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Sound - Noise Pollution and Control Measures

Grade 8CBSE

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

🔑Concepts

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Noise pollution is defined as the presence of excessive or unwanted sounds in the environment that may cause discomfort or health hazards.

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The unit used to measure the intensity or loudness of sound is the decibel, denoted as dBdB.

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Loudness of sound depends on its amplitude. Specifically, Loudness∝(Amplitude)2Loudness \propto (Amplitude)^2. If the amplitude is doubled, the loudness increases by a factor of 44.

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Sounds above 80 dB80 \text{ dB} are considered physically painful and contribute significantly to noise pollution.

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Health hazards of noise pollution include insomnia (lack of sleep), hypertension (high blood pressure), anxiety, and temporary or permanent hearing impairment.

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Control measures include the installation of silencing devices in aircraft engines, transport vehicles, and industrial machines.

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Environmental control measures involve planting trees along roads and residential areas (Green Belts) to absorb sound, and restricting the use of loudspeakers and horns near schools and hospitals.

📐Formulae

Loudness∝(Amplitude)2Loudness \propto (Amplitude)^2

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

Sound Level (in dB)=10log⁡10(II0)\text{Sound Level (in } dB) = 10 \log_{10}\left(\frac{I}{I_0}\right)

💡Examples

Problem 1:

If the amplitude of a sound wave is tripled (3×3 \times), by what factor will the loudness of the sound increase?

Solution:

The loudness will increase by a factor of 99.

Explanation:

Since Loudness∝(Amplitude)2Loudness \propto (Amplitude)^2, if the new amplitude is 3A3A, the new loudness will be (3A)2=9A2(3A)^2 = 9A^2. Therefore, the loudness becomes 99 times the original value.

Problem 2:

A vibrating object completes 4040 oscillations in 22 seconds. Calculate its frequency and determine if it falls within the audible range for humans.

Solution:

Frequency=20 HzFrequency = 20 \text{ Hz}. This is at the lower limit of the audible range.

Explanation:

Using the formula f=Total OscillationsTotal Timef = \frac{\text{Total Oscillations}}{\text{Total Time}}, we get f=402=20 Hzf = \frac{40}{2} = 20 \text{ Hz}. The human audible range is 20 Hz20 \text{ Hz} to 20,000 Hz20,000 \text{ Hz}, so this sound is just barely audible.