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Sound - Characteristics of Sound: Amplitude and Frequency

Grade 8CBSE

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

🔑Concepts

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Sound is produced by vibrating objects. Vibration is the rapid back-and-forth motion of an object about its mean position.

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Amplitude (AA) is the maximum displacement of a vibrating object from its central or mean position. It determines the loudness of the sound.

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Loudness is directly proportional to the square of the amplitude of vibration. If the amplitude is doubled, the loudness increases by a factor of 44. Loudness is expressed in the unit decibel (dBdB).

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Frequency (ff) is defined as the number of oscillations or vibrations per second. Its SI unit is Hertz (HzHz).

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The pitch or shrillness of a sound is determined by its frequency. A higher frequency results in a higher pitch (e.g., a bird's chirp), while a lower frequency results in a lower pitch (e.g., a lion's roar).

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Time Period (TT) is the time taken by a vibrating object to complete one full oscillation. It is the reciprocal of frequency.

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The range of audible frequencies for the human ear is roughly between 20 Hz20\text{ Hz} and 20,000 Hz20,000\text{ Hz}.

📐Formulae

f=Total Number of OscillationsTotal Time Taken (t)f = \frac{\text{Total Number of Oscillations}}{\text{Total Time Taken (t)}}

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

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

💡Examples

Problem 1:

A simple pendulum oscillates 4040 times in 44 seconds. Find its frequency and time period.

Solution:

f=404=10 Hzf = \frac{40}{4} = 10\text{ Hz} T=110=0.1 sT = \frac{1}{10} = 0.1\text{ s}

Explanation:

Frequency is calculated by dividing the total oscillations by the time in seconds. The time period is the inverse of the frequency.

Problem 2:

If the amplitude of a sound wave is increased from AA to 3A3A, how many times will the loudness increase?

Solution:

Initial Loudness (L1)∝A2\text{Initial Loudness } (L_1) \propto A^2 New Loudness (L2)∝(3A)2=9A2\text{New Loudness } (L_2) \propto (3A)^2 = 9A^2 Ratio L2L1=9A2A2=9\text{Ratio } \frac{L_2}{L_1} = \frac{9A^2}{A^2} = 9

Explanation:

The loudness of sound is proportional to the square of its amplitude. Therefore, tripling the amplitude increases the loudness by 32=93^2 = 9 times.

Problem 3:

A source produces sound with a frequency of 500 Hz500\text{ Hz}. How many times does it vibrate in 11 minute?

Solution:

f=500 Hzf = 500\text{ Hz} Time (t)=1 minute=60 seconds\text{Time } (t) = 1\text{ minute} = 60\text{ seconds} Total Vibrations=f×t=500×60=30,000\text{Total Vibrations} = f \times t = 500 \times 60 = 30,000

Explanation:

Frequency represents vibrations per second. To find total vibrations in a minute, multiply the frequency by 6060 seconds.