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Sound - Production and Propagation of Sound

Grade 7ICSE

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. A vibration is defined as a rapid back-and-forth motion of an object about its mean position.

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Sound requires a material medium (solid, liquid, or gas) for its propagation. It cannot travel through a vacuum because there are no molecules to transmit the vibrations.

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Sound travels in the form of longitudinal waves, where the particles of the medium vibrate parallel to the direction of the wave's propagation.

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The speed of sound depends on the nature of the medium: vsolid>vliquid>vgasv_{\text{solid}} > v_{\text{liquid}} > v_{\text{gas}}. For example, the speed of sound in air is approximately 330 m/s330\text{ m/s} to 340 m/s340\text{ m/s}, while in steel it is about 5000 m/s5000\text{ m/s}.

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Amplitude (AA) is the maximum displacement of a vibrating particle from its mean position. It determines the loudness of the sound (L∝A2L \propto A^2).

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Frequency (ff) is the number of vibrations per second, measured in Hertz (HzHz). It determines the pitch or shrillness of the sound.

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Time Period (TT) is the time taken by a vibrating body to complete one full vibration.

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The human audible range of frequencies is from 20 Hz20\text{ Hz} to 20,000 Hz20,000\text{ Hz} (or 20 kHz20\text{ kHz}). Sounds below 20 Hz20\text{ Hz} are called infrasonic, and sounds above 20,000 Hz20,000\text{ Hz} are called ultrasonic.

📐Formulae

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

v=f×λv = f \times \lambda

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

Frequency (f)=Total number of vibrationsTotal time in seconds\text{Frequency } (f) = \frac{\text{Total number of vibrations}}{\text{Total time in seconds}}

💡Examples

Problem 1:

A simple pendulum completes 4040 oscillations in 22 seconds. Find its time period and frequency.

Solution:

Given: Number of oscillations = 4040, Total time t=2 st = 2\text{ s}. Frequency f=Number of oscillationst=402=20 Hzf = \frac{\text{Number of oscillations}}{t} = \frac{40}{2} = 20\text{ Hz}. Time Period T=1f=120=0.05 sT = \frac{1}{f} = \frac{1}{20} = 0.05\text{ s}.

Explanation:

Frequency is the oscillations per unit time, and the time period is the inverse of the frequency.

Problem 2:

A person standing 660 m660\text{ m} away from a source of sound hears the sound after 22 seconds. Calculate the speed of sound in air.

Solution:

Given: Distance d=660 md = 660\text{ m}, Time t=2 st = 2\text{ s}. Using the formula: v=dtv = \frac{d}{t} v=6602=330 m/sv = \frac{660}{2} = 330\text{ m/s}.

Explanation:

The speed of sound is calculated by dividing the total distance traveled by the time taken.

Problem 3:

The frequency of a tuning fork is 256 Hz256\text{ Hz}. If the speed of sound is 340 m/s340\text{ m/s}, find the wavelength (λ\lambda) of the sound produced.

Solution:

Given: f=256 Hzf = 256\text{ Hz}, v=340 m/sv = 340\text{ m/s}. Using the wave equation: v=fλv = f \lambda λ=vf=340256≈1.328 m\lambda = \frac{v}{f} = \frac{340}{256} \approx 1.328\text{ m}.

Explanation:

Wavelength is the distance between two consecutive compressions or rarefactions, found by dividing speed by frequency.