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Sound - Energy of Sound Waves

Grade 9CBSE

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

🔑Concepts

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Sound is a form of energy produced by vibrations that travels as a mechanical longitudinal wave through a medium (solid, liquid, or gas).

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

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The energy of a sound wave is directly related to its amplitude; specifically, the energy is proportional to the square of the amplitude (E∝A2E \propto A^2).

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Intensity is defined as the amount of sound energy passing per unit time through a unit area held perpendicular to the direction of propagation. Its unit is W/m2W/m^2.

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Loudness is a physiological sensation that depends on the intensity of sound as well as the sensitivity of the human ear.

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The distance between two consecutive compressions or rarefactions is called the wavelength (λ\lambda), and the time taken for one complete oscillation is the time period (TT).

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Frequency (ν\nu) is the number of oscillations per unit time, measured in Hertz (HzHz).

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The speed of sound (vv) depends on the properties of the medium and is generally higher in solids than in liquids or gases.

📐Formulae

ν=1T\nu = \frac{1}{T}

v=νλv = \nu \lambda

v=dtv = \frac{d}{t}

I=PA=EA×tI = \frac{P}{A} = \frac{E}{A \times t}

Distanceecho=v×t2Distance_{echo} = \frac{v \times t}{2}

💡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 1.4 km1.4\text{ km}?

Solution:

Given: ν=2 kHz=2000 Hz\text{Given: } \nu = 2\text{ kHz} = 2000\text{ Hz} λ=35 cm=0.35 m\lambda = 35\text{ cm} = 0.35\text{ m} Distance (d)=1.4 km=1400 m\text{Distance } (d) = 1.4\text{ km} = 1400\text{ m} Step 1: Calculate Speed (v)\text{Step 1: Calculate Speed } (v) v=νλ=2000×0.35=700 m/sv = \nu \lambda = 2000 \times 0.35 = 700\text{ m/s} Step 2: Calculate Time (t)\text{Step 2: Calculate Time } (t) t=dv=1400700=2 st = \frac{d}{v} = \frac{1400}{700} = 2\text{ s}

Explanation:

First, convert all units to SI (HzHz, mm, and m/sm/s). Use the wave equation to find the velocity, then use the basic speed formula to find the time taken.

Problem 2:

A person claps their hands near a cliff and hears the echo after 2 s2\text{ s}. If the speed of sound is 346 m/s346\text{ m/s}, calculate the distance of the cliff from the person.

Solution:

Given: t=2 s,v=346 m/s\text{Given: } t = 2\text{ s}, v = 346\text{ m/s} Distance (d)=v×t2\text{Distance } (d) = \frac{v \times t}{2} d=346×22d = \frac{346 \times 2}{2} d=346 md = 346\text{ m}

Explanation:

In an echo, sound travels to the obstacle and back, covering the distance dd twice (2d2d). To find the one-way distance to the cliff, we divide the total distance covered by sound by 22.

Problem 3:

Calculate the frequency of a sound wave whose time period is 0.05 s0.05\text{ s}.

Solution:

Given: T=0.05 s\text{Given: } T = 0.05\text{ s} ν=1T\nu = \frac{1}{T} ν=10.05=1005\nu = \frac{1}{0.05} = \frac{100}{5} ν=20 Hz\nu = 20\text{ Hz}

Explanation:

Frequency is the reciprocal of the time period. Dividing 11 by the time period gives the frequency in Hertz.