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Sound - Production of Sound

Grade 9CBSE

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

🔑Concepts

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Sound is a form of energy that produces the sensation of hearing in our ears and is produced by vibrating objects.

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Vibration is defined as a rapid to-and-fro or back-and-forth motion of an object about its mean position.

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When an object vibrates, it transfers its energy to the surrounding particles of the medium, causing them to vibrate and propagate the sound wave.

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In humans, sound is produced by the vibration of the vocal cords located in the larynx (voice box).

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Stretched membranes (like in a drum), stretched strings (like in a guitar), and air columns (like in a flute) are common sources of sound production.

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The number of vibrations or oscillations per unit time is called the frequency (ff or ν\nu), measured in Hertz (HzHz).

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The time taken to complete one full vibration is called the Time Period (TT).

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The pitch of the sound depends on the frequency of vibration; a higher frequency ff results in a higher pitch.

📐Formulae

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

f=Total number of vibrationsTotal time taken (t)f = \frac{\text{Total number of vibrations}}{\text{Total time taken (t)}}

v=νλv = \nu \lambda

Amplitude (A)∝Loudness2\text{Amplitude (A)} \propto \text{Loudness}^2

💡Examples

Problem 1:

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

Solution:

Frequency f=402=20 Hzf = \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 number of oscillations per second, and Time Period is the reciprocal of the frequency.

Problem 2:

A tuning fork produces sound of frequency 512 Hz512\text{ Hz}. If another tuning fork produces 480 Hz480\text{ Hz}, calculate the difference in their frequencies using vertical subtraction.

Solution:

512−48032\begin{array}{r} 512 \\ - 480 \\ \hline 32 \end{array} The difference is 32 Hz32\text{ Hz}.

Explanation:

To find the difference in the pitch/frequency between two sound sources, we subtract the lower frequency from the higher frequency.

Problem 3:

If the time period of a vibrating string is 0.002 s0.002\text{ s}, calculate the frequency of the sound produced.

Solution:

f=1T=10.002=10002=500 Hzf = \frac{1}{T} = \frac{1}{0.002} = \frac{1000}{2} = 500\text{ Hz}

Explanation:

By applying the relationship between frequency and time period, f=1Tf = \frac{1}{T}, we find the vibrations per second.