This MCQ module is based on: Characteristics of Sound Waves and the Wave Equation
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Characteristics of Sound Waves and the Wave Equation
🎓 Class 9
Science
CBSE
Theory
Ch 10 — Sound Waves: Characteristics and Applications
⏱ ~16 min
🌐 Language: [gtranslate]
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10.5 Describing a Sound Wave
Although a sound wave in air is longitudinal — built from invisible compressions and rarefactions — we can plot the air pressure (or particle displacement) against position. The result looks just like a sine curve. The high crests of the curve correspond to the centres of compressions and the low troughs correspond to the centres of rarefactions. This convenient picture lets us define the same characteristics for a sound wave that we use for any other wave.
📐 Wave Property Decoder — Click each part of the wave to recall its meaning L1 Remember
Click the λ bracket, the A arrow, a compression tick, or a rarefaction tick to recall its definition, formula and SI unit.
Click any labelled part of the wave above to recall its definition, formula and SI unit.
Wavelength (λ)
The wavelength is the distance between two neighbouring compressions, or equivalently between two neighbouring rarefactions. The Greek letter λ (lambda) is used for it. The SI unit of wavelength is the metre (m).
Frequency (ν)
The number of complete oscillations made by a particle of the medium (or by the source) per second is called the frequency. Its symbol is ν (Greek letter "nu") and its SI unit is the hertz (Hz). A frequency of 50 Hz means 50 full oscillations every second. Larger units kilohertz (kHz = 10³ Hz) and megahertz (MHz = 10⁶ Hz) are also used.
Time period (T)
The time period is the time taken for one complete oscillation. It is the reciprocal of frequency:
\(T = \dfrac{1}{\nu}\) (SI unit: second, s)
If a tuning fork has a frequency of 200 Hz, its time period is \(1/200 = 0.005\) s, i.e. 5 ms.
Amplitude (A)
The amplitude measures how far each particle of the medium is pushed away from its rest position when the wave passes. A bigger amplitude means more energy in the wave.
10.6 Pitch, Loudness and Quality
The same musical note played on a flute and on a violin sounds clearly different even when both are at the same loudness. To explain such observations we use three perceptual qualities of sound — pitch, loudness, and quality — each linked to one physical property of the wave.
Pitch
Pitch is how high or low a sound seems to the ear. A whistle sounds shrill (high pitch); a bass drum sounds deep (low pitch). Pitch is decided by frequency: the higher the frequency, the higher the pitch. A bird's call has a high frequency, a lion's roar a low one.
Loudness
Loudness is how strong a sound seems to be. It is governed by amplitude. The bigger the amplitude (more displacement of air particles), the more energy carried, and the louder the sound. Loudness is roughly proportional to the square of amplitude. The unit used to express loudness is the decibel (dB).
Quality (Timbre)
The quality or timbre of a sound is what makes it possible to recognise the same note coming from different instruments or to identify the voice of a friend in a crowd. Quality depends on the shape of the wave — the mixture of additional frequencies (harmonics) that accompany the main note. A pure note from a tuning fork has a smooth sine shape; a violin's note is the same fundamental frequency but with a much richer wave shape.
| Perceived quality | Physical property |
|---|---|
| Pitch (shrill / deep) | Frequency, ν |
| Loudness (loud / soft) | Amplitude, A (and energy ∝ A²) |
| Quality / Timbre | Wave shape (mix of harmonics) |
10.7 Activity — Pitch from a Plastic Ruler
Activity 10.2 — Vary the Vibration, Vary the PitchL3 Apply
Predict first: Press a plastic ruler firmly on a desk so most of it sticks out, then twang the free end. What happens to the sound when you slide the ruler so that less of it is free to vibrate?
- Press one end of a 30 cm flexible ruler hard against the edge of a table; let about 25 cm hang free.
- Pluck the free end and listen.
- Slide the ruler so that only 15 cm hangs free; pluck again.
- Reduce to 5 cm free length and pluck once more.
Observations: The longer the free portion, the slower the back-and-forth motion and the deeper (lower) the sound. The shorter the free portion, the faster it vibrates and the higher (shriller) the sound becomes.
Conclusion: Shorter vibrating bodies → higher frequency → higher pitch. The same idea explains why a small tabla (dagga) sounds higher than a large one, and why a thin tight string on a sitar gives a sharper note than a loose thick one.
Conclusion: Shorter vibrating bodies → higher frequency → higher pitch. The same idea explains why a small tabla (dagga) sounds higher than a large one, and why a thin tight string on a sitar gives a sharper note than a loose thick one.
10.8 The Wave Equation: v = λν
For any periodic wave, the speed at which the wave travels is linked to its frequency and wavelength by a simple equation. In one time period T the wave moves forward by exactly one wavelength λ. So the speed of the wave is
\(v = \dfrac{\text{distance}}{\text{time}} = \dfrac{\lambda}{T} = \lambda \times \nu\)
v = λν (speed = wavelength × frequency)
v = λν (speed = wavelength × frequency)
Here \(v\) is in m/s, \(\lambda\) in m and \(\nu\) in Hz. Because v depends only on the medium (mostly on temperature for air), if the frequency goes up, the wavelength must go down so that the product stays the same.
Worked Example 1
A source produces sound waves of wavelength 0.85 m. The speed of sound in air is 340 m/s. Find the frequency.
\(\nu = v/\lambda = 340/0.85 = 400\) Hz.
\(\nu = v/\lambda = 340/0.85 = 400\) Hz.
Worked Example 2
A man hears a 256 Hz tuning fork. If sound travels at 332 m/s in the air around him, what is the wavelength?
\(\lambda = v/\nu = 332/256 ≈ 1.30\) m.
\(\lambda = v/\nu = 332/256 ≈ 1.30\) m.
Worked Example 3
A sound wave has frequency 2 kHz and wavelength 35 cm. How long will it take to cover 1.5 km?
v = λν = 0.35 × 2000 = 700 m/s.
t = d/v = 1500/700 ≈ 2.14 s.
v = λν = 0.35 × 2000 = 700 m/s.
t = d/v = 1500/700 ≈ 2.14 s.
Worked Example 4
A child produces sound waves of frequency 10 Hz by waving her hand. What is the time period?
T = 1/ν = 1/10 = 0.1 s. (Such a low frequency is below human hearing — it is infrasonic.)
T = 1/ν = 1/10 = 0.1 s. (Such a low frequency is below human hearing — it is infrasonic.)
Worked Example 5
A wave makes 300 vibrations in 5 s. Speed of sound = 340 m/s. Find frequency, time period and wavelength.
ν = 300/5 = 60 Hz. T = 1/60 ≈ 0.0167 s. λ = v/ν = 340/60 ≈ 5.67 m.
ν = 300/5 = 60 Hz. T = 1/60 ≈ 0.0167 s. λ = v/ν = 340/60 ≈ 5.67 m.
Worked Example 6
A man is exposed to a sound of 0.5 m wavelength. If the frequency is 660 Hz, find the speed of sound through that medium and identify whether it could be air.
v = λν = 0.5 × 660 = 330 m/s. This matches the speed of sound in air at room temperature, so yes — the medium can be air.
v = λν = 0.5 × 660 = 330 m/s. This matches the speed of sound in air at room temperature, so yes — the medium can be air.
10.9 Range of Hearing
Not every frequency is audible to a human ear. The healthy human ear responds only to frequencies between roughly 20 Hz and 20 000 Hz. This band is called the audible range. As we grow older, the upper limit usually drops.
Infrasonic and ultrasonic sounds
- Infrasonic sounds (below 20 Hz) cannot be heard by humans. Whales, elephants, rhinoceroses and earthquakes generate them. They warn many animals before earthquakes strike.
- Ultrasonic sounds (above 20 000 Hz, i.e. above 20 kHz) are also inaudible to us but are used for medical imaging, cleaning delicate parts and detecting cracks. Bats, dolphins and porpoises produce and hear them; dogs are sensitive to them as well.
Competency-Based Questions
In a music class, the teacher plays the same note (frequency 440 Hz) on a flute and on a tuning fork. The students agree both have the same pitch but the flute "sounds richer". Then she strikes the same fork softly and loudly so the students can compare loudness. The speed of sound in air that day was 350 m/s.
Q1. What is the wavelength of the 440 Hz note in air? L3
(b) λ = v/ν = 350/440 ≈ 0.795 m.
Q2. Why does the flute sound richer than the fork even though both produce the same frequency? L4
A tuning fork emits an almost pure sine wave at its fundamental frequency. A flute adds many higher harmonics (overtones), so its wave shape is richer. The pitch is the same but the quality (timbre) is different.
Q3. The teacher strikes the fork twice as hard. Which physical quantity changes and how does the listener perceive it? L2
The amplitude of vibration becomes larger. The listener hears a louder sound. Frequency and pitch are unchanged.
Q4. Fill in the blanks: A 50 Hz vibration has a time period of __________ s and is in the __________ range of human hearing. L1
T = 1/50 = 0.02 s. 50 Hz is well within the audible range (20 Hz – 20 kHz).
Q5. A bat sends a click of 50 kHz. Take v = 340 m/s. What is the wavelength? Is this audible to humans? L3
λ = v/ν = 340/50 000 = 6.8 × 10⁻³ m = 6.8 mm. 50 kHz is far above 20 kHz, so it is ultrasonic — humans cannot hear it.
Assertion–Reason Questions
Options: (A) Both A and R are true and R is the correct explanation of A. (B) Both true but R is not the correct explanation. (C) A true, R false. (D) A false, R true.
A: A note of higher frequency has a shorter wavelength in the same medium.
R: v = λν, and v is fixed for a given medium and temperature.
(A) Both true and R correctly explains A. With v constant, larger ν means smaller λ.
A: Two musical instruments playing the same note at the same loudness can be told apart by the ear.
R: Pitch and loudness are the only two characteristics of sound.
(C) Assertion is true (the ear distinguishes them by quality/timbre). Reason is false — quality is a third characteristic.
A: Bats emit ultrasonic sounds for navigation.
R: Frequencies above 20 000 Hz fall outside the audible range of humans.
(B) Both statements are true, but R only describes the upper limit of hearing — it does not explain why bats use ultrasound (they use it because the short wavelength reflects from small obstacles and gives them precise echolocation).
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