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Sound - Propagation of Sound

Grade 9CBSE

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

🔑Concepts

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Sound is a mechanical wave that requires a material medium (solid, liquid, or gas) for its propagation; it cannot travel through a vacuum.

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Sound travels through a medium as a series of compressions (high pressure/density) and rarefactions (low pressure/density).

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Sound waves are classified as longitudinal waves because the particles of the medium vibrate back and forth in the same direction in which the wave travels.

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The distance between two consecutive compressions or two consecutive rarefactions is called the Wavelength, denoted by the Greek letter λ\lambda (lambda). Its SI unit is meter (mm).

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Frequency (ν\nu) is the number of complete oscillations (or compressions and rarefactions) per unit time. Its SI unit is Hertz (HzHz).

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The 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}.

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The Speed of Sound (vv) depends on the properties of the medium through which it travels. Generally, vsolid>vliquid>vgasv_{\text{solid}} > v_{\text{liquid}} > v_{\text{gas}}.

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The speed of sound also increases with an increase in the temperature of the medium.

📐Formulae

ν=1T\nu = \frac{1}{T}

v=λνv = \lambda \nu

v=λTv = \frac{\lambda}{T}

Speed=DistanceTime\text{Speed} = \frac{\text{Distance}}{\text{Time}}

💡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}.

First, calculate the velocity (vv): v=νλv = \nu \lambda v=2000×0.35=700 m/sv = 2000 \times 0.35 = 700\text{ m/s}

Now, calculate the time (tt): t=dvt = \frac{d}{v} t=1500700t = \frac{1500}{700} t≈2.14 st \approx 2.14\text{ s}

Explanation:

We first convert all units to SI (Hz, meters, and seconds). Using the wave equation, we find the speed, and then use the basic definition of speed to find the time taken to cover the specified distance.

Problem 2:

Calculate the wavelength of a sound wave whose frequency is 220 Hz220\text{ Hz} and speed is 440 m/s440\text{ m/s} in a given medium.

Solution:

Given: Speed v=440 m/sv = 440\text{ m/s}, Frequency ν=220 Hz\nu = 220\text{ Hz}.

Using the formula v=νλv = \nu \lambda: λ=vν\lambda = \frac{v}{\nu} λ=440220\lambda = \frac{440}{220} λ=2 m\lambda = 2\text{ m}

Explanation:

The wavelength is the ratio of the speed of the wave to its frequency. By substituting the given values, we find the distance between consecutive compressions is 2 m2\text{ m}.