Sound Waves: Characteristics and Applications — Detailed Notes
Sound is a form of energy that produces the sensation of hearing. This chapter covers how sound is produced, how it travels, its wave nature, its characteristics, and its applications.
10.1 Production of sound
Sound is produced by vibrations (periodic to-and-fro motion) of an object — a plucked rubber band, a struck tuning fork, vibrating vocal cords in the larynx, an air column in a flute. The vibrating object is the source. When the vibration stops, the sound stops.
10.2 Propagation — sound needs a medium
Sound travels through solids, liquids and gases (a medium), but not through vacuum. The bell-jar experiment shows the sound fades as air is pumped out. That is why astronauts in space (near vacuum) cannot hear each other directly and use radio devices.
10.3 Sound waves
A vibrating source pushes air to make a high-density region (compression, C) and pulls back to make a low-density region (rarefaction, R). A travelling series of compressions and rarefactions is a sound wave. The particles only vibrate about their mean positions — they do not travel with the wave; it is the energy that is transferred. Sound is a longitudinal wave (particles vibrate parallel to the direction of propagation) and a mechanical wave (needs a medium). (In a transverse wave, particles vibrate perpendicular to propagation, e.g. light — which needs no medium.)
10.5 Graphical representation
Plotting density vs distance gives a wave: the maximum-density point is a crest (compression), the minimum-density point is a trough (rarefaction), about the average-density line.
10.6 Characteristics of a sound wave
• Wavelength (λ) — distance between two consecutive crests (or troughs); SI unit metre. • Frequency (ν) — number of oscillations per second; SI unit hertz (Hz). • Time period (T) — time for one oscillation; T = 1/ν. • Amplitude — the maximum change in density (bigger amplitude = more energy = louder). • Intensity — sound energy per unit area per unit time (decreases with distance as energy spreads). • Speed (v): v = λ × ν.
Speed of sound is fastest in solids, slower in liquids, slowest in gases (≈340 m s⁻¹ in air, ≈1500 in water, ≈5000 in steel). It increases with temperature and humidity (≈331 m s⁻¹ at 0 °C, ≈344 m s⁻¹ at 22 °C in air). For a given medium the speed is fixed — changing the frequency changes the wavelength, not the speed.
10.6.4 Human perception
Pitch = how we perceive frequency (high frequency → shrill/high pitch). Loudness = how we perceive amplitude (measured in decibels, dB). Audible range of humans = 20 Hz to 20,000 Hz. Below 20 Hz = infrasonic; above 20 kHz = ultrasonic. Timbre distinguishes the same note on different instruments.
10.7 Reflection of sound
Sound obeys the laws of reflection. An echo is a reflected sound heard separately — possible only if the reflected sound arrives at least 0.1 s after the original, so the reflecting surface must be at least 17 m away (distance = v×t/2 = 340×0.1/2). Reverberation is the persistence of sound by multiple reflections in a hall.
10.8 Ultrasound & infrasound applications
Ultrasound (>20 kHz): ultrasonography (imaging organs), breaking kidney stones, cleaning parts, detecting flaws in metals, SONAR (measuring sea depth / locating submarines). Echolocation — bats, dolphins locate objects using reflected ultrasound. Infrasound (<20 Hz): detecting earthquakes, volcanic eruptions, storms (elephants, whales use it).
Key Terms
vibration, medium, vacuum, compression, rarefaction, longitudinal/mechanical wave, crest, trough, wavelength (λ), frequency (ν), time period (T), amplitude, intensity, speed (v = λν), pitch, loudness, decibel, audible range, infrasonic, ultrasonic, echo, reverberation, echolocation, SONAR.
Revise, Reflect, Refine — All NCERT Exercise Answers
1Which observation best supports the idea that sound is a mechanical wave?
(ii) Sound needs a medium to propagate. A mechanical wave by definition requires a material medium (solid, liquid or gas) and cannot travel through vacuum. Reflection, frequency and carrying energy are properties of many waves (including light), so they alone do not prove it is mechanical.
2For a sound wave in a medium, increasing its frequency will increase its:
(iii) number of compressions per second. In a given medium the speed is fixed; since v = λν, raising the frequency decreases the wavelength (and decreases the time period) but produces more compressions each second.
3If 20 compressions pass a point in 4 seconds, the frequency is:
Frequency = number of oscillations ÷ time = 20 ÷ 4 = 5 Hz — option (ii).
4Reflected sound reaches the ear 0.05 s after production — echo or reverberation? Justify.
It is heard as reverberation, not an echo. An echo is heard as a separate sound only when the reflected sound arrives at least 0.1 s after the original. Here the gap is only 0.05 s (< 0.1 s), so the ear cannot separate the two — the sounds overlap and persist as reverberation.
5Two sound-wave graphs with the same scales — which has (i) greater wavelength (ii) smaller amplitude?
(i) The wave whose crests (or troughs) are farther apart has the greater wavelength. (ii) The wave whose curve rises/falls less from the average-density line (shorter height) has the smaller amplitude. (Compare the horizontal spacing for wavelength and the vertical height for amplitude on the two graphs.)
6Sources A (maximum frequency) and C (minimum frequency) are shown. Identify and mark A, B, C.
Since v = λν is the same medium, higher frequency means shorter wavelength. So A (maximum frequency) is the curve with the shortest wavelength (crests closest together), C (minimum frequency) is the curve with the longest wavelength (crests farthest apart), and B is the one in between.
7Draw a graph of a sound wave with density amplitude 3 units and wavelength 4 cm.
Plot density (y-axis) vs distance (x-axis). Draw a smooth wave about the average-density line so that each crest rises 3 units above and each trough falls 3 units below the average line (amplitude = 3 units), and the distance between two successive crests (or troughs) is 4 cm (wavelength = 4 cm). Mark compressions at crests (C) and rarefactions at troughs (R).
8A movie shows a spacecraft exploding in space with a flash of light AND sound at the same time. What errors are there?
Two errors: (1) Space is a near vacuum, so sound cannot travel there — the explosion should be silent. (2) Even if there were a medium, light travels far faster than sound, so the flash would be seen well before any sound was heard — they could not reach the viewer at the same time.
9A source produces a sound wave of wavelength 3.44 m travelling at 344 m s⁻¹. Find its time period.
Frequency ν = v/λ = 344 ÷ 3.44 = 100 Hz. Time period T = 1/ν = 1/100 = 0.01 s.
10A ship’s sonar detects an echo from a sunken ship after 5 s; speed in seawater = 1525 m s⁻¹. How deep is the wreck?
The signal travels down and back, so depth = (v × t) ÷ 2 = (1525 × 5) ÷ 2 = 7625 ÷ 2 = 3812.5 m.
11A parking sensor emits ultrasound; the obstacle is 1.2 m away. Time for the wave to go and return? (speed = 345 m s⁻¹)
Total path = 2 × 1.2 = 2.4 m. Time = distance ÷ speed = 2.4 ÷ 345 = 0.00696 s ≈ 6.96 ms.
12Speed of sound is 331 m s⁻¹ at 0 °C and 344 m s⁻¹ at 22 °C. How much extra time does thunder take over 1720 m when the air cools from 22 °C to 0 °C?
Time at 22 °C = 1720 ÷ 344 = 5.00 s. Time at 0 °C = 1720 ÷ 331 ≈ 5.196 s. Extra time ≈ 5.196 − 5.00 = ≈ 0.2 s (sound is slower in colder air).
13Density variation of a sound wave (speed 340 m s⁻¹) spans 8 cm (Fig. 10.32). Find the wavelength and frequency.
From the graph the wavelength λ = 8 cm = 0.08 m (the distance of one full density cycle). Frequency ν = v/λ = 340 ÷ 0.08 = 4250 Hz. (Read the exact length of one cycle from the graph and substitute.)
14Waves A and B travel at 345 m s⁻¹ (Fig. 10.33, marks at 2.5 and 5.0 cm). Find each wavelength and frequency.
If wave A has λ = 5.0 cm = 0.05 m → ν = 345 ÷ 0.05 = 6900 Hz; and wave B has λ = 2.5 cm = 0.025 m → ν = 345 ÷ 0.025 = 13,800 Hz. (The wave with the shorter wavelength has the higher frequency; read the exact wavelengths from the graph.)
15Two identical sources at A (in air) and B (in water) send sound to a cliff and back. Time for A is 4.5× that for B. Find the ratio of the speed of sound in air to that in water.
The horizontal distance to the cliff and back is the same for both, so time = 2d/v → time is inversely proportional to speed. t_A = 4.5 t_B means v is 4.5 times smaller in air: v_water/v_air = 4.5. Therefore v(air) : v(water) = 1 : 4.5 (i.e. 2 : 9).