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Physics - Sound (Production, Propagation, Amplitude, Pitch)

Grade 8ICSE

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

🔑Concepts

Sound is a form of energy produced by mechanical vibrations of an object. These vibrations are transmitted through a medium as longitudinal waves.

Sound requires a material medium (solid, liquid, or gas) for propagation and cannot travel through a vacuum (0 m/s0 \text{ m/s}).

The speed of sound depends on the nature of the medium. Generally, vsolids>vliquids>vgasesv_{\text{solids}} > v_{\text{liquids}} > v_{\text{gases}}. In air, the speed is approximately 330 m/s330 \text{ m/s} to 340 m/s340 \text{ m/s}.

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

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

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

The human audible range is between 20 Hz20 \text{ Hz} and 20,000 Hz20,000 \text{ Hz}. Sounds below 20 Hz20 \text{ Hz} are infrasonic, and above 20,000 Hz20,000 \text{ Hz} are ultrasonic.

📐Formulae

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

v=dtv = \frac{d}{t}

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

LoudnessA2\text{Loudness} \propto A^2

💡Examples

Problem 1:

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

Solution:

T=0.05 sT = 0.05 \text{ s}, f=20 Hzf = 20 \text{ Hz}

Explanation:

Frequency is calculated as f=oscillationstime=402=20 Hzf = \frac{\text{oscillations}}{\text{time}} = \frac{40}{2} = 20 \text{ Hz}. Time period is the reciprocal of frequency: T=1f=120=0.05 sT = \frac{1}{f} = \frac{1}{20} = 0.05 \text{ s}.

Problem 2:

Calculate the distance a sound wave travels in air in 5 s5 \text{ s} if the speed of sound is 330 m/s330 \text{ m/s}.

Solution:

d=1650 md = 1650 \text{ m}

Explanation:

Using the formula for distance d=v×td = v \times t, we substitute the given values: d=330 m/s×5 s=1650 md = 330 \text{ m/s} \times 5 \text{ s} = 1650 \text{ m}.

Problem 3:

If the amplitude of a sound is tripled, how will its loudness change?

Solution:

Loudness increases by 9 times\text{Loudness increases by } 9 \text{ times}

Explanation:

Loudness is directly proportional to the square of the amplitude (LA2L \propto A^2). If the new amplitude A=3AA' = 3A, then the new loudness L(3A)2=9A2L' \propto (3A)^2 = 9A^2. Therefore, the loudness becomes 99 times its original value.

Sound (Production, Propagation, Amplitude, Pitch) Revision - Class 8 Science ICSE