Chapter 10: Sound Waves: Characteristics and Applications
Production and propagation of sound, longitudinal waves, wavelength, frequency, amplitude, speed of sound, echo, reverberation and ultrasound.
Notes
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Production of sound
Sound is produced by vibrations: the periodic to and fro motion (oscillation) of an object. When the vibration stops, the sound stops.
The object that produces sound is the source. Examples: a plucked rubber band or string, a struck metal object, a vibrating air column in a bansuri, a stretched membrane.
Humans produce sound by the vibration of the vocal cords in the larynx; grasshoppers and crickets rub their wings or legs.
A tuning fork is a U-shaped metal bar with a stem. Its prongs (tines) vibrate when struck on a rubber pad. A vibrating prong touching water makes waves, showing that it vibrates.
Propagation of sound and the need for a medium
Sound travels through solids, liquids and gases. The material through which it travels is the medium.
Evidence: a knock on a desk is heard through the desk; submerged spoons tapped together are heard through water.
Vacuum is a space with no matter. In the bell jar experiment the electric bell is seen ringing but the sound fades as air is pumped out and returns when air is let in.
Sound cannot propagate in vacuum, so astronauts in space cannot hear each other directly and use special communication devices.
Sound waves: compressions and rarefactions
In a slinky, a push and pull makes regions where turns are closer together and regions where they are spread out travel along it. Each turn only oscillates about its own position.
In air, a forward-moving piston pushes particles together to form a compression (C), a region of higher than average density. A backward-moving piston produces a rarefaction (R), a region of lower density.
A sound wave is a series of alternating compressions and rarefactions travelling through a medium without the actual flow of the particles of the medium. The particles only vibrate about their mean positions.
The direction in which the wave travels is the direction of propagation.
From a point source in open space, sound spreads in all directions as spherical waves.
Longitudinal and mechanical waves
In sound, the particles vibrate parallel to the direction of propagation, so it is a longitudinal wave.
In a transverse wave the particles vibrate perpendicular to the direction of propagation.
Waves that need a material medium are mechanical waves. Sound is a mechanical wave. Light is not; it can travel through vacuum.
Energy of sound waves
Sound is a form of energy. A vibrating source transfers energy to the surrounding medium; the vibration of particles and their collisions carry the energy forward.
In the propagation of a sound wave it is the energy that is transferred, not the particles of the medium.
Demonstration: rice grains on a stretched sheet jump when a loud sound is made near it.
A microphone converts sound energy to an electrical signal; a speaker does the opposite.
Graph of a sound wave
Plot distance on the x-axis and density of the medium on the y-axis. The dashed horizontal line marks the average density.
The highest point (maximum density, a compression) is the crest. The lowest point (minimum density, a rarefaction) is the trough.
The graph can also be plotted as density against time at a fixed point.
Wavelength, frequency and time period
Wavelength (λ): distance between two consecutive crests or two consecutive troughs. SI unit: metre (m).
Frequency (ν): number of density oscillations at a fixed point per unit time. SI unit: hertz (Hz) = s⁻¹.
Time period (T): time for one complete oscillation. SI unit: second (s).
ν = 1/T (Eq. 10.1). A shorter time period means a higher frequency.
One complete oscillation: density goes from maximum to minimum and back to maximum.
Amplitude and intensity
Amplitude: the maximum change in the density of the medium in a compression (or rarefaction) compared with the average density.
A wave of larger amplitude carries more energy.
Intensity: the sound energy passing through a unit area perpendicular to the direction of propagation in unit time.
As the wave spreads over a larger area, the same energy is shared, so intensity decreases with distance from the source.
Speed of sound
The speed of sound is the distance travelled by a point on the wave, such as a crest, in unit time: v = λ/T, so v = ν × λ (Eq. 10.2).
It depends on the medium: fastest in solids, slower in liquids, slowest in gases. In water about 4–5 times faster than in air; in solids about 15–20 times faster than in air.
In air it increases with temperature and humidity: about 331 m s⁻¹ at 0 °C (dry air) and about 344 m s⁻¹ at 22 °C.
In most media the speed depends on the medium only, not on the source or the frequency. If the frequency changes, the wavelength changes and the speed stays the same.
Approximate speeds at 15 °C: steel 5000 m s⁻¹, water 1500 m s⁻¹, air 340 m s⁻¹.
Human perception: pitch and loudness
Pitch is how humans perceive frequency. High frequency means shrill (high pitch); low frequency means deep (low pitch).
Loudness is how humans perceive amplitude. Larger amplitude is heard louder. Loudness depends on the listener's hearing, whereas intensity is measurable.
Loudness is measured in decibels (dB). Normal conversation is about 60 dB.
Noise is unwanted or harmful sound. Prolonged exposure to loud sound causes hearing loss.
The audible range is 20 Hz to 20 kHz. Below 20 Hz: infrasonic (elephants can detect it). Above 20 kHz: ultrasonic (dogs, cats, bats, dolphins can detect it).
A tone has a single frequency; a musical note is a mixture of a fundamental and overtones. Timbre is the quality that distinguishes sounds of the same pitch and loudness from different sources. An octave is the interval between two notes where one has twice the frequency of the other.
Reflection of sound, echo and reverberation
Sound reflects from solids and liquids and follows the laws of reflection: the angle of incidence equals the angle of reflection, and the incident ray, reflected ray and normal lie in the same plane.
An echo is heard when reflected sound reaches us at least 0.1 s after the original. At 340 m s⁻¹, sound covers 34 m in 0.1 s (to the surface and back), so the minimum distance to the reflecting surface is 17 m.
Echoes are stronger from hard, smooth surfaces. Soft surfaces absorb sound; rough surfaces scatter it.
Reverberation: multiple reflections in a large hall make sound persist after the source stops (reflections arriving within 0.05 s). Concert halls are designed to have some; absorbing panels, curtains and upholstered chairs reduce it.
Ultrasonic and infrasonic waves and their uses
Infrasonic waves: detect earthquakes, volcanic eruptions and severe storms.
Ultrasonic waves: ultrasonography (imaging organs), breaking kidney stones, cleaning delicate machine parts, ultrasonic welding, detecting defects in metal blocks, locating objects by reflection.
Echolocation: bats, dolphins, whales and some birds send short bursts of ultrasonic waves and sense the reflected echoes to find obstacles and prey.
Sonar (sound navigation and ranging): ultrasonic waves sent into water are reflected; the time taken gives the distance of submarines, shipwrecks or the sea floor.
Distance by reflection: d = (v × t) / 2, where t is the total time for going and returning.
Numericals
Worked solutions for the examples and calculation questions in this chapter.
EQUATIONS
ν = 1/T
v = ν × λ = λ / T
Distance by echo or sonar: d = v × t / 2
Minimum echo distance = 340 × 0.1 / 2 = 17 m
EXAMPLE 10.1 · FREQUENCY AND TIME PERIOD
10 density oscillations in 2 s.
ν = 10 / 2 s = 5 Hz
T = 2 s / 10 = 0.2 s
EXAMPLE 10.2 · AUDIBLE RANGE WAVELENGTHS
v = 344 m s⁻¹.
λ = v/ν = 344 / 20 = 17.2 m (20 Hz)
λ = 344 / 20000 = 0.0172 m = 1.72 cm (20 kHz)
EXAMPLE 10.3 · THUNDER AND LIGHTNING
d = v × t = 340 × 5 = 1700 m (about 1.7 km)
EXAMPLE 10.4 · WAVE IN STEEL
λ = 50 m from the graph; v = 5000 m s⁻¹.
ν = v/λ = 5000 / 50 = 100 Hz
T = 1/ν = 0.01 s
EXAMPLE 10.5 · ECHO IN A CORRIDOR
d = v × t / 2 = 340 × 0.5 / 2 = 85 m
EXAMPLE 10.6 · NAVAL SONAR
d = 1530 × 0.90 / 2 = 1530 × 0.45 = 688.5 m
PAUSE AND PONDER 8 · 20 HZ PISTON
Oscillations per minute = 20 × 60 = 1200
PAUSE AND PONDER 9 · FIG. 10.19
From the graph, consecutive crests are 3.0 cm apart, so λ = 3.0 cm.
Half the wavelength = 1.5 cm
PAUSE AND PONDER 10 · RATIOS OF SPEEDS
(i) water / air = 1500 / 340 ≈ 4.4
(ii) steel / water = 5000 / 1500 ≈ 3.3
PAUSE AND PONDER 11 · FENCE AND GUNJAN
t(air) = 340 / 340 = 1 s
t(steel) = 340 / 5000 = 0.068 s
Difference = 1 − 0.068 = 0.932 s
Since 0.932 s is more than 0.1 s, she can tell the two sounds apart.
PAUSE AND PONDER 12 · MINIMUM DISTANCE
The echo must arrive 0.2 s after the sound (round trip).
d = 343 × 0.2 / 2 = 34.3 m
PAUSE AND PONDER 13 · OCEAN DEPTH
d = 1500 × 4 / 2 = 3000 m
REVISE, REFLECT, REFINE 3
ν = 20 / 4 s = 5 Hz
REVISE, REFLECT, REFINE 9
T = λ / v = 3.44 / 344 = 0.01 s
REVISE, REFLECT, REFINE 10 · SUNKEN SHIP
d = 1525 × 5 / 2 = 3812.5 m (about 3.8 km)
REVISE, REFLECT, REFINE 11 · PARKING SENSOR
t = 2 × 1.2 / 345 ≈ 0.007 s (about 7 ms)
REVISE, REFLECT, REFINE 12 · THUNDER AT 0 °C
t(22 °C) = 1720 / 344 = 5.0 s
t(0 °C) = 1720 / 331 ≈ 5.2 s
Extra time ≈ 0.2 s.
REVISE, REFLECT, REFINE 13 · FIG. 10.32
The 8 cm marked spans two wavelengths (compression to the second compression after it), so λ = 4 cm = 0.04 m.
ν = 340 / 0.04 = 8500 Hz
REVISE, REFLECT, REFINE 14 · FIG. 10.33
Wave A: λ = 2.5 cm; wave B: λ = 5.0 cm (from the graph).
ν(A) = 345 / 0.025 = 13800 Hz
ν(B) = 345 / 0.05 = 6900 Hz
REVISE, REFLECT, REFINE 15 · FIG. 10.34
The distance to the cliff is the same for both, so time ∝ 1/speed.
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Assertion and Reason
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NCERT Q&A
Answers to the Pause and Ponder questions and the Revise, Reflect, Refine exercises. Numerical working is in the Numericals tab.
Pause and Ponder
PAUSE AND PONDER 1
Explore various ways of producing sound.
Answer: Sound can be produced by plucking a stretched string or rubber band, striking a metal object or a tuning fork, blowing through a pipe (bansuri), beating a drum membrane, rubbing two surfaces (crickets) and by the vocal cords when we speak. In every case something vibrates.
PAUSE AND PONDER 2
Make a list of different types of musical instruments and identify their vibrating parts which produce sound.
Answer: Sitar, tanpura, veena, violin: vibrating strings. Tabla, mridangam, drum: vibrating stretched membrane. Bansuri, flute, shehnai: vibrating air column. Cymbals, taal, bells: the vibrating metal itself.
PAUSE AND PONDER 3
Assertion (A): We cannot hear the sound of a bell ringing in a closed jar after most of the air is pumped out. Reason (R): Sound requires a medium to travel.
Answer: (ii) Both A and R are true, and R is the correct explanation of A. With the air removed there is no medium, so the sound cannot reach us.
PAUSE AND PONDER 4
Assertion (A): Compressions and rarefactions move through the medium. Reason (R): Individual particles of the medium continuously move forward with the wave.
Answer: (iii) A is true, but R is false. The particles only vibrate about their mean positions; they do not move forward with the wave.
PAUSE AND PONDER 5
When sound travels from a tuning fork to your ear, which of the following actually reaches your ear?
Answer: (ii) Energy carried by sound waves. The air particles only vibrate about their mean positions.
PAUSE AND PONDER 6
Label compression and rarefaction on Fig. 10.17 (a) and (b), and draw the curves in (c) and (d).
Answer: Compressions (C) are the darker, crowded regions and rarefactions (R) are the lighter, sparse regions. In the graph, label the x-axis Distance and the y-axis Density. Draw a smooth wave about the dashed average-density line: a crest over each compression and a trough under each rarefaction. The wave with the closer-spaced compressions has the shorter wavelength.
PAUSE AND PONDER 7
Conduct Activity 10.1 with a thick rubber band and a thin one. Does the thin band vibrate faster? How do the frequency and time period differ?
Answer: Yes. The thin band vibrates faster, so its frequency is higher and its time period is shorter. It produces a higher pitched sound than the thick band.
PAUSE AND PONDER 8
If the frequency of a sound wave produced by an oscillating piston in a long tube filled with air is 20 Hz, how many oscillations does the piston complete per minute?
Answer: Oscillations per minute = 20 × 60 = 1200.
PAUSE AND PONDER 9
For the sound wave in Fig. 10.19, what is half of its wavelength?
Answer: The wavelength read from the graph is 3.0 cm (distance between consecutive crests), so half the wavelength is 1.5 cm.
PAUSE AND PONDER 10
Compare the speeds in different media by finding the ratio of (i) the speed of sound in water to that in air and (ii) the speed in steel to that in water.
Two friends stand 340 m apart along a steel fence. Calculate the time difference between the sound reaching Gunjan through air and through steel. Can she distinguish them?
Answer: Time in air = 340 / 340 = 1 s. Time in steel = 340 / 5000 = 0.068 s. Difference = 0.932 s. This is greater than 0.1 s, so she can hear the two sounds separately.
PAUSE AND PONDER 12
An experiment needs echoes to arrive at least 0.2 s after the emission of sound. What minimum distance should a reflecting surface be placed at? (v = 343 m s⁻¹)
Answer: d = v × t / 2 = 343 × 0.2 / 2 = 34.3 m.
PAUSE AND PONDER 13
A sonar signal sent to find the depth of the ocean takes 4 s to return. What is the depth if the speed of sound in seawater is 1500 m s⁻¹?
Answer: Depth = 1500 × 4 / 2 = 3000 m.
Revise, Reflect, Refine
QUESTION 1
Which observation best supports the idea that sound is a mechanical wave? (i) Sound shows reflection (ii) Sound needs a medium to propagate (iii) Sound has frequency (iv) Sound carries energy
Answer: (ii) Sound needs a medium to propagate. Mechanical waves are exactly those that need a material medium.
QUESTION 2
For a sound wave propagating in a medium, increasing its frequency will increase its (i) wavelength (ii) speed (iii) number of compressions per second (iv) time period
Answer: (iii) number of compressions per second. Frequency is the number of oscillations (compressions) per second. Speed stays the same and the wavelength and time period decrease.
QUESTION 3
If 20 compressions pass a point in 4 seconds, the frequency is (i) 80 Hz (ii) 5 Hz (iii) 10 Hz (iv) 0.2 Hz
Answer: (ii) 5 Hz, since 20 / 4 = 5 oscillations per second.
QUESTION 4
In a room, the reflected sound reaches the ear 0.05 s after its production. Will it produce an echo or reverberation? Justify.
Answer: The reflected sound arrives in less than 0.1 s, so it cannot be heard as a separate echo. Reflections arriving this quickly merge with the original sound and make it persist, which is reverberation.
QUESTION 5
Graphs of two sound waves are given in Fig. 10.30. Which has (i) greater wavelength and (ii) smaller amplitude?
Answer: Wave (a) has the greater wavelength (fewer waves in the same distance, with crests further apart). Wave (a) also has the smaller amplitude (its crests and troughs rise less far from the average line than those of wave (b)).
QUESTION 6
Three sources A, B and C are represented in Fig. 10.31. If the frequency of A is maximum and C is minimum, identify the curves.
Answer: A is the green curve (most waves in the same distance, shortest wavelength). B is the red curve (in between). C is the blue curve (fewest waves, longest wavelength).
QUESTION 7
Draw a graph to represent a sound wave for which the density amplitude is 3 units and wavelength is 4 cm.
Answer: Draw the x-axis as Distance (cm) and the y-axis as Density with a dashed average-density line. Draw a smooth wave that rises 3 units above the line at crests and falls 3 units below at troughs, with consecutive crests 4 cm apart (for example crests at 1 cm, 5 cm, 9 cm).
QUESTION 8
In a movie, the explosion of a spacecraft in space is shown with a flash of light and sound at the same time. What are the errors?
Answer: (1) Space is a near vacuum, so sound cannot propagate and no explosion sound could be heard. (2) Even in air, light travels far faster than sound, so the flash would be seen before the sound is heard, not at the same instant.
QUESTION 9
A source produces a sound wave of wavelength 3.44 m. If the wave travels at 344 m s⁻¹, find its time period.
Answer: T = λ / v = 3.44 / 344 = 0.01 s.
QUESTION 10
A ship searching for a sunken ship sent a sonar signal and detected an echo after 5 s. If ultrasonic waves travel at 1525 m s⁻¹ in seawater, how far down is the wreckage?
Answer: d = 1525 × 5 / 2 = 3812.5 m, about 3.8 km.
QUESTION 11
An ultrasonic parking sensor warns at 1.2 m from the obstacle. How much time does the wave take to go to the obstacle and come back? (v = 345 m s⁻¹)
Answer: Total distance = 2 × 1.2 = 2.4 m. t = 2.4 / 345 ≈ 0.007 s (about 7 ms).
QUESTION 12
The speed of sound in air is about 331 m s⁻¹ at 0 °C and 344 m s⁻¹ at 22 °C. How much extra time will thunder take to travel 1720 m if the temperature falls from 22 °C to 0 °C?
Answer: At 22 °C: t = 1720 / 344 = 5.0 s. At 0 °C: t = 1720 / 331 ≈ 5.2 s. Extra time ≈ 0.2 s.
QUESTION 13
The density variation for a sound wave travelling at 340 m s⁻¹ is shown in Fig. 10.32 (8 cm marked). Calculate the wavelength and frequency.
Answer: The 8 cm span covers two wavelengths, so λ = 4 cm = 0.04 m. ν = v / λ = 340 / 0.04 = 8500 Hz.
QUESTION 14
Two sound waves A and B travel at 345 m s⁻¹ (Fig. 10.33). Find the wavelength and frequency of each.
Answer: From the graph, λA = 2.5 cm and λB = 5.0 cm. νA = 345 / 0.025 = 13800 Hz; νB = 345 / 0.05 = 6900 Hz.
QUESTION 15
Two identical sources, one in air (A) and one in water (B), send sound to a cliff and back. The time at A is 4.5 times that at B. What is the ratio of the speeds of sound in air and water?
Answer: The distance is the same, so the speed is inversely proportional to the time. v(air) : v(water) = 1 : 4.5, i.e. about 0.22 (2 : 9).