CBSE Class 9 Science Chapter 10 Sound Waves Characteristics And Applications MCQs Set 03

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

Explanation:
A point source continuously producing sound causes compressions and rarefactions to spread through the surrounding medium in all directions as spherical waves. This explains why you can hear the tuning fork regardless of the orientation at which you hold it near your ear.

Answer: (B) Only the disturbance (compression and rarefaction regions) travels forward; particles themselves vibrate in place

Explanation:
The chapter emphasizes that in sound wave propagation, particles of the medium do not travel with the wave. They vibrate about their mean positions. The observed motion of the marked coil demonstrates this: the disturbance travels down the slinky while each turn moves back and forth parallel to the wave direction but does not move along with the wave itself.

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

Explanation:
This observation directly shows that sound cannot propagate in a vacuum. Without a material medium (air), even though the speaker continues to vibrate and produce disturbances, there is no medium through which compressions and rarefactions can travel to reach the observer. This is analogous to the vacuum bell jar experiment described in the chapter.

Answer: (C) 16 oscillations

Explanation:
Frequency defines the number of oscillations per unit time. If the frequency is 2 Hz, then 2 oscillations occur per second. In 8 seconds, the total number = 2 Hz × 8 s = 16 oscillations.

Answer: (C) 1020 m

Explanation:
Distance = speed × time = 340 m/s × 6 s = 2040 m. However, this is not the distance to the lightning; it is the distance sound traveled in 6 seconds. We can approximate the lightning strike distance as roughly 2 km away, since light reaches us almost instantaneously.

Answer: (B) Reverberation, where reflections blend with the original sound

Explanation:
For echoes to be heard as separate sounds, the time delay must be at least 0.1 s. A delay of 0.04 s is less than this threshold, so the reflected sound arrives too quickly to be distinguished from the original. Multiple reflections arriving with time differences less than 0.05 s create reverberation, where sounds blend together. This is why auditoriums experience reverberation rather than echo.

Answer: (C) Each instrument produces a unique pattern of overtones in addition to the fundamental frequency

Explanation:
The chapter explains that a musical note contains a fundamental frequency plus overtones. Different instruments produce different patterns and intensities of overtones based on their shape, material, and construction. This quality is called timbre. Even with identical fundamental frequencies and loudness, the different distribution of overtones makes each instrument sound unique.

Answer: (B) 1500 m

Explanation:
• The total time of 2.0 s includes travel to the object and back
• Time to reach the object = 2.0 s ÷ 2 = 1.0 s
• Distance = speed × time = 1500 m/s × 1.0 s = 1500 m

Answer: (C) 1.6 m

Explanation:
Wavelength is defined as the distance between two consecutive crests (compressions) or two consecutive troughs (rarefactions). One complete wavelength consists of one compression and one rarefaction region. If compressions are 0.8 m apart, this represents only half a wavelength. Therefore, the full wavelength = 0.8 m × 2 = 1.6 m.

Answer: (A) 0.33 seconds

Explanation:
Frequency = number of oscillations / time = 12 / 4 = 3 Hz. Time period T = 1 / frequency = 1 / 3 ≈ 0.33 s. The time period and frequency are inversely related, with shorter time periods corresponding to higher frequencies.

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

Explanation:
Shorter wavelengths allow for better spatial resolution when detecting objects through echolocation. At 40 kHz, the wavelength in air is much shorter than at 5 kHz (using v = λν). This shorter wavelength enables bats to detect and pinpoint the location of small flying insects with greater precision through reflected echoes.

Answer: (B) Wavelength increases as temperature increases

Explanation:
Using v = λν, wavelength λ = v / ν. At 0°C: λ = 331 / 172 ≈ 1.92 m. At 22°C: λ = 344 / 172 ≈ 2.0 m. Since frequency stays constant but speed increases with temperature, the wavelength increases proportionally. This illustrates that in a given medium at a given condition, the speed of sound depends only on the medium, not the frequency.

Answer: (B) Sound carries energy that can cause physical displacement of matter

Explanation:
The movement of grains shows that sound waves transfer energy to the medium. As sound propagates through air and reaches the sheet, it causes the sheet and grains to vibrate. This energy transfer is proportional to the amplitude and intensity of the sound—louder sounds cause larger displacements. The chapter notes this directly demonstrates that sound is a form of energy.

Answer: (B) Reverberation where the reflected sound blends with and enriches the original

Explanation:
Reverberation occurs when sound reflections arrive with time differences less than 0.05 s (distinct echoes require at least 0.1 s delay). A 0.025 s delay falls within the reverberation range, causing the reflected sound to blend with the original, creating a rich, sustained acoustic effect. This is why concert halls are deliberately designed with reverberation characteristics.

Answer: (B) Pitch relates to frequency; loudness relates to amplitude

Explanation:
The chapter explains that human perception of sound is subjective. High-pitch sounds correspond to high-frequency waves, while low-pitch sounds have lower frequencies. Loudness is perceived based on amplitude—larger amplitude produces louder sound, while smaller amplitude sounds softer. These perceptual qualities do not have simple mathematical relationships with physical properties but follow general trends.

Answer: (C) 6 cm

Explanation:
One complete wavelength consists of one compression plus one rarefaction. The compression spans 0 to 2 cm (2 cm) and the rarefaction spans 2 to 4 cm (2 cm). However, the pattern repeats, so the next compression would begin at 4 cm. The distance from the start of one compression to the start of the next compression is one complete wavelength: 2 cm (compression) + 2 cm (rarefaction) + 2 cm (next compression starts) = one full cycle. Thus wavelength = 4 cm represents half the full pattern shown; the complete wavelength is 4 cm if measuring crest-to-crest distance within the shown pattern, or continuing the repeating pattern, each complete C-R pair = 4 cm total spacing observed, making λ = 4 cm if that is crest-to-crest within the shown region. Re-examining: if C goes 0-2 and R goes 2-4, then the next C starts at 4. Distance from start of first C to start of next C = 4 cm. This is the wavelength.

Answer: (B) Sound waves spread outward as spherical waves from the vibrating source

Explanation:
The chapter explains that when a small sound source continuously produces sound in all directions, compressions and rarefactions spread through the surrounding medium in all directions as spherical waves. This happens because the vibrating particles of the source collide with surrounding particles in all directions, not just along a single path.

Answer: (B) Larger energy transfer corresponds to greater amplitude and energy in the sound wave

Explanation:
When the plate is struck harder, more energy is transferred to the particles of the surrounding medium, causing larger displacements from their mean positions. As a result, the sheet vibrates to a larger displacement and the grains jump higher. This shows amplitude and energy are directly related.

Answer: (B) The density variations (compressions and rarefactions) move while individual particles oscillate in place

Explanation:
The disturbance consisting of alternating compressions and rarefactions propagates through the medium without the actual flow of the particles themselves. Each part of the slinky oscillates about its own position while the wave pattern advances. This is the defining characteristic of how mechanical waves propagate.

Answer: (B) Different media have different amounts of resistance to vibration due to their molecular structure

Explanation:
Different media have different properties that affect how readily vibrations persist. Water has greater density and molecular interactions than air, which causes more rapid energy dissipation from the vibrating object. This relates to how sound propagates differently through solids, liquids, and gases based on their properties.

Answer: (A) P has the longest time period, Q is intermediate, and R has the shortest time period

Explanation:
The number of compressions passing a point per second equals the frequency. Frequency and time period are inversely related by the equation ν = 1/T. Wave P (10 Hz) has the longest time period of 0.1 s, Wave Q (40 Hz) has 0.025 s, and Wave R (80 Hz) has the shortest period of 0.0125 s.

Answer: (B) Sound requires a material medium to propagate and cannot travel through empty space

Explanation:
• Air is being removed → sound becomes fainter
• Near-vacuum is reached → almost no sound heard
• This directly demonstrates sound cannot propagate in vacuum and needs a medium.

Answer: (C) 200 Hz

Explanation:
Using v = λ × ν, the first wave gives v = 50 × 100 = 5000 m/s. Since the speed of sound in the medium does not change, the second wave must satisfy 5000 = 25 × ν, so ν = 200 Hz. When wavelength decreases by half while speed stays constant, frequency doubles.

Answer: (D) The first observer is approximately 1020 m closer to the lightning than the second observer

Explanation:
Distance to lightning = speed of sound × time delay. First observer: d₁ = 340 × 3 = 1020 m. Second observer: d₂ = 340 × 6 = 2040 m. The first observer is 2040 − 1020 = 1020 m closer to the lightning. The second observer is exactly twice as far away, but the first answer choice stating they are equidistant is incorrect.

Answer: (C) The pattern and intensity of overtones differ between the instruments, giving each its unique timbre

Explanation:
Timbre is the quality that makes different instruments sound unique even when playing the same note at the same loudness. This comes from their shape, material, and construction, which determine the pattern and intensity of the overtones that accompany the fundamental frequency. The chapter explicitly defines this for instruments like the flute, ektara, and tabla.