate parallel to the wave propagation direction

Explanation:
Sound waves are longitudinal mechanical waves in which particles of the medium vibrate back and forth parallel to the direction of wave propagation, creating alternating compressions and rarefactions along the direction the wave travels.

Answer: (B) The shape, material, and construction of instruments determine the pattern and intensity of overtones

Explanation:
The chapter explains that timbre—the quality that makes sounds unique—comes from the pattern and intensity of overtones produced by an instrument. This pattern depends on the instrument's shape, material, and construction. While the fundamental frequency may be identical across different instruments, the specific combination and strength of overtones differs, creating the distinctive character of each instrument.

Answer: (C) 1360 m

Explanation:
Light travels nearly instantaneously, so the 4-second delay represents the time for sound to reach you. Distance = speed × time = 340 m/s × 4 s = 1360 m. This principle applies because sound travels much slower than light, so we see the lightning before hearing the thunder.

Answer: (A) Reverberation caused by multiple reflections arriving within 0.05 seconds

Explanation:
In small enclosed spaces like bathrooms, sound reflects from multiple surfaces (walls, ceiling, floor) and reaches the listener with time gaps less than 0.05 seconds. This creates reverberation, where reflected sounds blend with and overlap the original sound, causing the voice to seem to echo without clear separation. A true echo requires at least 0.1 seconds delay and occurs in larger spaces.

Answer: (B) Ultrasonic waves have shorter wavelengths allowing detection of smaller objects

Explanation:
Ultrasonic waves have much higher frequencies than audible sound at the same speed, resulting in shorter wavelengths (λ = v/ν). Shorter wavelengths allow echolocation systems to detect smaller objects with better precision. This enables bats to locate tiny flying insects effectively and avoid obstacles in the dark.

Answer: (B) Sound requires a material medium to propagate and transfer energy

Explanation:
The progressive fading of sound as air is removed demonstrates that sound cannot propagate without a medium. Although the bell continues vibrating visibly in the near vacuum, almost no sound reaches the observer, proving sound is a mechanical wave requiring particles of matter to transmit the disturbance from source to listener.

Answer: (B) 0.6 m

Explanation:
Wavelength is defined as the distance between two consecutive crests (compressions) or two consecutive troughs (rarefactions). One complete compression-rarefaction pair represents one full cycle, which equals one wavelength. Therefore, λ = 0.6 m.

Answer: (C) Oscillate back and forth about their mean positions parallel to wave direction

Explanation:
In longitudinal waves like sound, particles of the medium vibrate back and forth parallel to the direction of wave propagation about their equilibrium positions. The particles do not travel with the wave; instead, the disturbance (compressions and rarefactions) propagates through the medium while each particle oscillates locally.

Answer: (B) The marked turn vibrates back and forth about its original position while regions of compressed and spread-out coils travel past it

Explanation:
In Activity 10.5, observers note that while the marked turn oscillates about its rest position, the disturbance itself (represented by closely spaced and spread-out regions) propagates along the slinky. This directly shows that the energy and pattern travel forward, but individual medium particles remain localized in their oscillatory motion.

Answer: (C) Sound spreads outward in all directions as spherical waves from the vibrating source

Explanation:
The chapter explains that when the medium is not confined to a tube, vibrating particles collide with surrounding particles in all directions. A small sound source produces compressions and rarefactions that spread through the medium in all directions as spherical waves, which is why the tuning fork can be heard regardless of its orientation relative to the listener.

Answer: (B) Sound is a form of mechanical energy that requires a material medium to propagate

Explanation:
The vacuum bell jar experiment directly demonstrates that sound requires a material medium. As air is removed, the sound diminishes because there are fewer particles to transmit the compressions and rarefactions. Without a medium, the vibrating bell produces no audible sound despite remaining visually visible.

Answer: (B) Differences in shape, material, and construction determine the pattern and intensity of overtones, which create the unique quality called timbre

Explanation:
The chapter explains that timbre—the unique quality of sound from different instruments—comes from their shape, material, and construction, which determine the pattern and intensity of the overtones produced alongside the fundamental frequency. Even with the same fundamental, different overtone combinations yield distinctly unique sounds.

Answer: (B) Distance = (1500 m/s × 2.4 s) / 2 = 1800 m

Explanation:
• Sound travels to the submarine and back—a round trip
• Total distance = speed × time = 1500 × 2.4 = 3600 m
• Submarine distance = total distance ÷ 2 = 1800 m

Answer: (C) wavelength

Explanation:
Wavelength is the distance between two consecutive crests (or troughs) and represents one complete compression-rarefaction cycle. Using the relationship v = λ × ν, the wavelength directly determines frequency for a given medium speed.

Answer: (A) Frequency = 3 Hz; Time period = 0.33 s

Explanation:
Frequency = number of oscillations ÷ time = 18 ÷ 6 = 3 Hz. Time period T = 1/ν = 1/3 ≈ 0.33 s. These are reciprocal quantities; a lower frequency corresponds to a longer time period between successive compressions.

Answer: (B) Reverberation, because the time delay is less than 0.05 s, causing sounds to blend

Explanation:
Reverberation occurs when sound reflections arrive with a time gap less than 0.05 s, causing them to blend with the original sound rather than being perceived as separate. A 0.03 s delay falls within this range, so the listener hears a persistence of sound rather than a distinct echo.

Answer: (A) 1.36 m

Explanation:
Total distance traveled = speed × time = 340 m/s × 0.008 s = 2.72 m. Since the sound travels to the insect and back, the insect's distance = 2.72 m ÷ 2 = 1.36 m.

Answer: (B) 1.04

Explanation:
Using v = λ × ν, rearranging gives λ = v/ν. At 22°C: λ₁ = 344/220. At 0°C: λ₂ = 331/220. The ratio λ₁/λ₂ = 344/331 ≈ 1.04, showing that wavelength increases slightly with temperature when frequency remains constant.