RowQ
The Vault
RowQ
The Vault
CBSE Class 9 Physics · 12 questions · 29 marks
Clap your hands and you set the air itself vibrating in a pattern that reaches every ear in the room a fraction of a second later. This chapter follows that disturbance: how a vibrating body produces compressions and rarefactions, why sound needs a material medium while light does not, and how pitch, loudness, and quality are decided. It ends with echoes and ultrasound, where a returning pulse becomes a measuring instrument.
An electric bell is ringing inside a sealed glass jar. As the air is slowly pumped out, what happens to the sound heard outside?
Answer
It grows fainter and finally cannot be heard at all — sound is a mechanical wave and needs particles of a medium to carry the compressions and rarefactions. As the air is removed there are fewer particles to transmit the vibrations, and in a near-vacuum no sound can reach the listener even though the hammer is still visibly striking the gong.
A tuning fork vibrates 340 times per second and the speed of sound in air is 340 m/s. What is the wavelength of the sound produced?
Answer
1 m — using v = fλ, we get λ = v/f = 340/340 = 1 m. Each complete compression-rarefaction pair therefore stretches over one metre of air.
Two sounds have the same frequency but one has a much larger amplitude. How do they differ to a listener?
Answer
The larger-amplitude sound is louder — loudness depends on the amplitude of the wave, since a bigger amplitude carries more energy to the ear. Pitch is decided by frequency, which is the same for both, so the two sounds are equally high or low in pitch but unequal in loudness.
Which of these frequencies would be classified as ultrasound for a human listener?
Answer
45000 Hz — the human audible range is about 20 Hz to 20000 Hz. Anything above 20000 Hz is ultrasound, so 45000 Hz qualifies. 12 Hz lies below the range and is infrasound, while 500 Hz and 16000 Hz are both ordinary audible sounds.
Assertion (A): You cannot hear a distinct echo when you shout at a wall standing only 5 m away from it. Reason (R): The sensation of a sound persists in the human ear for about 0.1 s, so a reflected sound arriving sooner than that merges with the original.
Answer
Both A and R are true and R is the correct explanation of A — from 5 m the sound covers only 10 m going and returning, taking about 10/344 ≈ 0.03 s. Since that is well under the 0.1 s persistence of hearing, the reflected sound overlaps the original and no separate echo is perceived. A distance of at least about 17.2 m is needed.
Explain why sound is called a longitudinal wave, and name one difference between it and a transverse wave.
Answer
Sound is called a longitudinal wave because the particles of the medium vibrate to and fro along the same line in which the wave travels, producing alternate compressions and rarefactions rather than crests and troughs. In a transverse wave, by contrast, the particles vibrate at right angles to the direction of propagation — as on a plucked string or the surface of water — and such waves can travel through solids and along liquid surfaces but not through the body of a gas.
A person standing between two parallel cliffs claps once and hears the first echo after 1.5 s and the second after 2.5 s. Taking the speed of sound as 340 m/s, find the distance of each cliff and the separation between them.
Answer
For an echo the sound travels to the cliff and back, so distance = (v × t)/2. Nearer cliff: d₁ = (340 × 1.5)/2 = 510/2 = 255 m. Farther cliff: d₂ = (340 × 2.5)/2 = 850/2 = 425 m. Since the person stands between the two cliffs, their separation = 255 + 425 = 680 m.
Why does a whisper sound different from a shout of the same pitch, and why do a violin and a flute sounding the same note still sound unlike each other?
Answer
A whisper and a shout of the same pitch have the same frequency but different amplitudes; the shout carries far more energy, so it is heard as much louder while remaining the same note. A violin and a flute playing the same note differ in quality (timbre), because each instrument produces a different mixture of additional overtones along with the main frequency, giving each a distinctive waveform and therefore a recognisable character.
A source vibrating at 250 Hz produces sound that travels 1500 m in 4.4 s through a liquid. Find the speed of sound in that liquid, its wavelength there, and the time period of the vibration.
Answer
Speed: v = distance / time = 1500/4.4 = 340.9 m/s (approximately 341 m/s). Wavelength: λ = v/f = 340.9/250 = 1.36 m (approximately). Time period: T = 1/f = 1/250 = 0.004 s. Note that the time period depends only on the source's frequency, so it stays 0.004 s no matter which medium the sound passes into, while the speed and wavelength both change with the medium.
(a) Describe how a vibrating tuning fork sets up a sound wave in the surrounding air, explaining compressions and rarefactions. (b) Explain why sound travels faster in steel than in air. (c) A hammer strike on a long steel rail is heard twice by a listener at the far end. Explain this observation.
Answer
(a) When a tuning fork is struck, its prongs vibrate rapidly to and fro. As a prong moves outward it pushes the layer of air next to it, crowding the molecules together and creating a region of higher pressure and density called a compression. As the prong swings back, it leaves behind a region where the molecules are spread out, of lower pressure and density, called a rarefaction. This alternate crowding and thinning is passed on from layer to layer, so a series of compressions and rarefactions travels outward through the air. The individual air molecules only oscillate about their own positions; it is the disturbance, not the air itself, that moves along. (b) Sound travels faster in steel because its particles are packed far more closely and are held by much stronger interatomic forces, making it highly elastic. A disturbance is therefore handed on from particle to particle almost immediately. In air the molecules are widely separated and interact weakly, so the disturbance is relayed much more slowly. (c) The hammer blow sends sound along two paths at once: through the steel rail and through the air. Because sound moves several times faster in steel, the vibration through the rail reaches the listener first, and the sound travelling through the air arrives a noticeable moment later. The listener therefore hears two separate sounds from a single blow.
(a) What is ultrasound? List three practical applications and explain how one of them works. (b) A ship's SONAR sends a pulse straight down and receives its echo from the seabed after 3.2 s. If the speed of sound in seawater is 1500 m/s, calculate the depth of the sea at that point.
Answer
(a) Ultrasound is sound of frequency higher than about 20000 Hz, which is above the upper limit of human hearing. Because it has a short wavelength, it can be sent out as a narrow, well-directed beam that reflects sharply from small objects. Three applications: (i) SONAR, to measure the depth of the sea and to locate submarines, shoals of fish, and submerged wrecks; (ii) medical ultrasonography, to image the heart, liver, kidneys, and a developing foetus; (iii) industrial use, to clean intricate parts such as spirals and electronic components, and to detect hidden cracks in metal blocks and pipelines. How flaw detection works: an ultrasonic pulse is sent through a metal casting. A sound, uniform block transmits the pulse right through to a detector on the other side. If there is an internal crack or air gap, the pulse is reflected back at that boundary and the detector receives a weakened or delayed signal, revealing the defect without cutting the block open. (b) The pulse travels down to the seabed and back, so the total path is twice the depth. Using d = (v × t)/2 = (1500 × 3.2)/2 = 4800/2 = 2400 m. The depth of the sea at that point is 2400 m.
Read the following and answer the questions that follow: A school auditorium has bare concrete walls, and students complain that speeches from the stage sound muddled. An acoustics consultant reports that sound from the stage reaches the rear wall 24 m away, reflects, and returns to the audience seated near the stage, producing overlapping repetitions. She recommends fixing thick curtains and perforated boards along the walls and ceiling. Take the speed of sound in air as 344 m/s. (a) Calculate the time between the direct sound and its reflection from the rear wall for a listener near the stage. (b) Will a listener hear this reflection as a separate echo? Justify your answer. (c) Explain how curtains and perforated boards fix the problem. (d) Name one other place where the same principle of sound reflection is deliberately put to use.
Answer
(a) The reflected sound travels to the rear wall and back, a total of 2 × 24 = 48 m, while the direct sound travels a negligible extra distance for a listener near the stage. Time = distance / speed = 48/344 = 0.14 s (approximately). (b) Yes. The gap of about 0.14 s is greater than the 0.1 s for which the sensation of sound persists in the human ear, so the reflection is heard as a distinct echo rather than merging with the original. This is exactly why the speech sounds muddled — every syllable is followed by a delayed copy of itself. (c) Curtains and perforated boards have soft, porous, uneven surfaces. Instead of reflecting sound cleanly the way hard concrete does, they trap the sound waves in their pores and fibres and absorb much of the energy, converting it to a tiny amount of heat. With far less sound reflected back, the echoes die away and speech from the stage stays crisp. (d) A stethoscope uses repeated reflection of sound along its tubes to carry the faint sounds of a patient's heart and lungs to the doctor's ears. Megaphones and the curved ceilings of concert halls also use deliberate reflection to send sound towards the audience.
RowQ generates fresh questions on Sound, marks your answers, and explains every step.
Start free