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Physics - Sound (Production, Propagation, Infrasonic, Ultrasonic)

Grade 9ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Sound is a form of energy produced by vibrating bodies and propagates as a mechanical longitudinal wave.

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A material medium (solid, liquid, or gas) is essential for the propagation of sound; it cannot travel through a vacuum.

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Longitudinal waves consist of regions of high pressure called compressions and regions of low pressure called rarefactions.

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The speed of sound depends on the elasticity and density of the medium. Generally, Vsolid>Vliquid>VgasV_{solid} > V_{liquid} > V_{gas}.

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Speed of sound in air is approximately 330 m s−1330 \text{ m s}^{-1} to 340 m s−1340 \text{ m s}^{-1} at standard temperature and pressure (0∘C0^\circ\text{C}).

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Audible Range: The human ear can perceive frequencies between 20 Hz20 \text{ Hz} and 20,000 Hz20,000 \text{ Hz}.

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Infrasonic Sound: Sound waves with frequencies below 20 Hz20 \text{ Hz}. These are produced by earthquakes, volcanic eruptions, and animals like elephants and whales.

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Ultrasonic Sound: Sound waves with frequencies above 20,000 Hz20,000 \text{ Hz}. These are used in SONAR, ultrasonography, and by bats for navigation.

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The relationship between velocity (VV), frequency (ff), and wavelength (λ\lambda) is given by V=fλV = f \lambda.

📐Formulae

f=1Tf = \frac{1}{T}

V=fλV = f \lambda

V=λTV = \frac{\lambda}{T}

d=V×t2 (For Echo/SONAR applications)d = \frac{V \times t}{2} \text{ (For Echo/SONAR applications)}

💡Examples

Problem 1:

A longitudinal wave travels at a speed of 340 m s−1340 \text{ m s}^{-1}. If its frequency is 20 kHz20 \text{ kHz}, calculate its wavelength in centimetres.

Solution:

Given: V=340 m s−1V = 340 \text{ m s}^{-1}, f=20 kHz=20,000 Hzf = 20 \text{ kHz} = 20,000 \text{ Hz}. Using the formula V=fλV = f \lambda, we get λ=Vf\lambda = \frac{V}{f}. λ=34020000=0.017 m\lambda = \frac{340}{20000} = 0.017 \text{ m}. To convert to cm: 0.017×100=1.7 cm0.017 \times 100 = 1.7 \text{ cm}.

Explanation:

The wavelength is the distance between two consecutive compressions or rarefactions, calculated by dividing the velocity by the frequency.

Problem 2:

Calculate the time period of a tuning fork vibrating at a frequency of 512 Hz512 \text{ Hz}.

Solution:

Given: f=512 Hzf = 512 \text{ Hz}. Using the formula T=1fT = \frac{1}{f}, T=1512≈0.00195 sT = \frac{1}{512} \approx 0.00195 \text{ s}.

Explanation:

The time period is the reciprocal of the frequency, representing the time taken for one complete vibration.

Problem 3:

A SONAR pulse is sent from a ship to the ocean floor. The signal is received back after 4 s4 \text{ s}. If the speed of sound in seawater is 1500 m s−11500 \text{ m s}^{-1}, find the depth of the ocean.

Solution:

Given: t=4 st = 4 \text{ s} (total time for 'to and fro'), V=1500 m s−1V = 1500 \text{ m s}^{-1}. Depth d=V×t2d = \frac{V \times t}{2} d=1500×42=3000 md = \frac{1500 \times 4}{2} = 3000 \text{ m}.

Explanation:

In SONAR, the sound travels twice the distance (down and up), so we divide the total distance by 2 to find the depth.

Sound (Production, Propagation, Infrasonic, Ultrasonic) Revision - Class 9 Science ICSE