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Sound - Production and Propagation of Sound

Grade 7ICSE

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

🔑Concepts

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.

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.

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

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}.

Amplitude (AA) is the maximum displacement of a vibrating particle from its mean position. It determines the loudness of the sound (LA2L \propto A^2).

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

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

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=3402561.328 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.

Production and Propagation of Sound - Revision Notes & Key Formulas | ICSE Class 7 Science