Acoustics — Class 10 Science (Samacheer Kalvi)
TN State Board (Samacheer Kalvi) Class 10 Science, Physics — Chapter 5. The science of sound — how it travels, reflects, and is used.
1. About this chapter
This chapter studies sound waves, their reflection (echo, reverberation), useful applications like SONAR, the audible range, and ultrasonic sound with its many uses.
2. Nature of sound
- Sound is a longitudinal mechanical wave — it needs a material medium (cannot travel through vacuum).
- It travels as compressions and rarefactions.
- Speed of sound is greatest in solids, less in liquids, least in gases (≈ 340 m s⁻¹ in air).
3. Reflection of sound
- Echo: a sound heard again after reflection. To hear a distinct echo the reflecting surface must be at least 17 m away (so the gap ≥ 0.1 s). Echo formula: 2d = v × t → distance d = v t / 2.
- Reverberation: persistence of sound by repeated reflections in a hall; reduced using sound-absorbing materials.
- Applications of reflection: SONAR, stethoscope, megaphone, sound boards in auditoriums.
4. SONAR and the audible range
- SONAR (Sound Navigation and Ranging) sends ultrasonic pulses and times the echo to find depth/distance: d = v t / 2.
- Audible range for humans: 20 Hz to 20,000 Hz.
- Below 20 Hz → infrasonic; above 20 kHz → ultrasonic.
- Uses of ultrasound: SONAR, cleaning fine machinery, detecting flaws in metals, and medical imaging (ultrasound scan).
5. Worked examples
Example 1. An echo returns in 2 s. If the speed of sound is 340 m s⁻¹, how far is the wall? d = v t / 2 = 340 × 2 / 2 = 340 m.
Example 2. A SONAR pulse returns from the seabed in 4 s; speed of sound in water = 1500 m s⁻¹. Find the depth. d = v t / 2 = 1500 × 4 / 2 = 3000 m.
Example 3. Why can't sound travel through vacuum? Because sound is a mechanical wave that needs particles of a medium to transfer compressions and rarefactions.
6. Common mistakes
- Mistake: Forgetting the factor 2 in the echo formula. Fix: Sound travels to the surface and back: 2d = vt, so d = vt/2.
- Mistake: Calling sound a transverse wave. Fix: Sound is a longitudinal wave.
- Mistake: Confusing ultrasonic and infrasonic. Fix: Ultrasonic > 20 kHz; infrasonic < 20 Hz.
7. Practice (book-back style)
- Why is sound a mechanical wave?
- State the conditions to hear a distinct echo.
- Write the echo/SONAR distance formula.
- A sound reflects from a cliff in 3 s (v = 340 m s⁻¹). Find the distance.
- Define the audible range and name two uses of ultrasound.
8. Answer key
- It needs a material medium to propagate (via compressions and rarefactions).
- Reflecting surface at least 17 m away; the echo must arrive at least 0.1 s after the original sound.
- d = v t / 2 (since 2d = vt).
- d = vt/2 = 340 × 3 / 2 = 510 m.
- 20 Hz to 20,000 Hz; uses — SONAR, ultrasound scanning, cleaning, flaw detection (any two).
9. Quick revision
- Physics Ch 5 · sound, reflection, echo, SONAR, ultrasound.
- Sound = longitudinal wave, needs a medium; ≈ 340 m s⁻¹ in air.
- Echo/SONAR: d = v t / 2 (factor 2 for to-and-fro).
- Distinct echo: surface ≥ 17 m, gap ≥ 0.1 s.
- Audible 20 Hz–20 kHz; ultrasonic > 20 kHz; infrasonic < 20 Hz.
