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Sound - Production and Propagation of Sound

Grade 9CBSE

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

🔑Concepts

Sound is a form of energy produced by vibrations of objects. These vibrations travel through a medium as a mechanical wave.

A medium is necessary for the propagation of sound. Sound cannot travel through a vacuum because there are no molecules to compress or rarefy.

Sound waves are longitudinal waves. In these waves, the particles of the medium vibrate back and forth in the same direction as the wave propagation.

Compressions (CC) are regions where particles are crowded together (high pressure/density), and Rarefactions (RR) are regions where particles are spread apart (low pressure/density).

Wavelength (λλ) is the distance between two consecutive compressions or two consecutive rarefactions. Its SI unit is meter (mm).

Frequency (νν) is the number of complete oscillations per unit time. Its SI unit is Hertz (HzHz).

Time Period (TT) is the time taken by two consecutive compressions or rarefactions to cross a fixed point. It is the reciprocal of frequency: T=1νT = \frac{1}{\nu}.

Amplitude (AA) is the magnitude of the maximum disturbance in the medium on either side of the mean value. It determines the loudness of the sound.

The speed of sound depends on the properties of the medium (density and elasticity) and the temperature. Speed increases as the temperature of the medium increases.

The speed of sound is highest in solids, followed by liquids, and lowest in gases.

📐Formulae

Speed(v)=DistanceTime\text{Speed} (v) = \frac{\text{Distance}}{\text{Time}}

ν=1T\nu = \frac{1}{T}

v=νλv = \nu \lambda

Speed=Frequency×Wavelength\text{Speed} = \text{Frequency} \times \text{Wavelength}

💡Examples

Problem 1:

A sound wave has a frequency of 2 kHz2 \text{ kHz} and a wavelength of 35 cm35 \text{ cm}. How long will it take to travel 1.5 km1.5 \text{ km}?

Solution:

Given: Frequency, ν=2 kHz=2000 Hz\nu = 2 \text{ kHz} = 2000 \text{ Hz}. Wavelength, λ=35 cm=0.35 m\lambda = 35 \text{ cm} = 0.35 \text{ m}. Distance, d=1.5 km=1500 md = 1.5 \text{ km} = 1500 \text{ m}. Step 1: Calculate speed v=νλ=2000 Hz×0.35 m=700 m/sv = \nu \lambda = 2000 \text{ Hz} \times 0.35 \text{ m} = 700 \text{ m/s}. Step 2: Calculate time t=dv=1500 m700 m/s2.14 st = \frac{d}{v} = \frac{1500 \text{ m}}{700 \text{ m/s}} \approx 2.14 \text{ s}.

Explanation:

First, convert all units to SI (HzHz, mm, ss). Use the wave equation to find the velocity, then use the basic speed formula to find the time taken for that velocity to cover the given distance.

Problem 2:

A person clapped his hands near a cliff and heard the echo after 2 s2 \text{ s}. What is the distance of the cliff from the person if the speed of sound, vv, is taken as 346 m/s346 \text{ m/s}?

Solution:

Given: Speed of sound v=346 m/sv = 346 \text{ m/s}. Time for echo t=2 st = 2 \text{ s}. In an echo, the sound travels twice the distance (2d2d). 2d=v×t2d = v \times t 2d=346 m/s×2 s=692 m2d = 346 \text{ m/s} \times 2 \text{ s} = 692 \text{ m} d=692 m2=346 md = \frac{692 \text{ m}}{2} = 346 \text{ m}.

Explanation:

The sound travels from the person to the cliff and back to the person. Thus, the total distance covered is twice the actual distance between the observer and the reflecting surface.

Production and Propagation of Sound - Revision Notes & Key Formulas | CBSE Class 9 Science