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Sound - Characteristics of Sound (Loudness, Pitch, Quality)

Grade 7ICSE

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

🔑Concepts

Loudness: It is the characteristic of sound that distinguishes a loud sound from a faint one. It depends on the Amplitude (AA) of vibration. Loudness is directly proportional to the square of the amplitude (LA2L \propto A^2). The SI unit for loudness level is the decibel (dBdB).

Pitch (Shrillness): It is the characteristic of sound that distinguishes a shrill sound from a grave (flat) sound. It depends on the Frequency (ff) of vibration. High-frequency vibrations produce high-pitched sound (e.g., a whistle), while low-frequency vibrations produce low-pitched sound (e.g., a drum).

Quality (Timbre): It is the characteristic that allows us to distinguish between two sounds of the same loudness and same pitch produced by different instruments (e.g., a flute and a violin). It depends on the waveform and the presence of subsidiary vibrations called overtones.

Frequency (ff): The number of vibrations or oscillations completed by a body in one second. It is measured in Hertz (HzHz).

Time Period (TT): The time taken by a vibrating body to complete one full vibration. It is measured in seconds (ss).

📐Formulae

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

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

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

💡Examples

Problem 1:

An object vibrates 5050 times in 22 seconds. Calculate its frequency and time period.

Solution:

f=25 Hzf = 25\text{ Hz}, T=0.04 sT = 0.04\text{ s}

Explanation:

Frequency is calculated as f=Number of vibrationsTime=502=25 Hzf = \frac{\text{Number of vibrations}}{\text{Time}} = \frac{50}{2} = 25\text{ Hz}. The time period is the reciprocal of frequency: T=1f=125=0.04 sT = \frac{1}{f} = \frac{1}{25} = 0.04\text{ s}.

Problem 2:

If the amplitude of a vibrating string is doubled, how will the loudness of the sound produced change?

Solution:

The loudness will become 44 times its original value.

Explanation:

Since loudness (LL) is proportional to the square of the amplitude (AA), LA2L \propto A^2. If the new amplitude A=2AA' = 2A, then the new loudness L(2A)2=4A2L' \propto (2A)^2 = 4A^2. Therefore, L=4LL' = 4L.

Problem 3:

Two sound waves, XX and YY, have frequencies of 256 Hz256\text{ Hz} and 512 Hz512\text{ Hz} respectively. Which sound is shriller?

Solution:

Sound wave YY is shriller.

Explanation:

Shrillness is another term for pitch. Pitch depends on frequency. Since the frequency of sound YY (512 Hz512\text{ Hz}) is higher than that of sound XX (256 Hz256\text{ Hz}), sound YY has a higher pitch and is shriller.