CBSE Weightage:

CBSE Class 9 Science — Sound

CBSE Class 9 Science — Sound — chapter overview, key concepts, solved examples, and exam strategy.

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Chapter Overview & Weightage

Sound is one of the high-scoring chapters in CBSE Class 9 Science. It regularly contributes 5-7 marks in the annual exam, spread across MCQs, short-answer questions, and occasionally a 5-mark question on the properties of sound or the human ear.

In recent CBSE Class 9 papers, questions from this chapter focus on: (1) echo and reverberation calculations, (2) the human ear diagram with labelling, (3) numerical problems on speed of sound, and (4) the difference between audible and ultrasonic frequencies.

YearMarks in ExamFocus Area
20236Echo + Human ear
20225Speed of sound + Frequency
20217Properties of sound + Human ear diagram
20205Ultrasound + Applications

Key Concepts You Must Know

Production of sound: Sound is produced by vibration of objects. When a sitar string vibrates, it produces compressions and rarefactions in the surrounding air — these pressure variations travel as a longitudinal wave.

Nature of sound waves: Sound is a mechanical, longitudinal wave. It requires a material medium to travel (cannot travel in vacuum). The particles of the medium vibrate parallel to the direction of wave propagation.

Characteristics of sound:

  • Frequency (ff): Number of vibrations per second. Unit: Hertz (Hz). Determines pitch.
  • Amplitude (AA): Maximum displacement of particles from equilibrium. Determines loudness.
  • Wavelength (λ\lambda): Distance between two consecutive compressions (or rarefactions).
  • Speed (vv): v=fλv = f \lambda. In air at 25°C, sound travels at ~346 m/s.

Audible range: 20 Hz to 20,000 Hz (20 kHz). Below 20 Hz = infrasound; above 20 kHz = ultrasound.

Pitch vs Loudness vs Quality (Timbre):

  • Pitch depends on frequency (high frequency = shrill/high pitch)
  • Loudness depends on amplitude (high amplitude = loud)
  • Quality/timbre depends on the waveform shape (distinguishes instruments)

Important Formulas

v=fλ(speed = frequency × wavelength)v = f \lambda \quad \text{(speed = frequency }\times\text{ wavelength)}

Speed of sound in air at 0°C ≈ 332 m/s

Speed increases with temperature: vTv \propto \sqrt{T} (approximately +0.6 m/s per °C)

Speed in water ≈ 1500 m/s; in steel ≈ 5100 m/s

Sound travels fastest in solids and slowest in gases.

For an echo to be heard, the minimum distance of the reflecting surface from the source:

d=v×t2d = \frac{v \times t}{2}

where tt = minimum time for the ear to distinguish the echo from the original sound = 0.1 seconds

With v=340v = 340 m/s:

dmin=340×0.12=17 md_{min} = \frac{340 \times 0.1}{2} = 17 \text{ m}

Loudness is measured in decibels (dB). The human ear can detect sounds from 0 dB (threshold of hearing) to 120 dB (threshold of pain).

L=10log10(II0) dBL = 10 \log_{10}\left(\frac{I}{I_0}\right) \text{ dB}

(Not required for Class 9 CBSE exams, but useful context)


Solved Previous Year Questions

PYQ 1 — Echo Calculation

Q: A person claps his hands near a cliff and hears an echo after 4 seconds. If speed of sound is 340 m/s, how far is the cliff?

Solution:

The sound travels from the person to the cliff and back — total distance = 2d2d.

2d=v×t=340×4=1360 m2d = v \times t = 340 \times 4 = 1360 \text{ m} d=13602=680 md = \frac{1360}{2} = \mathbf{680 \text{ m}}

PYQ 2 — Speed and Wavelength

Q: A sound wave has frequency 200 Hz. If the speed of sound is 340 m/s, calculate the wavelength.

Solution:

λ=vf=340200=1.7 m\lambda = \frac{v}{f} = \frac{340}{200} = 1.7 \text{ m}

PYQ 3 — Ultrasound Application

Q: Give two medical applications of ultrasound. How is ultrasound different from audible sound?

Solution:

Medical applications:

  1. Ultrasonography: Ultrasound waves reflect off organs and are used to create images of the foetus, heart, kidney, etc.
  2. Breaking kidney stones: High-intensity focused ultrasound breaks kidney/gall bladder stones into fine sand-like particles that pass out naturally.

Difference: Ultrasound has frequency above 20,000 Hz, beyond the human audible range. Audible sound ranges from 20 Hz to 20,000 Hz.


The Human Ear — Diagram and Function

The human ear has three main parts: outer ear, middle ear, and inner ear.

Outer ear (Pinna + Ear canal): The pinna (external flap) collects sound waves. The ear canal directs them to the eardrum.

Middle ear (Eardrum + Three ossicles): The eardrum (tympanic membrane) vibrates when sound hits it. Three tiny bones — malleus (hammer), incus (anvil), and stapes (stirrup) — amplify the vibrations and transmit them to the inner ear.

Inner ear (Cochlea + Auditory nerve): The cochlea is a fluid-filled spiral cavity containing hair cells that convert mechanical vibrations into electrical nerve impulses. The auditory nerve carries these to the brain, which interprets them as sound.

The CBSE board has asked students to draw and label the human ear in 5-mark questions multiple times. Memorise: pinna, ear canal, eardrum, three ossicles (malleus/incus/stapes), cochlea, auditory nerve, Eustachian tube (connects middle ear to throat — equalises pressure). That’s 9 labels — enough for full marks.


Difficulty Distribution

DifficultyType of QuestionsExamples
Easy (40%)Define terms, state propertiesDefine pitch, loudness, ultrasound
Medium (40%)Numerical problems, comparisonsEcho calculations, wavelength from v and f
Hard (20%)Application-based, diagram + explanationHuman ear diagram + how it works

Expert Strategy

Start with definitions. Every sound-related term has a precise definition. Pitch, loudness, quality, echo, reverberation, ultrasound, infrasound — learn all definitions before touching numericals. Definitions alone can fetch 2-3 marks in short-answer questions.

For echo problems, the “factor of 2” is crucial. Sound goes from the source to the wall and returns — so total distance = 2 × (distance to wall). Students forget to divide by 2 and lose a mark.

For the human ear question: Draw a clear, labelled diagram. CBSE gives 1 mark for the diagram + marks for labelling + marks for explaining the function. If you can’t draw it neatly, write the function of each part clearly in bullet points — the function explanation often carries more marks than the diagram.

Application questions (uses of ultrasound, uses of infrasound) are high-probability. Prepare 3 points for each:

  • Ultrasound: medical imaging, kidney stone removal, cleaning delicate equipment, detecting cracks in metals (SONAR for submarines)
  • Infrasound: used by elephants and whales for communication, weather prediction (associated with earthquakes/volcanoes)

Common Traps

Trap 1: “Sound travels faster in air than in water.” The opposite is true — sound travels faster in denser media (water: ~1500 m/s vs air: ~340 m/s). Students confuse speed of sound with speed of light (which does slow down in denser media).

Trap 2: Forgetting the 0.1 s minimum time for echo detection. The standard question asks for the minimum distance to hear an echo. With v = 340 m/s and t = 0.1 s, the minimum distance is 17 m. If you use any other time, you’ll get the wrong answer.

Trap 3: Confusing echo and reverberation. Echo = distinct repetition heard after the original sound, requires minimum 17 m distance. Reverberation = persistence/prolongation of sound due to multiple reflections in an enclosed space, heard almost simultaneously with original. Concert halls are designed to reduce reverberation; SONAR uses echo.

Trap 4: Writing that ultrasound “cannot be heard by humans because it is too quiet.” That’s wrong. Ultrasound has a frequency above 20,000 Hz — it is outside the human audible frequency range. Loudness is not the issue; frequency is.