Chapter 8 · ICSE Class 9 Physics (Selina Concise)

Propagation of Sound Waves

How sound travels, what frequency and wavelength mean, and how echoes and sonar work.

🔊 Chapter 8

Wave terms and the wave equation A sine wave with wavelength marked crest to crest and amplitude marked, with formulas for wave speed, echo distance and sonar depth. wavelength λ (crest to crest) amplitude v = f × λ = 500 Hz × 0.68 m = 340 m/s. Time period T = 1/f = 0.002 s. Echo: 2 s after the shout, distance = 340 × 2 ÷ 2 = 340 m. Minimum distance for a clear echo = 340 × 0.1 ÷ 2 = 17 m. Sonar: echo in 0.4 s at 1500 m/s means depth 300 m. Audible range 20 Hz to 20,000 Hz.

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8.1

Sound as a wave

Sound is produced by vibrating bodies and needs a material medium; it cannot travel in a vacuum. In air it moves as a longitudinal wave: layers of compressions and rarefactions travel along the direction of the vibration. Terms: amplitude (largest displacement, related to loudness), frequency f (vibrations per second, in hertz, related to pitch), time period T = 1/f, wavelength λ (distance between two crests or compressions). The wave equation: v = f × λ.

1

A wave has frequency 500 Hz and wavelength 0.68 m. Find its speed.

v = 500 × 0.68 = 340 m/s.

2

Find the time period of a 500 Hz sound.

T = 1 ÷ 500 = 0.002 s.

3

Why can we not hear sound on the Moon?

There is no air, and sound needs a medium to travel.

8.2

Speed of sound and the audible range

The speed of sound is greatest in solids, less in liquids and least in gases: about 340 m/s in air at room temperature, about 1500 m/s in water and about 5000 m/s in steel. It rises with temperature and is unaffected by pressure. Humans hear 20 Hz to 20,000 Hz. Below 20 Hz is infrasound, above 20,000 Hz is ultrasound.

1

Find the wavelength at 20 Hz and at 20,000 Hz in air (340 m/s).

At 20 Hz: 340 ÷ 20 = 17 m. At 20,000 Hz: 340 ÷ 20000 = 0.017 m.

2

Arrange in order of speed: air, water, iron.

Air < water < iron.

3

Which has the higher pitch: 100 Hz or 1000 Hz?

1000 Hz; higher frequency means a higher pitch.

8.3

Echo, reverberation and ultrasound

An echo is sound reflected from a distant hard surface and heard separately. Because the ear holds a sound for about 0.1 s, the reflecting surface must be at least v × 0.1 ÷ 2 away. Distance = v × t ÷ 2. Many echoes overlapping in a hall cause reverberation, reduced by curtains and acoustic tiles. Ultrasound is used in sonar, medical scanning and cleaning. Bats find insects with echoes.

1

An echo returns after 2 s. Find the distance to the cliff (v = 340 m/s).

d = 340 × 2 ÷ 2 = 340 m.

2

Find the minimum distance for a distinct echo.

d = 340 × 0.1 ÷ 2 = 17 m.

3

A sonar pulse returns from the sea bed in 0.4 s (v = 1500 m/s). Find the depth.

d = 1500 × 0.4 ÷ 2 = 300 m.

Practice

Exercises

1. What is a longitudinal wave?

2. Define frequency and its unit.

3. Write the wave equation.

4. A source of 340 Hz makes sound of wavelength 1 m. Find the speed.

5. Why does sound not travel in a vacuum?

6. Give the audible range of humans.

7. What is the minimum distance for an echo?

8. An echo is heard after 1 s. Find the distance of the reflector (v = 340).

9. What does pitch depend on?

10. Name two uses of ultrasound.

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